Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 10 Recruitment Methods Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 10 Recruitment Methods

12th Commerce Guide Recruitment Methods Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
Recruitment is the process of identifying ………………..
a) Right Man for Right Job
b) good performer
c) Right job
d) All of the above
Answer:
a) Right Man for Right Job

Question 2.
Recruitment bridges gap between …………. and ………..
a) job seeker and job provider
b) job seeker and agent
c) job provider and owner
d) owner and servant
Answer:
a) job seeker and job provider

Question 3.
Advertisement is a …………………… source of recruitment.
a) internal
b) external
c) agent
d) outsourcing
Answer:
b) external

Question 4.
Transfer is an …………… source of recruitment.
a) internal
b) external
c) outsourcing
d) None of of the above
Answer:
a) internal

Question 5.
E – Recruitment is possible only through ……………….. facility.
a) Computer
b) Internet
c) Broadband
d) 4G
Answer:
b) Internet

II. Very short answer questions.

Question 1.
Give the meaning of Recruitment.
Answer:
Recruitment is the process of finding suitable candidates for the various posts in an organisation.

Question 2.
What is promotion?
Answer:
Based on the Seniority and merits of the employees they are given the opportunity to move up in the organisational hierarchy.

Question 3.
Write any two internal sources of recruitment.
Answer:
Benefits of an internal source of recruitment:

1. The internal source will reduce the cost and expenses of recruitment.
2. It is very useful, by way of selecting from the existing and retiring employees.

Question 4.
What is meant by campus recruitment?
Answer:
The organisations visit the educational institutions to identify and recruit suitable candidates.

Question 5.
What is meant by poaching?
Answer:
Organizations instead of training and developing their own employees hire employees of other competitive companies by paying them more both financial and non-financial benefits. It is also called raiding.

III. Short Answer Questions.

Question 1.
Define the term Recruitment.
Answer:
According to Edwin B. Flippo, “It is a process of searching for prospective employees and stimulating and encouraging them to apply for jobs in an organisation.”

Question 2.
What is meant by unsolicited applicants?
Answer:
These are the applications of job seekers who voluntarily apply for the vacancies not yet notified by the organisations.

Question 3.
What is meant by job portals?
Answer:

•  A job portal is a web that links the gap between the employer and job seekers.
•  Because of IT development, the employer can select the abled from throughout the world.
•  Using Internet job portals organisation can screen for prospective candidates and fill up their vacancies.

Question 4.
State the steps in the Recruitment process, outsourcing.
Answer:

1. Planning recruitment
2. Determining vacancies
3. Identifying the sources
4. Drafting information for advertisement
5. Selecting the suitable mode of advertisement
6. Facilitating selection process
7. Evaluation and control

IV. Long Answer Questions.

Question 1.
Explain the internal sources of recruitment (any 5).
Answer:
Internal Sources:
The following are the Internal Sources of recruitment.
(i) Transfer:
The simplest way by which employee recruitment can be filled is through the transfer of employees from one department with surplus staff to that of another with deficit staff.

(ii) Upgrading:
Performance appraisal helps in the process of moving employees from a lower position to a higher position.

(iii) Promotion:
Based on the seniority and merits of the employees they are given the opportunity to move up in the organisational hierarchy.

(iv) Demotion:
The movement of an employee from a higher position to a lower position because of poor performance continues to make him realise the significance of the performance.

(v) Job Rotation:

•  One single employee managing to learn how to perform in more than one job on rotation.
•  This familiarises the employees with all kinds of jobs performed and becomes a source.

Question 2.
Explain the External sources of recruitment. (any 5)
Answer:
External Sources:
a. Direct – Sources: (WAR)
Walk-ins: Walk-in applicants with suitable qualifications and requirements can be another source of recruitment.

Advertisement:

•  The employer can advertise in periodicals – about the vacancies in the organisation.
•  Specifying nature of work and vacancy.
•  Qualification and experience required.
•  Salary offered.
•  Mode of applying – The last date to apply.

Rival firms:

•  Where the efficient employers of rival companies are drawn to the organisation by higher pay and benefits.
• It is known as “Poaching” or “Attracting” or “Raiding” or “Hunting”.
• And other Direct – Sources are campus and e-recruitment, unsolicited applicants, Factory Gate recruitment, etc.

b. Indirect – Sources : [JEEP]
Job portals: Using internet job portals organisations can screen for prospective candidates and fillup their vacancies.

Employee referral:
The existing employees of the organisation may recommend some of their relatives or known people who will be suitable for the existing vacancies.

Employment Agencies :
It is like a Public Employment Exchange, owned by the private parties and connect the job providers and the job seeker.

Public Employment Exchange :
It is established by the Government and bridging the gap between the Employer and the job seeker and makes available the information through a database. And other indirect sources are Labour Contractors Poaching Outsourcing, Deputation, etc. Word of mouth.

Question 3.
What are the Recent Trends in Recruitment?
Answer:
The recent methods of recruiting by organisations include the following methods:
1 . Outsourcing – There are outsourcing firms that help in the process of recruiting through screening of applications and finding the right person for the job for which job they are paid service charges.
Recruitment Process Outsourcing:

2. Poaching – Organisations instead of training and developing their own employees hire employees of other competitive companies by paying them more both financial and non-financial benefits. It is also called raiding.

12th Commerce Guide Recruitment Methods Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
Promotion is a …………………….. source of requirement.
a) internal
b) external
c) modern
d) NOTA
Answer:
a) internal

Question 1.
The internal source of recruitment are _________
(i) promotion
(ii) e-recruitment
(iii) retention
(iv) advertisements
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)
Answer:
(b) (i) and (iii)

Question 3.
Poaching is also known as ………………
a) Selection
b) Yield
c) Success
d) Raiding
Answer:
d) Raiding

Question 4.
The process of attracting potential people to apply for a job in an organisation is ………………
a) Selection
b) Recruitment
c) HR
d) NOTA
Answer:
c) E-recruitment

Question 5.
…………………………….. helps in the process of moving employees from a lower position to a higher position.
a) Appraisal
b) Knowledge
c) Skill
d) All of these
Answer:
d) All of these

Question 6.
Movement of an employee from a higher position to a lower position because of poor performance is known as ………………
a) Promotion
b) Demotion
c) Rotation
d) Retention
Answer:
b) Demotion

Question 7.
The organisation visits the Educational institutions to recruit suitable candidates is known as……………
a) Job Portals
b) Campus
c) Word of Mouth
d) Online
Answer:
b) Campus

Question 8.
Why vacancy arises?
a) Death
b) Retirement
c) Resignation
d) All of these
Answer:
d) All of these

Question 9.
Applicants with suitable qualification requirement can be known as …………..
a) Walk-in
b) Campus
c) Advertisement
d) NOTA
Answer:
a) Walk-in

Question 10.
Pick the odd one out:
a) Rival firms
b) Walk-ins
c) Campus
d) Deputation
Answer:
d) Deputation

Question 11.
Pick the odd one out:
a) Word of mouth
b) Job portals
c) Labour contractors
d) Campus
Answer:
d) Campus

Question 12.
Pick the odd one out:
a) Transfer
b) Promotion
c) Demotion
d) Advertisement
Answer:
d) Advertisement

Question 13.
Pick the odd one out:
a) Planning
b) Determining
c) Identifying
d) Upgrading
Answer:
d) Upgrading

Question 14.
Which one of the following is correctly matched?
a) E-Recruitment – Online methods
b) Unsolicited Applicants – Applications of job seekers
c) Employee referral – Existing employees recommendation
d) Private Employment Agencies – Established by the Government
Answer:
d) Private Employment Agencies – Established by the Government

Question 15.
Which is the correct statement?
The vacancy arises due to:
(i) Retirement or Death of an employee.
(ii) Resignation of an employee.
(iii) Disablement and Dismissal of an employee.
a) (i) is correct
b) (ii) is correct
c) (iii) is correct
d) All (i), (ii) and (iii) are correct
Answer:
d) All (i), (ii) and (iii) are correct

II. Match The Following

Question 1.
Match List I with List – II

Answer:
a) (i) 4, (ii) 3, (iii) 1, (iv) 2

Question 2.

Answer:
b) (i) 3, (ii) 4, (iii) 1, (iv) 2

Question 3.

Answer:
a) (i) 2,(ii) 1,(iii) 4,(iv) 3

III Assertion And Reason.
Question 1.
Assertion (A): One single employee managing to learn how to perform in more than one job on rotation.
Reason (R): It familiarises the employees with all kinds of jobs performed and becomes a source. ,
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are True
d) Both (A) and (R) are False
Answer:
c) Both (A) and (R) are True

IV. Very Short Answer Questions.

Question 1.
What is Retention?
Answer:
The retiring employees can be used to meet the requirement after extension as per management discretion.

Question 2.
How the Dependants are appointed?
Answer:
The legal heirs or The Dependant employees may be given a chance to replace the deceased.

Question 3.
What is meant by retention?
Answer:
Retention is a method of an internal source of recruitment. The retiring employees can be used to meet the requirement after superannuation as per management discretion.

Question 4.
What is Campus Recruitment?
Answer:
The organisations visit the Educational Institutions to identify and recruit suitable candidates.

Question 5.
What is E-Recruitment?
Answer:
The organisations which carryout recruitment online method is said to follow E-recruitment.

Question 6.
What is Deputation?
Answer:
A person who is already an employee of an organisation can be deputed for a specific job for a specified period to another organisation as a short-term solution.

Question 7.
What is word of mouth?
Answer:
The information relating to job seekers is collected through repute who pass on the message about the vacancy to their known people.

Question 8.
What is outsourcing?
Answer:
There are Outsourcing firms that help in the process of recruiting through screening of applications and finding the right Peron for the job for Which job they are paid service charge.

Question 9.
Mention any two Features of campus recruitment.
Answer:

•  Most of the applicants are assembled in one place.
•  The company can arrange it in a short period.

Question 10.
List the benefits of external sources of recruitment.

• The company receives new abilities and skills.
• New employees can change the habit of old employees.

V. Short Answer Questions.

Question 1.
What do you mean by poaching?
Answer:
Poaching means that the company can hire employees from other companies by paying them more. So the company can reduce the expense of giving training and developing their own employees. It is also called raiding.

Question 2.

VI. Long Answer Questions.

Question 1.
What are the Internal Sources of Recruitment? [PUT] [DR]
Answer:

• Promotion: Based on seniority and merits of employees moving from lower position to higher position. [organisational hierarchy].
• Upgrading: Performance appraisal helps in the process of moving employees from lower positions to higher positions.
• Transfer: Transfer of an employee from one department with surplus staff to that of another with deficit staff.
• Demotion: Movement of employees from a higher position to a lower position because of poor performance.
• Retired employees: The employees who have already retired can be called to fill the vacancy as they have the required qualification and experience.

Question 2.
Explain the different methods of recruitment.
Answer:
There are basically two ways (methods) by which an organisation can recruit its employees.

• Internal Source.
• External Source.

External Sources can be further classified into:

• Direct Sources
• Indirect Sources

Internal Sources:

• Promotion – Job rotation
• Upgrading – Acquisition and Mergers
• Transfers – Retired employees
• Demotion – Dependents
• Recommodation by existing employees.

External Sources:
(a) Direct Sources: [WAR] [CUF] [E]

• Walk-ins – Campus Recruitment
• Advertisement – Unsolicited Applicants
• Rival firms – Factory gate Recruitments.
• E-recruitment

(b) Indirect Sources: [JEEP] [LOW] [PGD}

• Job portals
• Employee referral
• Employment – Consultancies – Agencies
• Professional Association
• Labour contractors
• Outsourcing
• Word of mouth
• Poaching [Hunting – Attracting Raiding]
• Government (public) Employment exchange Deputation.

Question 3.
Elaborate on the factors affecting recruitment.
Answer:
Internal Factors: SIR IF
Internal Factors are the factors within the organisation that affect recruiting the personnel in the organisation.
They are:

• Size of the organisation
• Image of the organisation
• Recruitment policy
• Image of job
• Future plans

External Factors: LULLS
External Factors are the factors outside the organisation influence recruiting the personnel in the organisation.
They are:

• Labour Market.
• Unemployment situation.
• Labour Laws.
• Legal Consideration.
• Socio-economic factors.

Question 4.
Discuss the importance of Recruitment.
Answer:

•  Recruitment is the process of finding suitable candidates for various posts.
•  It is the process of attracting people to apply for a job.
•  It is bridging the gap between job provider and job seeker.
•  It encourages prospective employees to apply for a job.
•  It involves the communication of vacancies.
•  It is an economical method.
•  Its process is very simple.
•  It is a positive one.
•  No contractual relationship is established.
•  It requires less time.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 8 Ionic Equilibrium Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

12th Chemistry Guide Ionic Equilibrium Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 x 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is ………………
(a) 2.42 x 10^-8 mol³ L^-3
(b) 2.66 x 10^-12 12 mol³ L^-3
(c) 45 x 10^-11 mol³ L^-3
(d) 5.619 x 10^-12 mol³ L^-3
Answer:
(d) 5.619 x 10^-12 mol³ L^-3
Solution:
Ag₂C₂O₄ 2Ag⁺ + C₂ O₄^2-
[Ag⁺] = 2.24 x 10^-4 mol L^-1
[C₂ O₄^2-] = \(\frac{2.24 \times 10^{-4}}{2}\) mol L^-1
= 1.12 x 10^-4 mol L^-1
K_sp = [Ag⁺]² [C₂O₄^2-]
=(2.24 x 10^-4 mol L^-1)² (1.12 x 10^-4 mol L^-1)
=5.619 x 10^-12 mol³ L^-3

Question 2.
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
(i) 60 mL \(\frac { M }{ 10 }\) HCI + 40 mL \(\frac { M }{ 10 }\) NaOH
(ii) 55 mL \(\frac { M }{ 10 }\) HCl + 45 mL \(\frac { M }{ 10 }\) NaOH
(iii) 75 mL \(\frac { M }{ 5 }\) HCI +25 mL \(\frac { M }{ 5 }\) MNaOH
(iv) 100 mL \(\frac { M }{ 10 }\) HCI+ 100 mL \(\frac { M }{ 10 }\) NaOH

pH of which one of them wilt be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii) 75 mL \(\frac { M }{ 5 }\) HCI + 25 mL \(\frac { M }{ 5 }\) NaOH
No of moles of HCl = 0.2 x 75 x 10^-3 = 15 x 10^-3
No of moles of NaOH = 0.2 x 25 x 10^-3 = 5 x 1o^-3
No of moles of HCl after mixing = 15 x 10^-3 – 5 x 10^-3
∴ Concentration of HCl
\(\frac{\text { No of moles of }}{\mathrm{HCl} \text { Vol inlitre }}\)
for (iii) solution, pH of 0.1 M HCI = – 1og₁₀ (0.1) = 1.

Question 3.
The solubility of BaSO₄ in water is 2.42 x 10^-3 gL^-1 at 298K. The value of its solubility product
(K_sp) will be …………………..
(Given molar mass of BaSO₄ = 233g mol^-1)
(a) 1.08 x 10^-14 mol²L²
(b) 1.08 x 10^-12 mol²L²
(c) 1.08 x 10^-10 mol² L²
(d) 1.08 x 10^-8 mol²L^-2
Answer:
(c) 1.08 x 10^-10 mol² L²
Solution:
BaSO₄ \(\rightleftharpoons\) Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 x 10^-3g L^-1)²
= \(\left(\frac{2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}}{233 \mathrm{~g} \mathrm{~mol}^{-1}}\right)^{2}\)
= (0.01038 x 1o^-3)²
= (1.038 x 10^-5)²
= 1.077 x 10^-10
= 1.08 x 10^-10 mol² L²

Question 4.
pH of a saturated solution of Ca(OH)₂ is 9. The Solubility product (K_sp) of Ca(OH)₂
(a) 0.5 x 10^-15
(b) 0.25 x 10^-10
(c) 0.125 x 10^-15
(d) 0.5 x 10^-10
Answer:
(a) 0.5 x 10^-15
Solution:
Ca(OH)₂ \(\rightleftharpoons\) Ca²⁺ + 2OH^–
Given that pH = 9
pOH = 14 – 9 = 5
[p0K = – 1og₁₀[OH^–]]
[OH^–] = 10^-pOH
[OH^–] =10^-5M
K_sp = [Ca²⁺] [OH^–]²
= \(\frac{10^{-5}}{2} \times\left(10^{-5}\right)^{2}\) = 0.5 x 10^-15

Question 5.
Conjugate base for bronsted acids H₂O and HF are ………………
(a) OH^– and H₂FH⁺, respectively
(b) H₃O⁺ and F^–, respectively
(c) OH^– and F^–, respectively
(d) H₃O⁺ and H₂F⁺, respectively
Answer:
(c) OH^– and F^–, respectively
Solution:

∴ Conjugate bases are OH^– and F^– respectively

Question 6.
Which will make basic buffer?
(a) 50 mL of 0.1M NaOH + 25mL of 01M CH₃COOH
(b) 100 mL of 0.1M CH₃COOH + 100 mL of 0.1M NH₄OH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
(d) 100 mL of 0.1M HCI + 100 mL of O.1 M NaOH
Answer:
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
Solution:
Basic buffer is the solution which has weak base and its salt

Question 7.
Which of the following fluro – compounds is most likely to behave as a Lewis base?
(a) BF₃
(b) PF₃
(c) CF₄
(d) SiF₄
Answer:
(b) PF₃
Solution:
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CF₄ → neutral → neither lewis acid nor base
SiF₄ → neutral → neither lewis acid nor base

Question 8.
Which of these is not likely to act as lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F^–
Answer:
(a) BF₃
Solution:
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F^– → unshared pair of electron → lewis base

Question 9.
What is the decreasing order of strength of bases?
OH^–, NH₂^–, H – C ≡ C^– and CH₃ – CH₂^–
(a) OH^– > NH₂^– > H – C ≡ C > CH₃ – CH₂^–
(b) NH₂^– > OH^– > CH₃ – CH₂^– > H – C ≡ C^–
(c) CH₃ – CH₂^–, > NH₂^– > H – C ≡ C^– > OH^–
(d) OH^– > H – C ≡ C^– > CH₃ – CH₂^– > NH₂^–
Answer:
(c) CH₃ – CH₂^–, > NH₂^– > H – C ≡ C^– > OH^–
Solution:
Acid strength decreases in the order
HOH > CH ≡ CH > NH₃ > CH₃CH₃
Its conjucate bases arc in the reverse order
CH₃ – CH₂^– > NH₂^– > H – C ≡ C^– > OH^–

Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
Answer:
(b) basic, acidic, basic
Solution:

Question 11.
The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10M aqueous pyridine solution (Kb for C₅H₅N = 1.7 x 10^-9) iS ……………..
(a) 0.006%
(b) 0.013%
(c) 0.77%
(d) 1.6%
Answer:
(b) 0.013%

Percentage of dissociation
= \(\sqrt { 1.7 }\) x 10^-4 x 100 = 1.3 x 10^-2 = 0.013%

Question 12.
Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
(a) 37 x 10^-2
(b) 10^-6
(c) 0.111
(d) none of these
Answer:
(a) 3.7 x 10^-2
pH = – log₁₀ [H⁺]
[H⁺] = 10^-pH
Let the volume be x mL
V₁M₁ + V₂M₂ + V₃M₃ = VM
x mL of 10^-1M + x mL of 10^-2M + x mL of 10^-3 M
= 3 x mL of [H⁺]
= 3 x mL of [H⁺]

= 0.037 = 3.7 x 10^-2

Question 13.
The solubility of AgCl (s) with solubility product 1.6 x 10^-10 in 0. 1 M NaCl solution would be ………….
(a) 1.26 x 10^-5 M
(b) 1.6 x 10^-9 M
(c) 1.6 x 10^-11 M
(d) Zero
Answer:
(b) 1.6 x 10^-9 M
AgCl (s) \(\rightleftharpoons\) Ag⁺(aq) + Cl^–(aq)

Question 14.
If the solubility product of lead iodide is 3.2 x 10^-8, its solubility will be …………..
(a) 2 x 10^-3M
(b) 4 x 10^-4 M
(c) 1.6 x 10^-5 M
(d) 1.8 x 10^-5 M
Answer:
(a) 2 x 10^-3M
Solution:
PbI₂ (s) \(\rightleftharpoons\) Pb²⁺ (aq) + 2I^– (aq)
K_sp = (s) (2s)²
3.2 x 10^-8 = 4s³
S = \(\left(\frac{3.2 \times 10^{-8}}{4}\right)^{1 / 3}\)
= (8 x 10^-9)^1/3 = 2 x 10^-3MNaOH

Question 15.
Using Gibb’s free energy change, ∆G⁰ = 57.34 KJ mol^-1, for the reaction, X₂Y_(s) \(\rightleftharpoons\) 2X⁺ + Y^2-(aq), calculate the solubility product of X₂Y in water at 300K (R = 8.3 J K^-1 Mol^-1)
(a) 10^-10
(b) 10^-12
(c) 10^-14
(d) can not be calculated from the given data
Answer:
(a) 10^-10
Solution:

Question 16.
MY and NY₃, are insoluble salts and have the same K_sp values of 6.2 x 10^-13 at room temperature. Which statement would be true with regard to MY and NY₃?
(a) The salts MY and NY₃ are more soluble in O.5 M KY than in pure water
(b) The addition of the salt of KY to the suspension of MY and NY₃ will have no effect on
(c) The molar solubities of MY and NY₃ in water are identical
(d) The molar solubility of MY in water is less than that of NY₃
Answer:
(d) The molar solubility of MY in water is less than that of NY₃
Solution:
Addition of salt KY (having a common ion Y^–) decreases the solubility of MY and NY₃ due to common ion effect. Option (a) and (b) are wrong.
For salt MY, MY \(\rightleftharpoons\) M⁺ + Y^–
K_sp = (s) (s)
6.2 x 10^-13 = s²
therefore \(\mathrm{s}=\sqrt{6.2 \times 10^{-13}} \simeq 10^{-7}\)
for salt NY₃,
NY₃ ⇌ N³⁺ + 3Y^–
K_sp = S (3S)³
K_sp = 27s⁴
\(S=\left(\frac{6.2 \times 10^{-13}}{27}\right)^{1 / 4} \mathrm{~s} \simeq 10^{-4}\)
The molar solubility of MY in water is less than of NY₃

Question 17.
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
Answer:
(d) 12.65
Solution:
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 x x x 10^-3 = 0.l x x 10^-3
No. of moles of HCl = 0.01 x x x 10^-3 = 0.01 x x 10^-3
No. of moles of NaOH after mixing = 0.1 x x 10^-3 – 0.01x  x 10^-3
= 0.09x x 10^-3
Concentration of NaOH = \(\frac{0.09 x \times 10^{-3}}{2 x \times 10^{-3}}=0.045\)
[OH^–] = 0.045
p^OH = – log (4.5 x 10^-2)
= 2 – log 4.5
= 2 – 0.65 = 1.35
pH = 14 – 1.35 = 12.65

Question 18.
The dissociation constant of a weak acid is 1 x 10^-3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ………………..
(a) 4:3
(b) 3:4
(c) 10:1
(d) 1:10
Answer:
(d) 1:10
Solution:
K_a = 1 x 10^-3 ; pH = 4

Question 19.
The pH of 10^-5 M KOH solution will be …………..
(a) 9
(b) 5
(c) 19
(d) none of these
Answer:
(a) 9
Solution:

[OH^–] = 10^-5M.
pH = 14 – pOH .
pH = 14 – ( – log [OH^–])
= 14 + log [OH^–]
= 14 + log 10^-5
= 14 – 5 = 9

Question 20.
H₂PO₄^– the conjugate base of …………….
(a) PO₄^3-
(b) P₂O₅
(c) H₃PO₄
(d) HPO₄^2-
Answer:
(c) H₃PO₄
Solution:

Question 21.
Which of the following can act as lowery – Bronsted acid well as base?
(a) HCl
(b) SO₄^2-
(c) HPO₄^2-
(d) Br^–
Answer:
(c) HPO₄^2-
Solution:
HPO₄^2- can have the ability to accept a proton to form HPO₄^–
It can also have the ability to donate a proton to form PO₄^-3.

Question 22.
The pH of an aqueous solution is Zero. The solution is ……………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(b) strongly acidic
Solution:
pH = – log₁₀[H⁺]
[H⁺] =10^-pH
= 10⁰ = 1
[H⁺] = 1 M
The, solution is strongly acidic

Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by ………………

Answer:
a
Solution:
According to Henderson equation

Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?

Answer:
c) h = \(\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}\)
Solution:
h = \(\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}\)

Question 25.
Dissociation constant of NH₄OH is 1.8 x 10^-5 the hydrolysis constant of NH₄Cl would be …………….
(a) 1.8 x 10^-19
(b) 5.55 x 10^-10
(c) 5.55 x 10^-5
(d) 1.80 x 10^-5
Answer:
(b) 5.55 x 10¹⁰
Solution:
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{b}}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\)
= 0.55 x 10^-9
= 5.5 x 10^-10

II. Answer the following questions

Question 1.
What are lewis acids and bases? Give two example for each.
Answer:

Question 2.
Discuss the Lowry – Bronsted concept of acids and bases.
Answer:

• Lowry Bronsted acids – proton donors
• Lowry Bronsted bases – proton acceptors
• An acid₁ after donating a proton becomes a base₁ (conjugate base)
• A base₂ after accepting a proton becomes an acid₂ (conjugate acid)
• In general Lowry – Bronsted acid — base reaction is
  
• In the above reaction
  
• A conjugate acid – base pair differs by a proton.

Limitation : BF₃, AlCl₃ etc., that do not donate protons are known to behave as acid s.

Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.
i) HS^– (aq) + HF \(\rightleftharpoons\) F^– (aq) + H₂S (aq)
ii) HPO₄^2- + SO₃^2- \(\rightleftharpoons\) PO₄^3- + HSO₃^–
iii) NH₄⁺ + CO₃^2- \(\rightleftharpoons\) NH₃ + HCO₃^–
Answer:

Question 4.
Account for the acidic nature of HClO₄. In terms of Bronsted – Lowry theory, identify its conjugate base.
Answer:
HClO₄ \(\rightleftharpoons\) H⁺ + ClO₄^–
1. According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

2. Let us consider the stabilities of the conjugate bases ClO₄^– , ClO₃^–, CIO₂^– and ClO^– formed from these acid HClO₄, HClO₃, HCIO₂, HOCI respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently, the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

3. The charge stabilization mercases in the order, ClO^– < ClO₂^– < ClO₃^– < ClO₄^– .

This means ClO₄^– will have maximum stability and therefore will have a minimum attraction for W. Thus CIO₄^– will be weakest base and its conjugate acid HCIO₄ is the strongest acid.

4. CIO₄^– is the conjugate base of the acid HClO₄.

Question 5.
When aqueous ammonia is added to CuSO₄ solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H₂O)₆]²⁺(aq) + 4NH₃ (aq) \(\rightleftharpoons\) [Cu(NH₃)₄]²⁺ (aq), among H₂O and NH₃ which is stronger Lewis base.
Answer:

• [Cu(H₂O)₆]²⁺(aq) + 4NH₃ (aq) \(\rightleftharpoons\) [Cu(NH₃)₄]²⁺ (aq) + H₂
• Nitrogen less electronegative than oxygen and donates its lone pair of electrons readily. Hence NH₃ is a stronger Lewis base.
• If a better Lewis base (ligand) is available, a Lewis acid (central metal ion) will react (Ligand exchange reaction)
• In this reaction, H₂O is exchanged with NH₃.
• The Lewis acid Cu²⁺ exchanges the Lewis base H with better Lewis base N-H₃ to form [Cu(NH₃)₄]²⁺
• Hence NH₃ is a stronger Lewis base than H₂O in this reaction.

Question 6.
The concentration of hydroxide ion in a water sample is found to be 2.5 x 10^-6 M. Identify the nature of the solution.
Answer:
The concentration of OH ion in a water sample is found to be 2.5 x 10 M
pOH = – log₁₀ [OH^– ]
pOH = – 1og₁₀ [2.5 x 10^-6]
= – log₁₀ [2.5] – log₁₀ [10^-6]
= – 0.3979 – ( – 6)
= – 0.3979 + 6
pOH = 5.6

We know that,
pH + pOH = 14
pH + 5.6 = 14
pH = 14 – 5.6
pH = 8.4
pH = 8.4, shows the nature of the solution is basic.

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H₃O⁺] = 4 x 10⁵ M. Is the solution neutral (or) acidic (or) basic.
Answer:
[H₃O⁺] = 4 x 10^-5M
pH = – log₁₀ [H₃O⁺]
pH = – 1og₁₀[4 x 10⁵]
pH = – log₁₀ [4] – log₁₀ [10^-5]
pH = – 0.6020 – ( – 5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.
Since pH is less than 7, the solution is acidic

Question 8.
Calculate the pH of 0.04 M HNO₃ solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = – 1og[H₃O⁺]
= – log (0.04)
= – log(4 x 10^-2)
= 2 – log4 = 2 – 0.6021
= 1.3979
= 1.40

Question 9.
Define solubility product.
Answer:
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.

Question 10.
Define ionic product of water. Give its value at room temperature.
Answer:
1. The product of the concentration of H⁺ and OH^– ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature (25°C) is,
K_w = [H⁺] [OH⁺] (or)
K_w= [H₃O⁺] [OH⁺]
K_w =(1 x 10^-7) (1 x 10^-7)
K_w= 1 x 10^-14 mol² dm^-6

Question 11.
Explain common ion effect with an example.
Answer:
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists.
CH₃COOH (aq) \(\rightleftharpoons\) H⁺(aq)+ CH₃COO^– (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO ion.
CH₃COONa (aq) → Na⁺ (aq) + CH₃COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s priñciple that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H ions to produce much more unionized CH₃COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12.
Derive an expression for Ostwald’s dilution law.
Answer:

• Ostwald’s dilution law relates the dissociation constant of the weak acid (Ka) with its degree of dissociation (a) and the concentration ( C)
• Ostwald’s dilution law states that, when dilution increases, the degree of dissociation of weak electrolyte also increases
• Degree of dissociation
  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
• Consider the dissociation of a weak acid (CH₃COOH)

Question 13.
Define pH.
Answer:
pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = – log₁₀ [H₃O⁺] (or) pH = – log₁₀ [H⁺]

Question 14.
Calculate the pH of 1.5 x 10^-3 M solution of Ba(OH)₂
Answer:
Acidity of Ba(OH)₂ in 2
∴ Normality = Molarity x acidity
= 15 x 10^-3 x 2
= 3 x 10^-3
So, [OH^–] = Normality = 3 x 10^-3
pOH = – log [OH^–]
= -log3 x 10^-3
= -log3 – log10^-3
= log 3 +3 log 10
= 3 – log3
= 3 – 0.4771
pOH = 2.5229
pH+ pOH = 14
pH = 14 – pOH
=14 – 2.5229
= 11.4771
pH = 11.48

Question 15.
50 ml of 0.05 M HNO₃ is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution:
Number of millinioles VrnI x Molarity
Millimoles of HNO₃ = 50 x 0.05 = 2.5
Millimoles of KOH = 50 x 0.025 = 1.25
After mixing millimoles of
HNO₃ remaining = 2.5 – 1.25 = 1.25
Total volume = 50 + 50 = 100 ml
Molarity = \(\frac{\text { Number of millimoles }}{\mathrm{Vml}}\)
= 1.25/100
Molarity = 1.25 x 10^-2
HNO₃ is mono basic.
∴ Normality Molarity x basicity
= 1.25 x 10^-2 x 1
= 1.25 x 10^-2
∴[H₃O⁺] = 1.25 x 10^-2
pH = – log[H₃O⁺]
= – log 1.25 x 10^-2
= – log 1.25 – log 10^-2
= – log 1.25 + 2 log 10
=2-log 1.25 = 2-0.0969
pH = 1.9031

Question 16.
The K_a value for HCN is 10^-9. What is the pH of 0.4 M HCN solution?
Answer:
K_a =10^-9
c = O.4M
pH = – log₁₀ [H₃O⁺]

∴ pH = – log(2 x 10^-5)
= – log 2 – log (10^-5)
= – 0.3010 + 5
pH = 4.699

Question17.
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that.
K_a = K_b = 1.8 x 10^-5
Solution:
K_a = K_b = 1.8 x 10^-5

Question 18.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Answer:
Consider the reaction between the strong acid HCl and the weak base NH4OH to form the salt NH₄Cl and water.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

In aqueous solution NH4C1 is completely dissociated as follows:
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4}{ }^{+}{ }_{(\mathrm{aq})}+\mathrm{Cl}^{-}{ }_{(\mathrm{aq})}\)

NH₄⁺ is the strong conjugate acid of the weak base NH₄OH.
Hence it has a tendency to react with 0H from water to form unionised NH4OH.
\(\mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}+\mathrm{H}^{+}(\mathrm{aq})\)
No such tendency is shown by Cl^– the weak conjugate base of the strong acid HCI.

In the above reaction 11+ ion is formed.
∴ [H⁺] > [OH^–] and the solution is acidic and the pH is less than 7.
Equilibrium constant (hydrolysis constant) for the above equilibrium is

Question 19.
Solubility product of Ag₂CrO₄ is 1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.01 M AgNO₃ solution?
Answer:


[Ag+] = 2 S + 0.01
0.01 > > 2 S, 2 S can be neglected.
∵ [Ag⁺] = 0.01 = 1 x 10^-2
[CrO₄^–] = S; K_sp= 1 x 10-12
∴ K_sp = [Ag⁺]² [CrO₄^2-]
1 x 10^-12 = (1 10^-2)² x s
∴ S = \(\frac{1 \times 10^{-12}}{1 \times 10^{-4}}\) ⇒ S = 1 x 10^-8M

Question 20.
Write the expression for the solubility product of Ca₃(PO₄)₂
Answer:
Ca₃(PO₄)₂ (s) \(\rightleftharpoons\) 3Ca²⁺ (3s) + 2PO₄^3- (2s)
Solubility of Ca₃(PO₄)₂ is,
K_sp = [Ca²⁺]³ . [PO₄^3-]²
K_sp = (3s)³ . (2s)²
K_sp= 27 s³ . 4s²
K_sp = 108s⁵.

Question 21.
A saturated solution, prepared by dissolving CaF₂(s) in water, has [Ca²⁺] = 3.3 x 10^-4 M. What is the K_spof CaF₂?
Answer:
CaF_{2 (s)} \(\rightleftharpoons\) Ca²⁺_(aq) + 2F^–_(aq)
[F^–] = 2 [Ca²⁺] = 2 x 33 x 10^-4 M
= 6.6 x 10^-4 M
= [Ca²⁺] [F^–]²
= (3.3 x 10^-4) (6.6 x 10^-4)²
= 1.44 x 10^-10

Question 22.
K_sp of AgCl is 1.8 x 10^-10. Calculate molar solubility in 1 M AgNO₃
Answer:

[Cl^–] = S
K_sp = [Ag⁺] [Cl^–]
1.8 x 10^-10 = (1) (S)
S = 1.8 x 10^-10M

Question 23.
A particular saturated solution of silver chromate Ag₂CrO₄ has [Ag⁺] = 5 x 10^-5 and [CrO₄]^2- = 4.4 x 10 M. What is the value of for Ag₂CrO₄?
Answer:
Ag₂CrO₄ (s) \(\rightleftharpoons\) 2Ag⁺(aq) + CrO₄^2-(aq)
K_sp = [Ag⁺]² [CrO₄^2-]
= (5 x 10^-5)² (4.4 x 10^-4)
= 1.1 x 10^-12

Question 24.
Write the expression for the solubility product of Hg₂CI₂.
Answer:

Question 25.
K_sp of Ag₂CrO₄ is 1.1 x 10^-12 What is the solubility of Ag₂CrO₄ in 0.1M K₂CrO₄.
Answer:

Question 26.
Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO₃)₂ and 0.100 L of 0.2 M NaCl are mixed? (PbCI₂) = 1.2 x 10^-5.
Answer:
Total volume = 0.150 + 0.100 = 0.250L

Number of moles of Pb2 = V x M
= 0.150 x 0.1

Number of moles of Cl^– = V x M = 0.100 x 0.2

Ionic Product = [Pb²⁺] [Cl^–]²
= (0,06) (0.08)²
= 3.84 x 10^-4
Given K_sp= 1.2 x 10^-5
Ionic product > K_sp
∴ Precipitation of PbCl₂ will occur

Question 27.
of Al(OH)₃ is 1 x 10^-15 M. At what pH does 1.0 x 10^-13 M AI³⁺ precipitate on the addition of buffer of NH₄CI and NH₄OH solution?
Answer:
Al(OH)₃ Al³⁺ (aq) + 3OH^– (aq)
K_sp = [Al³⁺] [OH^–]³
Al(OH)₃ precipitates when
[Al³⁺] [OH^–]³ > K_sp
(1 x 10^-3)[OH^–]³ > K_sp
[OH^–]³ > 1 x 10^-12
[OH^–] > 1 x 10^-4M
[OH^–] = l x 10^-4 M
pOH = – 1og₁₀[OH^–] = – log (1 x 10^-4) = 4
pH = 14 – 4 = 10
Thus, Al (OH)₃ precipitates at a pH of 10

III. Evaluate Yourself

Question 1.
Classify the following as acid (or) base using Arrhenius concept

1. HNO₃
2. Ba(OH)₂
3. H₃PO₄
4. CH₃COOH

Answer:
1. HNO₃:
Nitric acid, dissociates to give hydrogen ions in water.
HNO₃ is acid.

2. Ba(OH)₂:
Barium hydroxide dissociates to give hydroxyl ions in water.
Ba(OH)₂ is base.

3. H₃PO₄:
Orthophosphoric acid dissociates to give hydrogen ions in water.
H₃PO₄ is acid.

4. CH₃COOH:
Acetic acid dissociates to give hydrogen ions in water.
CH₃COOH is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid-base pairs.
i) NH₄
ii) H₂SO₄
iii) CH₃COOH.
Evaluate.
Answer:

Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.
(i). CaO + CO₂ → CaCO₃
ii) CH₃ – O – CH₃ – AlCl₃ →

Answer:
(i). CaO + CO₂ → CaCO₃
CaO – Lewis base – All metals oxides are Lewis bases
CO₂ – Lewis acid – CO₂ contains a polar double bond.

ii)

Question 4.
H₃BO₃ accepts hydroxide ion from water as shown below
Answer:
H₃BO₃ (aq) + H₂O(l) = B(OH)₄^– + H⁺
Predict the nature of H₃BO₃ using Lewis concept.

Question 5.
At a particular temperature, the K_w of a neutral solution was equal to 4 x 10^-14. Calculate the concentration of [H₃O⁺] and [OH^–].
Answer:
Given solution is neutral
[H₃O⁺] = [OH^–]
Let [H₃O⁺] = x ; then [OH^–] = x
K_w = [H₃O⁺] [OH^–]
4 x 10^-14 = x . x
x² = 4 x 10^-14
X = \(\sqrt{4 \times 10^{-14}}=2 \times 10^{-7}\)

Question 6.
a) Calculate pH of 10^-8 M H₂SO₄
b) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4
c) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M NaOH
Answer:
a.

In this case the concentration of H₂SO₄ is very low and hence [H₃O] from water cannot be neglected
∴ [H₃O⁺] = 2 x 10^-8 (from H₂SO₄) + 10^-7 (from water)
= 10^-8(2+ 10)
= 12 x 10^-8 = 1.2 x 10^-7
pH = – log₁₀[H₃O⁺]
= – log₁₀( 1.2 x 10^-7)
= 7 – log₁₀1.2
= 7 – 0.0791 = 6.9209

b. pH of the solution = 5.4
[H₃O⁺] = antilog of (- pH)
= antilog of (- 5.4)
= antilog of (-6 + 0.6) = \(\overline{6} .6\)
= 3.981 x 10^-6
i.e., 3.98 x 10^-6 mol dm^-3

c. No. of moles of HCl = 0.2 x 50 x 10^-3 = 10 x 10^-3
No. of moles of NaOH =0.1 x 50 x 10^-3 = 5 x 10^-3
No. of moles of HCl after mixing = 10 x 10^-3 – 5 x 10^-3
= 5 x 10^-3
after mixing total volume = 100 mL
∴ Concentration of HCl in moles per litre
\(=\frac{5 \cdot 10^{-3} \mathrm{~mole}}{100 \cdot 10^{-3} \mathrm{~L}} ;\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-2} \mathrm{M}\)
[H₃O⁺] = 5 x 10^-2 M
pH = – log ( 5 x 10^-2)
= 2 – log 5
= 2 – 0.6990
= 1.30

Question 7.
K_b for NH₄OH is 1.8 x 10^-5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution.
Answer:

Question 8.
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.
2. Calculate the pH of a buffer solution consisting of 0.4M CH₃COOH and 0.4 M CH₃COONa. What is the change in the pH after adding 0.01 mol of HCI to 500m1 of the above buffer solution.
Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105).
Answer:
1. Dissociation of buffer components
NH₄OH (aq) \(\rightleftharpoons\) NH₄⁺ (aq) + OH^– (aq)
NH₄CI → NH₄⁺ + Cl^–
Addition of OH^–
The added H⁺ ions are neutralized by NH₄OH and there is no appreciable decrease in pH.
NH₄OH(aq) + H⁺ \(\rightleftharpoons\) NH₄⁺(aq) + H₂O (1)
Addition of
NH₄^– (aq) + OH^– (aq) → NH₄OH (aq)
The added OH ions react with NH₄ to produce unionized NH₄OH. Since NH₄OH is a weak base, there is no appreciable increase in pH.

2. pH of buffer

Addition of 0.01 mol HCI to 500 ml of buffer
Added [H⁺]
= \(\frac{0.01 \mathrm{~mol}}{500 \mathrm{~mL}}=\frac{0.01 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}\)
= 0.02

∴ pH = – log (1.99 x 10^-5)
= 5 – log 1.99
= 5 – 0.30
= 4.70

Question 9.
1. How can you prepare a buffer solution of pH9. You are provided with 0.1 M NH₄OH solution and ammonium chloride crystals. (Given: pKb for NH₄OH is 4.7 at 25°C)

2. What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pK_a for formic acid is 3.75.)
Answer:
1. We know that pH + pOH = 14
9 + pOH = 14
= pOH = 14 – 9 = 5

[NH₄Cl] = 0.1 M x 1.995
= 0. 1995 M
=0.2 M
Amount of NH₄CI required to prepare 1 litre 0.2 M solution = Strength of NH₄CI x molar
mass of NH₄CI
= 0.2 x 535
= 10.70 g
10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH₄OH solution and the prepared NH₄CI solution will give a buffer solution with the required pH value (pH = 9).

2.

[Sodium formate] = number of moles of HCOONa
= 0.6 x V x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
= 80 x 10^-3
4 = 3.75 + log \(\frac { 0.6V }{ 80 }\)
0.25 = log \(\frac { 0.6V }{ 80 }\)
antilog of 0.25 = \(\frac { 0.6V }{ 80 }\)
0.6V = 1.778 x 80
= 1.78 x 80
= 142.4
V = \(\frac { 142.4 mL }{ 0.6 }\) = 237.33 mL

Question 10.
Calculate the
i) hydrolysis constant
ii) degree of hydrolysis and
iii) pH of 0.05M sodium carbonate solution pK_a for HCO₃^– is 10.26.
Answer:
Sodium carbonate is a salt of weak acid, H₂CO₃ and a strong base, NaOH and hence the solution is alkaline due to hydrolysis.

Given
K_w = 1 x 10^-14
c = 0.05 M
PK_a = 10.26
K_a = – log K_a
K_a = antilog of (- pK_a)
K_a = antilog of (- 10.26)
K_a = 5.49 x 10^-11

h = 6.034 x 10^-2

12th Chemistry Guide Ionic Equilibrium Additional Questions and Answers

I. Choose the best answer

Question 1.
The Latin word acidus means
a) bitter
b) sour
c) sweet
d) salty
Answer:
b) sour

Question 2.
According to Arrhenius concept, an acid is
a) hydrogen ion donor
b) hydrogen ion acceptor
c) hydroxyl ion donor
d) electron donor
Answer:
a) hydrogen ion donor

Question 3.
According to Arrhenius concept, a base is
a) hydrogen ion donor
b) hydroxyl ion acceptor
c) hydroxyl ion donor
d) electron acceptor
Answer:
c) hydroxyl ion donor

Question 4.
Arrhenius theory does not explain the behaviour of acids and bases in
a) aqueous solvents
b) non-aqueous solvents
c) water
d) None of the above
Answer:
b) non-aqueous solvents

Question 5.
According to Lowry-Bronsted theory, an acid is
a) proton donor
b) proton acceptor
c) hydroxyl ion donor
d) electron donor
Answer:
a) proton donor

Question 6.
According to Lowry-Bronsted theory, a base is
a) proton donor
b) proton acceptor
c) hydroxyl ion acceptor
d) electron donor
Answer:
b) proton acceptor

Question 7.
According to Lowry-Bronsted theory, an equilibrium exists between an acid and its
a) base
b) conjugate acid
c) conjugate base
d) water
Answer:
c) conjugate base

Question 8.
A conjugate acid-base pair differs only by
a) an electron
b) a proton
c) a hydroxyl ion
d) none of the above
Answer:
b) a proton

Question 9.
Lowry-Bronsted theory could not explain the acidic behaviour of
a) BF₃
b) AlCl₃
c) SO₂
d) all the above
Answer:
d) all the above

Question 10.
According to Lewis concept an acid is
a) proton donor
b) proton acceptor
c) electron pair acceptor
d) electron-pair donor
Answer:
c) electron pair acceptor

Question 11.
According to Lewis concept a base is
a) hydroxyl ion donor
b) hydroxyl ion acceptor
c) electron pair acceptor
d) electron-pair donor
Answer:
d) electron-pair donor

Question 12.
In a coordination complex the central metal ion acts as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
Answer:
d) Lewis acid

Question 13.
Which of the following is true for acidic solutions?
a) [H₃O⁺] > [OH^–]
b) [H₃O⁺] < [OH^–]
c) [H₃O⁺] = [OH^–]
d) [H₃O⁺] < [OH^–]
Answer:
a) [H₃O⁺] > [OH^–]

Question 14.
Electron deficient molecules can act as
a) Arrhenius base
b) Lewis base
c) Lowry-Bronsted base
d) Lewis acid
Answer:
d) Lewis add

Question 15.
Molecules with one or more lone pairs of electrons can act as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
Answer:
b) Lewis base

Question 16.
Which among the following is a Lewis base?
a) BF₃
b) SO₃
C) SF₄
d) CaO
Answer:
d) CaO

Question 17.
Which among the following is not a Lewis base?
a) MgO
b) CO₂
c) H₂O
d) Na₂O
Answer:
b) CO₂

Question 18.
Which among the following is a Lewis acid?
a) NH₃
b) CaO
c) RNH₂
d) FeCl₃
Answer:
d) FeCl₃

Question 19.
Which among the following is not a Lewis acid?
a) SiF₄
b) CH₂=CH₂
c) BeF₂
d) Fe³⁺
Answer:
b) CH₂=CH₂

Question 20.
The conjugate acid of H2O is
a) HCl
b) OH^–
c) H₃O⁺
d) HSO₄^–
Answer:
c) H₃O⁺

Question 21.
The conjugate base of H2O is
a) HCl
b) OH^–
c) H₃O⁺
d) HSO₄^–
Answer:
b) OH^–

Question 22.
Which of the following can act both as Bronsted acid and Bronsted base?
a)Cl^–
b) H₃O⁺
c) HCO₃^–
d) CO₃^2-
Answer:
c) HCO₃^–

Question 23.
In which of the following cases, the sparingly soluble salt solution is unsaturated?
a) Ionic product > solubility product (Ksp)
b) Ionic product < solubility product (Ksp)
c) Ionic product = solubility product (KcD)
d) Both (a) and (b)
Answer:
b) Ionic product > solubility product (Ksp)

Question 24.
Which of the following is the strongest conjugate base?
a) Cl^–
b) SO₄^2-
c) CH₃COO^–
d) NO₃^–
Answer:
c) CH₃COO^–
Reason: The conjugate base of a weak acid is a stronger base.

Question 25.
Among the following the Ka value for a strongest acid is
a) 1.8 x 10^-4
b) 1.8 x 10^-5
c) 1.8 x 10⁵
d) 1.8 x 10⁴
Answer:
c) 1.8 x 10⁵

Question 26.
Which among the following is the strongest base?
a) HSO₄^–
b) H₂O
C) F^–
d) H^–
Answer:
d) H^–
H₂ is the weakest acid and its conjugate base
H^– is the strongest base.

Question 27.
Which of the following salts do not undergo salt hydrolysis?
a) Sodium acetate
b) Ammonium acetate
c) Ammonium chloride
d) Sodium nitrate
Answer:
d) Sodium nitrate

Question 28.
If the hydrogen ion concentration of the solution is 10^-5M, its hydroxyl ion concentration is
a) 10^-5 M
b) 10^-9M
c) 10^-14M
d) 10^-7M
Ans:
b) 10^-9M

Question 29.
If the hydrogen ion concentration of a solution is 10~SM, its pOH is
a) 5
b) 9
c) 14
d) 7
Answer:
b) 9 Reason: pH = -log[H₃0⁺] = -log 10^-5 = 5
pH + pOH = 14; pOH = 14 – pH = 14 – 5 = 9

Question 30.
If the pH of a solution is 9, the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic
Answer:
c) basic
Reason:.pH < 7; pOH > 7 = Acidic
pH > 7; pOH < 7 = Basic

Question 31.
If the pH of a solution is zero the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic
Answer:
d) strongly addic

Question 32.
The pOH of IN HCI is
a) 0
b) 1
c) 7
d) 14
Answer:
d) 14
Reason: pH = -log[H₃O⁺] = -log 1 = 0
pOH = 14 – pH = 14 – 0 = 14

Question 33.
As [H₃O⁺] of a solution increases, its pH
a) increases
b) deposes
c) remains the same
d) becomes
Answer:
b) decreases

Question 34.
The concentration of a solution of acetic acid changes from 10^-4 M to 10^-2 M, its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
b) decreases
Reason: As dilution increases (Concentration decreases) degree of dissociation increases. Here concentration increases. Hence degree of dissociation decreases.

Question 35.
Which of the following represents Ostwald’s dilution law for a binary electrolyte whose degree of disscociation is a and concentration C

Answer:
c

Question 36.
NH₄OH is a weak base because
a) It has a low vapour pressure
b) It is completely ionised
c) it is only partially ionised
d) It has low density
Answer:
c) it is only partially ionised

Question 37.
The concentration of acetic acid changes from 10^-2M to 10^-4 M its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
a) increases

Question 38.
For a weak acid the hydrogen ion concentration is given as

Answer:
d) both (a) & (b)

Question 39.
When ammonium chloride is added to ammonium hydroxide, the degree of dissociation of ammonium hydroxide
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
b) decreases
Reason: Common ion effect

Question 40.
The relationship between the solubility product (Ksp) and molar solubility (S) for Ag2 (OO4) is
a) K_sp = S³
b) K_sp = S²
c) K_sp = 4S³
d) K_sp = 3S²
Answer:
c) K_sp = 4S³

Question 41.
The decrease in degree of dissociation of HF on addition of NaF is known as
a) buffer action
b) neutralization
c) common ion effect
d) hydrolysis
Answer:
c) common ion effect

Question 42.
Which is not a buffer solution?
a) CH₃COOH + CH₃COONa
b) HCl + NaCl
c) NH₄OH + NH₄Cl
d) H₂CO₃+NaHCO₃
Answer:
b) HCl + NaCl
Reason: A strong acid and its salt is not a buffer solution. A weak acid or base and its salt is a buffer solution.

Question 43.
Which among the following does not undergo hydrolysis?
a) NH₄Cl
b) CH₃COONa
c) NaCl
d) CH₃COONH₄
Answer:
c) NaCl
Reason: NaCt is a salt of strong acid and strong base.

Question 44.
Hydrolysis of CH3COONa gives
a) acidic solution
b) basic solution
c) neutral solution
d) No solution
Answer:
b) basic solution
Reason: CH₃COONa is a salt of weak acid and strong base. Hence CH₃COO^– undergoes hydrolysis giving a basic solution.

Question 45.
The pH of the solution resulting from the hydrolysis of NH₄Cl is
a) 7
b) greater than 7
c) less than 7
d) 14
Answer:
c) less than 7
Reason: NH₄Cl is a salt of strong acid and weak base. NH₄C⁺ undergoes hydrolysis giving an acidic solution. Hence pH is less than 7.

Question 46.
For the hydrolysis of salt of weak acid and weak base which among the following is true?
a) K_h K_a = K_w
b) K_h K_b = k_w
c) K_a K_b K_h = K_w
d) K_h= \(\frac{K_{1} K_{b}}{K_{w}}\)
Answer:
c) K_a K_b K_h = K_w

Question 47.
The pH of the solution obtained from the hydrolysis of salt of strong acid and weak base

Answer:
b

Question 48.
\(X_{m} Y_{n(s)} \stackrel{H_{2} O}{\rightleftharpoons} m X_{(a q)}^{n+}+n Y_{(a q)}^{m-}\)
The solubility product of the salt X_mY_n is given as
a) K_sp = [X^m+]ⁿ[Y^n-]^m
b) K_sp = m[Xⁿ⁺] n[Y^m-]
c) K_sp = [Xⁿ⁺]^m [Y^m-]ⁿ
d) K_sp =[X_mY_n]^m+n
Answer:
c) K_sp = [Xⁿ⁺]^m [Y^m-]ⁿ

Question 49.
The aqueous solution sdium acetate, ammonium chloflde, sodi maitrate are respectively. PTA -61
a) Neutral,acidic,basic
b) acidic, basic, neutral
c) basic, acidic, neutral
d) basic, aidk, basic
Answer:
d) basic, acidic, basic

Question 50.
The relationship between the solubility product and molar solubility of Al₂(SO₄)₃ is
a) S²
b) 4S³
c) 108S⁵
d) 27S⁵
Answer:
c) 108S⁵
Reason:

II. Pick out the correct statements

Question 1.
i) Ka measures the strength of an acid
ii) Larger the pK_a value, stronger is the acid.
iii) Weak acids undergo complete ionisation
iv) Acids with K_a value greater than 10 are considered as strong acids.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement:
ii) Larger the pK_a value, weaker is the acid, (or) Lower the pKa value, stronger is the acid.
iii) Weak acids undergo partial ionisation, (or) strong acids undergo complete ionisation.

Question 2.
i) Degree of dissociation
\(=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
ii) dilution law is applicable to strong acids.
iii) Strong acids undergo complete dissociation
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.
a) (i) & (ii) b) (ii) & (iii)
c) (ii) & (iv) d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statement:
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement:
ii) Ostwald’s dilution law is applicable to strong acids.
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.

Question 3.
i) A buffer solution is a mixture of a weak acid and its conjugate base
ii) Common ion effect is based on Le Chatelier’s principle.
iii) Blood contains a buffer solution of HCl and NaCl
iv) For a buffer solution
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
\(\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{acid}]}{[\mathrm{salt}]}\)
Answer:
a) (i) & (ii)
Correct statement:
iii) Blood contains a buffer solution of H₂CO₃ and NaHCO₃
iv) For a buffer solution
pH = pKa + log\(\frac{[\text { acid }]}{[\text { salt }]}\)

Question 4.
i) The solution obtained from the hydrolysis of CH₃COONa is acidic
ii) The solution obtained from the hydrolysis of NaCl is neutral
iii) The nature of the solution obtained from the hydrolysis of a salt of weak acid and weak base depends on the strength of acid or base.
iv) The solution obtained from the hydrolysis of NH4C/ is basic
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
i) The solution obtained from the hydrolysis of CH₃COONa is basic
iv) The solution obtained from the hydrolysis of NH₄Cl is acidic

III. Pic out the correct statements

Question 1.
For the equilibrium
\(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{ClO}_{4}^{-}\)
Which are the incorrect statements?
i) HClO₄ is the conjugate acid of H₂O
ii) H₂O is the conjugate base of H₃O⁺
iii) H₃O⁺ is the conjugate base of H₂O
iv) \(\mathrm{ClO}_{4}^{-}\) is the conjugate base of HClO₄
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement:
i) HClO₄ is the conjugate acid of \(\mathrm{ClO}_{4}^{-}\)
iii) H₃O⁺ is the conjugate acid of H₂O

Question 2.
i) All metal ions are Lewis acids.
ii) All metal oxides are Lewis acids.
iii) All anions are Lewis bases
iv) Molecules that contain polar double bond
are Lewis bases
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (ìv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statement:
ii) All metal oxides are Lewis bases.
iv) Molecules that contain polar double bond are Lewis acids.

Question 3.
i) The conjugate base of OH^– is O^2-
ii) The conjugate acid of \(\mathbf{N H}_{2}^{-}\) is NH₃
iii) The conjugate acid of H₂SO₄ is \(\mathbf{H S O}_{4}^{-}\)
iv) The conjugate base of H₂O is H₃O⁺
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement:
iii) The conjugate acid of H₂SO₄ is \(\mathbf{H S O}_{4}^{-}\)
iv) The conjugate base of H₂O is H₃O⁺

Question 4.
i) If the hydrogen ion concentration of a solution is 10^-9 M, it is basic
ii) 1f the pOH of a solution is 14, it is strongly basic.
iii)As pH of a solution increases, its pOH also increases.
iv) At 25°C the sum of pH and pOH of a solution
is 14.
a) (i) & (ii) b) (ii) & (iii) c) (iii) & (iv)
d) (j) & (iv) Ans: b) (ii) & (iii)
Correct statement:
ii) If the pOH of a solution is 14 it is strongly acidic. (or) 1f the pH of a solution is 14, it is strongly basic
iii) As pH of a solution increases, its pOH decreases. (or) As pH of a solution decreases, its pOH increases.

IV. Assertion and reason

Question 1.
Assertion (A): Ionic product of water K_w increases with increase in temperature
Reason (R): Dissociation of water is an exothermic reaction.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong.
Correct R: Dissociation of water is an endothermic reaction.

Question 2.
Assertion (A): Carbanion is a Lewis base
Reason (R): Carbanion can donate a pair of electron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A& R are correct, R explains A Correct statement:

Question 3.
Assertion (A): An aqueous solution of IICI is acidic
Reason (R): An aqueous solution of HCl
contains less H30* than OH- ions.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong.
Correct R: An aqueous solution of HCl contains more H₃O⁺ than 0H ions.

Question 4.
Assertion (A): When sodium acetate is added to acetic acid, the dissociation of acetic acid increases.
Reason (R): The addition of CH₃COO^– ion shifts the equilibrium of dissociation of acetic acid to the left.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
d) A is wrong, but R is correct.
Correct A: When sodium acetate is added to acetic acid, the dissociation of acetic acid decreases.

V. Match the following:

Question 1.

Answer:
i) c
ii) d
iii) a
iv) b

Question 2.

Answer:
i) c
ii) d
iii) b
iv) a

VII. Two Mark Questions

Question 1.
What are Arrhenius acids and base? Give examples.
Answer:

Question 2.
What are the limitations of Arrhenius concept of acids and bases?
Answer:

• Does not explain the behaviour of acids and bases in non-aqueous solvents such as acetone, THF etc.,
• Does not account for the basic nature of the substances like ammonia which do not contain a hydroxyl group.

Question 3.
How does BF₃ act as Lewis acid?
Answer:

Boron has a vacant 2p-orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. Hence BF₃ acts as a Lewis acid.

Question 4.
Arrange HCl, HCOOH and CH3COOH in their increasing order of acid strength if their K_a values at 25°C are 2 x 10⁶, 1.8 x 10^-4 and 1.8 x 10^-5 respectively.
Answer:

• As K_a value increases, the acid strength increases.
• Increasing order of Ka value is 1.8 x 10^-5 < 1.8 x 10^-4 < 2 x 10⁶
• The increasing order of acid strength is CH₃COOH < HCOOH < HCl.

Question 5.
What is neutralization?
Answer:
The reaction in which an acid reacts with a base to form salt and water is called neutralization.

Question 6.
What do yo mean by salt hydrolysis?
Answer:
Salt hydrolysis is the reaction of the cation or the anion or both the ions of the salt with water to produce either acidic or basic solution.

Question 7.
What is Buffer index (β) ?
Answer:
It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pFl by unity.
\(\beta=\frac{\mathrm{dB}}{\mathrm{d}(\mathrm{pH})}\)
dB = number of gram equivalents of acid/base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.

Question 8.
How solubility product is determined by molar solubility?
Answer:
Solubility product can be calculated from the molar solubility i.e,, the maximum number of moles of solute that can be dissolved in one litre of the solution.

Question 9.
Define pOH
Answer:
pOH of a solution is defined as the negative logarithm of base 10 of the molar concentration of hydroxyl ions present in the solution. pOH = -log₁₀[OH^–]

Question 10.
Write the pH value of the following substances.
(a) Vinegar
(b) Black coffee
(c) Baking soda
(d) Soapy water
Answer:

VII.Three Mark Questions

Question 1.
Derive the relation between pH and pOH
Answer:

Question 2.
What are buffer solutions? Explain their types with examples.
Answer:
Solutions which resist drastic changes in its pH upon the addition of small amount of acids or bases are called buffer solutions.
A buffer solution consists of a mixture of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt)

Types:
Acidic buffer: Solution of acetic acid and sodium, acetate
Basic buffer: Solution of ammonium hydroxide and ammonium chloride.

Question 3.
What are conjugate acid-base pairs? Give example.
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
The species that remains after the donation of
a proton is a base (Base}) and is called the conjugate base of the Bronsted acid (Acid}). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs.
Conjugate acid – base pair

VIII. Five Mark Questions.

Question 1.
Write a note on Lewis’s concepts of acids and bases ?
Answer:

Question 2.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of weak acid and strong base. Answer:
Consider the reaction between the weak acid CH₃COOH and the strong base NaOH to form the salt CH₃COONa and water.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aqq}}+\mathrm{NaOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{I})}\)

In aqueous solution CH3COONa is completely dissociated as follows,
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}+\mathrm{Na}_{(l)}^{+}\)
CH₃ COO^– is the strong conjugate base of the weak acid CH₃COOH
\(\mathrm{CH}_{3} \mathrm{COO}_{\text {(aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{\text {(ач) }}^{-}\)
No such tendency is shown by Na+, the weak conjugate acid of the strong base NaOH.
In the above reaction OH^– ion is formed.
[OH^–] > [H⁺] and the solution is basic and the pH is greater than 7.
Equilibrium constant (hydrolysis constant) for the above

Dissociation of the weak acid is CH3COOH(aq) \(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}{ }_{(\mathrm{aq})}+\mathrm{H}^{+}{ }_{(a q)}\)
Dissociation constant of the weak acid

Question 3.
Explain buffer action of acidic buffer?
Answer:

• The chemical reaction which is responsible to resist the pH change of a buffer solution is called its buffer action.
• To resist changes in its pH on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components.
• These components neutralise the effect of added acid or base.
• At the same time, these components should not consume each other.
• Consider an acidic buffer containing CH₃COOH and CH₃COONa.

When an acid is added to this buffer, the added H⁺ ions are consumed by the conjugate base CH₃COO^– to form undissociated CH₃COOH. This keeps the H+ concentration and the pH unchanged.
\(\mathrm{CH}_{3} \mathrm{COO}_{(a \mathrm{q})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{ay}}\)

When a base is added to this buffer, the added OH- ions are consumed by H30+. To maintain the equilibrium more CH3COOH is dissociated keeping the H+ Concentration and the pH unchanged

Question 4.
Derive Henderson-Hasselbalch equation.
Answer:
Consider an acidic buffer

• The weak acid is dissociated only to a small extent.
• Due to the common ion effect, the dissociation is further suppressed.
• Hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid.
• Similarly, the equilibrium concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

Equations (4) & (5) are known as
Henderson-Hasselbaich equations.

Question 5.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of a weak acid and weak base.
Answer:

• Consider the reaction between the weak acid CH₃COOH and the weak base NH₄OH to form the salt CH₃COONH₄ and water.
  
• \(\mathrm{NH}_{4}^{+}\) is the strong conjugate acid of the weak base NH₄OH.
• CH₃COO^– is the strong conjugate base of the weak acid CH₃COOH.
• Hence both \(\mathrm{NH}_{4}^{+}\) and CH₃COO^– have the tendency to react with water.
• The nature of the solution depends on the strength of acid or base.
• if K_a >K_b then the solution is acidic and pH > 7.
• if K_a < K_b then the solution is basic and pH > 7.1
  
• if K_a = K_b then the solution is neutral and pH = 7.

Consider the hydrolysis of both \(\mathrm{NH}_{4}^{+}\) and CH₃COO^–

Dissociation of a weak acid

Dissociation of weak base

Question 6.
0.1 M Solution of HF is a weak acid. But 5M solution of HF is a stronger acid. Why?
Answer:

• Hydrofluoric acid is a much stronger acid when it is concentrated than when it is diluted. As the concentration of hydrofluoric acid approaches 100 percent.
• It’s acidity increases because of home association, where a base and conjugate acid form a bond.
• 3HF ⇌ H₂F⁺ + HF₂^–
• The FHF-bifluoride anion is stabilized by a strong hydrogen bond between hydrogen and fluorine.
• The stated ionization constant of hydrofluoric acid, 10 – 3.15, does not reflect the true acidity of concentrated HF solutions.
• Hydrogen bonding also accounts for the higher boiling point of HF compared to other hydrogen halides.

IX. Additional problems:

Problems based on pH.

Question 1.
What is the pH of 0.001M NaOH?
Solution:
[OH^–] = Normality = Molarity x acidity
= 0.001 x 1 = 1 x 10^-3
pOH = -log 10 [OH^–] = -log 1 x 10^-3
= -log 1 – log 10^-3 [log 1 = 0]
= 0 + 3log10 [log10 = 1]
pOH = 3
pH + pOH = 14
pH = 14 – pOH
= 14-3
pH = 11

Question 2.
What is the pH of a solution with hydronium ion concentration 6.2 x 10^-9 M.
Solution:
[H₃O⁺] = 6.2 x 10^-9
pH = -log10[H₃O⁺]
= – log 6.2 x 10^-9
= – log 6.2 – log 10^-9
= – log 6.2 + 9 log 10
= 9 – log 6.2
= 9 -0 .7924
pH = 8.2076

Question 3.
What is the pH of 10^-2 M Ca(OH)₂?
Solution:
[OH^–] = Normality = Molarity x acidity
= 10^-2 x 2
[OH^–] = 2 x 10^-2
pOH = -log10[OH^– ]
= -log 2 x 10^-2
= -log 2 – log 10^-2
= -log 2 + 21ogl0
= 2 – log 2
= 2 – 0.3010
pOH = 1.6990
pH+pOH = 14
pH = 14 -pOH
= 14 -1.6990
pH = 12.3010

Question 4.
Calculate the hydroxyl ion concentration of a solution with pH = 5
Solution:
[H₃O⁺] = 10^pH
[H₃O⁺] = 10^-5
[H₃O⁺][OH^–] = 10^-14
[OH^–] = \(\frac{10^{-14}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\)
= \(\frac{10^{-14}}{10^{-5}}\)
[OH^–] = 10^-9

Question 5.
Calculate the pH of 10^-8M HNO₃
[H₃O⁺] = 10-8 which is less than 10 7
[H₃O⁺] from water should be added.
[H₃O⁺] = 10 “(from water) + 10-8(from HN03)
[H₃O⁺] = 10-7 (1+10-1)
= 10-7 (1+0.1)
[H₃O⁺] = 1.1 x 10^-7
pH = – log10 [H₃O⁺]
= – log 1.1 x 10^-7
= – log 1.1 – log 10^-7
= – log 1.1 + 7 log 10
= 7 – log 1.1
= 7 – 0.0414
pH = 6.9586

Question 6.
The dissociation constant of 0.1 M weak acid is 1 x 10^-5. Calculate its pH.
Answer:

Question 7.
If a solution with pH = 5 is diluted 100 times. What will be the pH of the resulting solution?
Solution:
pH = 5; [H₃O⁺] = 10-pH
[H₃O⁺] = 10^-5
After dilution
[H₃O⁺] = \(\frac{10^{-5}}{100}=10^{-7} \mathrm{M}\)
As [H₃O⁺] = 10^-7 M, the [H₃O⁺] contributed
by water should be added.
Final [H₃O⁺] = 10^-7 (from water) + 10^-7 (from the solution
= 10^-7(1+1)
[H₃O⁺] = 2 x 10^-7M
= – log [H₃O⁺]
= – log 2 x 10^-7
= – log 2 – log 10^-7
= – log 2 + 7 log 10
= 7 – log 2
= 7 – 0.3010
= 6.6990

Question 8.
Calculate the hydrogen ion concentration in human blood. (pH of human blood is 7.4)
Solution:
pH = 7.4
pH = – logio [H₃O⁺]
log [H₃O⁺] = -pH
[H30+] = Antilog of (-pH)
= Antilog of (-7.4)
= Antilog of (-7-0.4)
= Antilog of (-7-0.4+1-1)
= Antilog of (-8+0.6)
= Antilog of (8.6)
[H₃O⁺] = 3.98 x 10^-8

Question 9.
Calculate the pH of a solution obtained by mixing 50ml of 0.4 N HC1 and 50ml of 0.2N NaOH.
Solution:
Milli equivalent of 50 ml, 0.4N HCl
= V_m1 x N = 50 x 0.4 = 20
Milli equivalent of 50 ml, 0.2N NaOH
= V_m1 x N = 50 x 0.2 = 10
Remaining milli equivalent of Hcl = 20-10 =10

Question 10.
Calculate the pH of 10^-7M HCl.
Answer:
If we do not consider [H₃O⁺] from the ionisation of H₂O, then [H₃O⁺] = [HCl]= 10^-7M)
In this case the concentration of the acid is very low (10^-7M). Hence, the [H₃O⁺] (10^-7M) formed due to the auto ionisation of water cannot be neglected.
[H₃O⁺] = 10^-7 (from HCl) + 10^-7 (from H₂O)
= 10^-7 (1+1) = 2 x 10^-7
pH = -log₁₀ [H₃O⁺]
= log₁₀[2 x 10^-7] = – log 2 + log 10^-7
= -log 2 – (-7) log10¹⁰
= 7 – log2 .
= 7 – 0.3010=6.6990
= 6.70

Problems based on Henderson – Hasselbaich equations

Question 1.
A buffer solution contains equal volumes of 0.2 M NH₄OH and 0.02 M NH₄Cl. Kb of NH₄OH is 1 x 10^-5. Calculate the pH of the buffer.
Solution:
[Base] = 0.2 M; [salt] = 0.02 M
K_b = 1 x 10^-5
pK_b = -log K_b = – log 1 x 10^-5 = 5
For a basic buffer
\(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\text { Base }]}\)
= 5+log 0.02/0.2
= 5+1og10^-1
= 5 – 1
pOH = 4
pH + pOH = 14
pH = 14 – pOH
= 14 -4
pH = 10

Question 2.
Find the pH of a buffer solution containing 0.20 mole per litre CH₃COONa and 0.15 mole per litre CH₃COOH, K_a for acetic acid is 1.8 x 10^-5
Solution:
[acid] = 0.15 M; [salt] = 0.20 M
K_a = 1.8 x 10^-5
PK_a = -log Ka = – log L8 x 10^-5
= – log 1.8 – log10^-5
= – log 1.8 + 5 log 10
= 5 – log 1.8
= 5 – 0.2553 pKa
PK_a = 4.7447
For an acidic buffer
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
= 4.7447 + log 0.20/0.15
= 4.7447 + log 4/3
= 4.7447 + log 4 – log 3
= 4.7447+0.6021-0.4771
pH = 4.8697

Question 3.
Calculate the pKb of NH₄OH, if the pH of a buffer solution containing 0.1N NH₄OH and 0.1 N NH₄Cl is 9.25.
Solution;)
pH of basic buffer = 9.25
∴ pOH = 14 – pH
= 14 – 9.25
pOH = 4.75
[Base] = 0.1N ; [Salt] = 0.1 N
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
PK_b = 4.75 – log M = 4.75 – log 0.1/0.1 = 4.75 – 0 = 4.75

Question 4.
Calculate the pH of the solution by mixing 1.5 mole of HCN and 0.15 mole of KCN in water and making up the total volume to 0.5 litre.
Solution:
K_a = 5 x 10^-10;
fAcid] = 1.5/0 5 – 3.0
[Salt] = 0.15/0.5 = 0.3
pK_a = – log K_a
= – log 5 x 10^-10
= – log 5 – log 10^-10
= – log 5 + 10 log^-10
= 10 – log 5
= 10 – 0.6990
pK^-10 = 9.3010
\(\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\)
= 9.3010 + log 0.3/3.0
= 9.3010 + log 10^-10
= 9.3010 -1 log 10
pH = 8.3010

Problems based on Ostwald’s dilution law

Question 1.
Calculate the degree of dissociation of 0.1 N CH₃COOH if Ka for CH₃COOH is 1 x 10-5.
Solution:

Question 2.
What is the dissociation constant of 0.1 M HCN which is 0.01% ionised?
Solution:
C = 0.1M ; α = 0.01% = \(\frac{0.01}{100}\) = 1 x 10 ^-4
Ka = ?
Ka = C α² = 0.1 x (1 x 10^-4)² = 1 x 10^-9

Question 3.
A weak monobasic acid is 1% ionised in 0.1M solution at 25°C. What is the percentage of ionisation in its 0.025M Solution?
Solution:
C =0.1 M; α = 1% = 1/100 = 10^-2 Ka = ?
C =0.025 M % a = ?
Ka = C α² = 0.1 x (10^-2)² = 10^-5
\(\alpha=\sqrt{\frac{K a}{C}}=\sqrt{\frac{10^{-5}}{0.025}}=0.02\)
% α =100 x α = 100 x 0.02 = 2%

Question 4.
A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solution?
Solution:
C = 0.005 M
pH = 5
∴ [H₃O⁺] = 10^-ph = 10^-5
HA → H⁺ + A^–
ka = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
[H⁺] = [A^–] = 10^-5
[HA] = 0.005M

α = 2 x 10^-3
% α = 100 x α
= 100 x 2 x 10^-3 = 0.2%

Problems based on Solubility Product

Question 1.
Calculate the solubility product of AgCl if the solubility of AgCl is 1.3 x 10^-5 moles per litre at 25°C.
Solution:
S = 1.3 x 10^-5 M ; Ksp = ?

Question 2.
Calculate the solubility of Calcium fluoride in a saturated solution if its solubility product is 3.2 x 10^-11 M³.
Solution:
Ksp = 3.2 x 10^-11 M³; S = ?

Question 3.
The solubility of AgCl is 0.0014 g per litre at 18°C. Calculate its solubility product at 18°C. Molecular weight of AgCl is 143.5. Solubility in
Answer:

Ksp = [Ag⁺][Cl^–]
Ksp = S x S
Ksp = S²
Ksp = (9.756 x 10^-6M)²
Ksp = 95.1795 x 10^-12
Ksp = 9.518 x 10^-11 M²

Question 4.
The solubility product of a salt having general formula MX₂ in water is 4 x 10^-12. What is the concentration of M²⁺ ions in the aqueous solution of the salt?
Solution:

Question 5.
Solid Ba (N0₃)₂ is gradually dissolved in 1 x 10^-4 m Na₂CO₃ solution. At what concentration of Ba²⁺ will a precipitate begin to form?
(Ksp for BaCO₃ = 5.1 x 10^-9 )
Solution:

Ionic Product > Ksp, Precipitation will occur.
∴ After attaining 5.1 x 10^-5 M [Ba²⁺] concentration, a precipitate will begin to form.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.6

Choose the most suitable answer from the given four alternatives:

Question 1.
Let X be random variable with probability density function

Which of the following statement is correct?
(a) Both mean and variance exist
(b) Mean exists but variance does not exist
(c) Both mean and variance do not exist
(d) Variance exists but Mean does not exist
Solution:
(b) Mean exists but variance does not exist
Hint:

Question 2.
A rod of length 2l is broken into two pieces at random. The probability density function of the shorter of the two pieces is

The mean and variance of the shorter of the two pieces are respectively
(a) \(\frac { l }{ 2 }\), \(\frac { l^2 }{ 3 }\)
(b) \(\frac { l }{ 2 }\), \(\frac { l^2 }{ 6 }\)
(c) l, \(\frac { l^2 }{ 12 }\)
(d) \(\frac { l }{ 2 }\), \(\frac { l^2 }{ 12 }\)
Solution:
(d) \(\frac { l }{ 2 }\), \(\frac { l^2 }{ 12 }\)
Hint:

V(X)= E(X²) – [E(x)]²
\(\frac { l^2 }{ 3 }\) – \(\frac { l^2 }{ 4 }\) = \(\frac { l^2 }{ 12 }\)

Question 3.
Consider a game where the player tosses a six-sided fair die. If the face that comes up is 6, the player wins Rs 36, otherwise he loses Rs k², where k is the face that comes up k = {1, 2, 3, 4, 5}. The expected amount to win at this game in Rs is
(a) \(\frac { 19 }{ 6 }\)
(b) –\(\frac { 19 }{ 6 }\)
(c) \(\frac { 3 }{ 2 }\)
(d) –\(\frac { 3 }{ 2 }\)
Solution:
(b) –\(\frac { 19 }{ 6 }\)
Hint:
Probability mass function

Question 4.
A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Hint:
X^-1 (7) = {(3,4), (4, 3), (5, 2), (6, 1)} = 4

Question 5.
A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
(d) 2
Hint:
n = 25, p = \(\frac { 4 }{ 5}\), q = \(\frac { 1 }{ 5 }\)
SD = \(\sqrt { npq }\)
= \(\sqrt { 25 × \frac { 4 }{ 5} × \frac { 1 }{ 5} }\)
= 2

Question 6.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. Then the possible values of X are
(a) i + 2n, i = 0, 1, 2 …, n
(b) 2i – n, i = 0, 1, 2 …, n
(c) n – i, i = 0, 1, 2 …, n
(d) 2i + 2n, i = 0, 1, 2 …, n
Solution:
(b) 2i – n, i = 0, 1, 2 …, n
Hint:
if n = 1 then x = 0,1
if n = 2 then x = -2, 0, 2
if n = 3 then x = -3, -1, 1, 3
Let x = 2i – n
when n = 2
x = 2i – 2; i = 0, 1, 2
i = 0 ⇒ x = -2
i = 1 ⇒ x = 0
i = 2 ⇒ x = 2
x = -2, 0, 2
when n =3
x = 2i – 3; i = 0, 1, 2, 3
i = 0 ⇒ x = 3
i = 1 ⇒ x = -1
i = 2 ⇒ x = 1
i = 3 ⇒ x = 3
x = -3, -1, 1, 3
∴ 2i – n, i = 0, 1, 2, ……. n.

Question 7.
If the function f(x) = \(\frac { 1 }{ 12 }\) for a < x < b represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
(a) 0 and 12
(b) 5 and 17
(c) 7 and 19
(d) 16 and 24
Solution:
(d) 16 and 24
Hint:

Question 8.
Four buses carrying 160 students from the same school arrive at a football stadium. The buses carry, respectively, 42, 36, 34, and 48 students. One of the students is randomly selected. Let X denote the number of students that were on the bus -carrying the randomly selected student. One of, the 4 bus drivers is also randomly selected. Let Y denote the number of students on that bus. Then E[X] and E[Y] respectively are
(a) 0.11
(b) 1.1
(c) 11
(d) 1
Solution:
(b) 1.1
Hint:

Question 9.
Two coins are to be flipped. The first coin will land on heads with probability 0.6, the second with Probability 0.5. Assume that the results of the flips are independent and let X equal the total number of heads that result. The value of E[X] is
(a) 50, 40
(b) 40, 50
(c) 40.75, 40
(d) 41, 41
Solution:
(c) 40.75, 40
Hint:
P(H₁) = 0.6 P(H₁) = 0.4
P(H₂) = 0.5 P(H₂) = 0.5
Random variable x = 0, 1, 2
P(x = 0) = P(H₁) P(H₂)
= 0.4 × 0.5 = 0.2
P(x = 1) = P(H₁).P(H₂) + P(H₂).P(H₁)
= 0.6 × 0.5 + 0.5 × 0.4
= 0.3 + 0.2 = 0.5
P(X = 2) = P(H₁).P(H₂)
= 0.6 × 0.5 = 0.3

E(x) = 0 × 0.2 + 1 × 0.5 + 2 × 0.3
= 0 + 0.5 + o.6
= 1.1

Question 10.
On a multiple-choice exam with 3 possible destructives for each of the 5 questions, the probability that a student will get 4 or more correct answers just by guessing is
(a) \(\frac { 11 }{ 243 }\)
(b) \(\frac { 3 }{ 8 }\)
(c) \(\frac { 1 }{ 243 }\)
(d) \(\frac { 5 }{ 243 }\)
Solution:
(c) \(\frac { 11 }{ 243 }\)
Hint:

Question 11.
If P(X = 0) = 1 – P(X = 1). If E[X] = 3 Var(X), then P(X = 0) is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 2 }{ 5 }\)
(c) \(\frac { 1 }{ 5 }\)
(d) \(\frac { 1 }{ 3 }\)
Solution:
(d) \(\frac { 1 }{ 3 }\)
Hint:
P(X = 0) = 1 – P(x = 1)
Let P(x = 0) = a
1 – P(X = 1) = a
P(X = 1) = 1 – a

E(x) = 0 × a + 1 × 1 – a
= 0 + 1 – a
= 1 – a
E(x²) = 0² × a + 1² × 1 – a
= 1 – a
Var x = E(x²) – (E(x))²
= 1 – a – (1 – a)²
= 1 – a – (1 – 2a + a²)
= 1 – a – 1 + 2a – a²
= a – a²
E(x) = 3 var (x)
1 – a = 3 (a – a²)
1 – a = 3a – 3a²
3a² – 4a + 1 = 0
(3a – 1)(a – 1) = 0
a = \(\frac { 1 }{ 3 }\) a = 1
P(X = 0) = a = \(\frac { 1 }{ 3 }\)

Question 12.
If X is a binomial random variable with I expected value 6 and variance 2.4, Then P(X = 5) is

Solution:
(d)
Hint:
E(X) = np = 6, Var(x) = npq = 2.4

Question 13.
The random variable X has the probability density function
and E(X) = \(\frac { 7 }{ 12 }\) then a and b are respectively
(a) 1 and \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 2 }\) and 1
(c) 2 and 1
(d) 1 and 2
Solution:
(a) 1 and \(\frac { 1 }{ 2 }\)
Hint:

Question 14.
Suppose that X takes on one of the values 0, 1 and 2. If for some constant k, P(X = i) = kP(X = i – 1) for i = 1, 2 and P(X = 0) = \(\frac { 1 }{ 7 }\). Then the value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
P(X = 0) = \(\frac { 1 }{ 7 }\)
P(X = i) = k P(X = i – 1)
i = 1
P(X = 1) = k P(X = 0)
= \(\frac { k }{ 7 }\)
i = 2
P(X = 2) = k P(X = 1)
= k(\(\frac { k }{ 7 }\)) = \(\frac { k^2 }{ 7 }\)
Probability mass function

Σ p(x) = 1
\(\frac { 1 }{ 7 }\) + \(\frac { k }{ 7 }\) + \(\frac { k^2 }{ 7 }\) = 1
1 + k + k² = 7
k² + k – 6 = 0
(k + 3)(k – 2) = 0
k = -3 (Not possible); k = 2
∴ k = 2

Question 15.
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only
(c) III only
(d) II and III
Solution:
(a) I and II

Question 16.
If is a probability density function of a random variable, then the value of a is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1
Hint:
f is p.d.f \(\int_{0}^{a}\)
\(\int_{0}^{a}\) 2x dx = 1
[x²]\(_{0}^{a}\) = 1
a² = 1
a = 1

Question 17.
The probability mass function of a random variable is defined as:

Then E(X) is equal to:
(a) \(\frac { 1 }{ 15 }\)
(b) \(\frac { 1 }{ 10 }\)
(c) \(\frac { 1 }{ 3 }\)
(d) \(\frac { 2 }{ 3 }\)
Solution:
(d) \(\frac { 2 }{ 3 }\)
Hint:
f(x) is p.m.f, Σ f(x) = 1
k + 2k + 3k + 4k + 5k = 1
15k = 1
k = \(\frac { 1 }{ 15 }\)
E(X) = -2k -2k + 0 + 4k + 10k
= 10 k
= 10 × \(\frac { 1 }{ 15 }\) = \(\frac { 2 }{ 3 }\)

Question 18.
Let X have a Bernoulli distribution with a mean of 0.4, then the variance of (2X – 3) is
(a) 0.24
(b) 0.48
(c) 0.6
(d) 0.96
Solution:
(d) 0.96
Hint:
p = 0.4
q = 1 – p
= 0.6
Var (x) = pq
= 0.4 × 0.6
= 0.24
Var (2x – 3) = 2² var(x)
= 4 var(x)
= 4(0.24)
= 0.96

Question 19.
If in 6 trials, X is a binomial variable which follows the relation 9P(X = 4) = P(X = 2), then the probability of success is
(a) 0.125
(b) 0.25
(c) 0.375
(d) 0.75
Solution:
(b) 0.25
Hint:
9 P(X = 4) = P(X = 2)
9 x ⁶C₄ p⁴ q² = ⁶C₂ p² q⁴
9p² = q²
9p² = 1- 2p + p²
8P² + 2p – 1 = 0
p = \(\frac { 1 }{ 4 }\) = 0.25

Question 20.
A computer salesperson knows from his past experience that he sells computers to one in every twenty customers who enter the showroom. What is the probability that he will sell a computer to exactly two of the next three customers?
(a) \(\frac { 57 }{ 20^3 }\)
(b) \(\frac { 57 }{ 20^2 }\)
(c) \(\frac { 19^3 }{ 20^3 }\)
(d) \(\frac { 57 }{ 20 }\)
Solution:
(a) \(\frac { 57 }{ 20^3 }\)
Hint:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5

Question 1.
Compute P(X = k) for the binomial distribution, B(n, p) where
(i) n = 6, p = \(\frac { 1 }{ 3 }\), k = 3
(ii) n = 10, p = \(\frac { 1 }{ 5 }\), k = 4
(iii) n = 9, p = \(\frac { 1 }{ 2 }\), k = 7
Solution:

Question 2.
The probability that Mr. Q hits a target ait any trial is \(\frac { 1 }{ 4 }\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) atleast one time
Solution:
Let p be the probability of hitting the target
n = 10, p = \(\frac { 1 }{ 4 }\), X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
q = 1 – p = \(\frac { 3 }{ 4 }\)
(i) Probability of hitting the target exactly 4 times

(ii) Probability of hitting atleast one time
P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
= 1 – ¹⁰C₀ × (\(\frac { 1 }{ 4 }\))° × (\(\frac { 3 }{ 4 }\))¹⁰
= 1 – \(\frac { 3^{10} }{ 4^{10} }\)

Question 3.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times and X denotes the number of heads.
(ii) A fair die is tossed 240 times and X denotes the number of times that four appeared.
Solution:
(i) Let p be the probability of getting heads
⇒ p = \(\frac { 1 }{ 2 }\)
n = 100, p = \(\frac { 1 }{ 2 }\), X ~ B(n, p)
q = 1 – p = \(\frac { 1 }{ 2 }\)
Mean = np
= 100 × \(\frac { 1 }{ 2 }\)
= 50
Variance = npq
= 50 × \(\frac { 1 }{ 2 }\)
= 25

(ii) Let p be the probability of getting 4 when a die is thrown
n = 240, p = \(\frac { 1 }{ 6 }\), X ~ B(n, p)
q = 1 – p = \(\frac { 5 }{ 6 }\)
Mean = np
= 240 × \(\frac { 1 }{ 6 }\) = 40
Variance = npq
= 40 × \(\frac { 5 }{ 6 }\) = \(\frac { 100}{ 3 }\)

Question 4.
The probability that a certain kind of component will survive a electrical test is \(\frac { 3 }{ 4 }\). Find the probability that exactly 3 of the 5 components tested survive.
Solution:
Let p be the probability of a certain component survive in an electrical test.
n = 5, p =\(\frac { 3 }{ 4 }\), X ~ B(n, p), k = 3
q = 1 – p = \(\frac { 1 }{ 4 }\)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n

Question 5.
A retailer purchases a certain kind of electronic, device from a manufacturer. The manufacturer indicates that the defective rate of the device 5 is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be
(i) atleast one defective item
(ii) exactly two defective items?
Solution:
Let p be the probability that indicates the defective rate of an electronic device
p = 5%
= 0.05
q = 1 – p = 0.95
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 – (X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀ (0.05)° (0.95)¹⁰
= 1 – (0.95)¹⁰

(ii) Exactly two defective items
P(X = 2) = ¹⁰C₂ (0.05)² (0.95)⁸

Question 6.
If the probability that fluorescent light has a useful life of atleast 600 hours is 0.9, find the probabilities that among 12 such lights
(i) exactly 10 will have a useful life of atleast 600 hours
(ii) At atleast 11 will have a useful life of atleast 600 hours
(iii) At atleast 2 will not have a useful life of at least 600 hours.
Solution:
Let p be the probability of the useful life hours of a fluorescent light.
n = 12, p = 0.9, X ~ B(n, p)
q – 1 – p = 0.1
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
(i) Exactly 10
P(X = 10) = ¹²C₁₀ (0.9)¹⁰ (0.1)²

(ii) Atleast 11
P(X ≥ 11) = P(X = 11) + P(X = 12)
= ¹²C₁₁ (0.9)¹¹ (0.1)¹ + ¹²C₁₂ (0.9)¹² (0.1)°
= 12 × (0.9)¹¹ × 0.1 + 1 × (0.9)¹² × 1
= (0.9)¹¹ (12 × 0.1 × 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (2.1) (0.9)¹¹

(iii) Atleast 2 will not have a useful
P(X ≤ 10) = 1 – P(X > 10)
= 1 – [P(X = 11) + P(X = 12)]
= 1 – (2.1)(0.9)¹¹

Question 7.
The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find
(i) The probability mass function
(ii) P(X = 3)
(iii) P(X ≥ 2).
Solution:
X ~ B(n, p)
Given, Mean = 6, Standard deviation = 2
np = 6, \(\sqrt { npq }\) = 2
npq = 4
\(\frac { npq }{ np }\) = \(\frac { 4 }{ 6 }\)
q = \(\frac { 2 }{ 3 }\)
p = 1 – q = \(\frac { 1 }{ 3 }\)
npq = 4
n × \(\frac { 1 }{ 3 }\) × \(\frac { 2 }{ 3 }\) = 4
n = 18
(i) Probability mass function
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n

Question 8.
If X ~ B(n, p) such that 4P(X = 4) = P(x = 2) and n = 6. Find the distribution, mean and standard deviation of X.
Solution:
X ~ B(n, p)
Given, 4P(X = 4) = P(X = 2), n = 6
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
⇒ 4 ⁶C₄ p⁴ q² = ⁶C₂ p² q⁴
⇒ 4 p² = q²
4 (1 – q)² = q²
4(1 – 2q + q²) = q²
3q² – 8q + 4 = 0
(q – 2) (3q – 2) = 0
q = \(\frac { 2 }{ 3 }\) (q ≠ 2)
Distribution
P(X = x) = ⁶C_x (\(\frac { 1 }{ 3 }\))^x (\(\frac { 2 }{ 3 }\))^6-x
x = 0, 1, 2, 3, 4, 5, 6
Mean = np = 6 × \(\frac { 1 }{ 3 }\) = 2
Variance = npq = 2 × \(\frac { 2 }{ 3 }\) = \(\frac { 4 }{ 3 }\)

Question 9.
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable.
Solution:
n = 5, X ~ B(n, p)
Given, P(X = 1) = 0.4096
P(X = 2) = 0.2048
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
⇒ ⁵C₁ p q⁴ = 0.4096
⇒ ⁵C₂ p² q³ = 0.2048
5 p q⁴ = 0.4096 …….. (1)
10 p² q³ = 0.2048 ………. (2)
(1) divided by (2),
⇒ \(\frac { 5pq^4 }{ 10p^2q^3 }\) = 2
\(\frac { p }{ q }\) = 4
q = 4p
q = 4 (1 – q)
q = 4 – 4q
5q = 4
q = \(\frac { 4 }{ 5 }\)
p = 1 – q = \(\frac { 1 }{ 5 }\)
Mean = np
= 5 × \(\frac { 1 }{ 5 }\) = 1
Variance = npq
= 1 × \(\frac { 4 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Distribution
P(X = x) = ⁵C_x (\(\frac { 1 }{ 5 }\))^x (\(\frac { 4 }{ 5 }\))^5-x
x = 0, 1, 2, 3, 4, 5.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.4

Question 1.
For the random variable X with the given prob-ability mass function as below, find the mean and variance.

Solution:


Var(X)= E(X²) – [E(X)]²
= 8.1 – (2.3)²
= 8.1 – 5.29
= 2.81

Var (X) = E(X²) – [E(X)]²
= 3.33 – (1.67)²
= 3.33 – 2.79
= 0.54




Var (X) = E(X²) – [E(X)]²
= 8 – 4
= 4

Question 2.
Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let X be the possible outcomes drawing red balls. Find the probability mass function and mean for X.
Solution:
Let X be the random variable that denotes the number of red balls.
X = {0, 1, 2}
Sample space consist of ⁷C₂ = 21
X= 0, X^-1 (BB) = ³C₂
X= 1, X^-1 (BR) = ³C₁ × ⁴C₁ = 12
X = 2, X^-1 (RR) ⁴C₂ = 6

Question 3.
If µ and σ² are the mean and variance of the discrete random variable X and E(X + 3) = 10 and E(X + 3)² = 116, find µ and σ²
Solution:
Given E(X + 3) = 10
E(aX + b) = aE(X) + b
E(X + 3) = 10 ⇒ E(X) + 3 = 10
⇒ µ = E(X) = 7
E(X + 3)² = 116
E(X² + 6X + 9) = 116
E(X²) + 6E(X) + 9 = 116
E(X²) + 6(7) + 9 = 116
E(X²) = 65
σ² = Var (X) = E(X²) – [E(X)]²
= 65 – 49
= 16

Question 4.
Four fair coins are tossed once. Find the probability mass function, mean and variance for a number of heads that occurred.
Solution:
Let X be the random variable that denotes the number of heads when four coins are tossed once.
X ={0, 1, 2, 3}
n(S) = 16

Var (X) = E(X²) – [E(X)]²
= 5 – 4
= 1

Question 5.
A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let X denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the pdf of X is

Obtain and interpret the expected value of the random variable X.
Solution:

The average waiting time for the student is 15 minutes.

Question 6.
The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function

Find the expected life of this electronic equipment.
Solution:

Expected life of the electronic equipment = \(\frac { 1 }{ 3 }\)

Question 7.
The probability density function of the random variable X is given by

find the mean and variance of X.
Solution:
Mean:

Var (X) = E(X²) – [E(X)]²
\(\frac { 3 }{ 8 }\) – \(\frac { 1 }{ 4 }\)
= \(\frac { 1 }{ 8 }\)

Question 8.
A lottery with 600 tickets gives one prize of Rs 200, four prizes of Rs 100 and six prizes of Rs 50. If the ticket costs is Rs 2, find the expected winning amount of a ticket.
Solution:
Let X be the random variable denotes the amount win in the lottery

= 0.5
Loss = Rs 0.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 9 Human Resource Management Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 9 Human Resource Management

12th Commerce Guide Human Resource Management Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question. 1
Human resource is a …………. asset.
a) Tangible
b) Intangible
c) Fixed
d) Current
Answer:
b) Intangible

Question 2.
Human Resource management is both ………………….. and ………………..
a) Science and art
b) Theory and practice
c) History and Geography
d) None of the above
Answer:
a) Science and art

Question 3.
Planning is a …………. Function.
a) selective
b) pervasive
c) both a and b
d) none of the above
Answer:
b) pervasive

Question 4.
Human resource management determines the ……………………….. relationship.
a) internal, external
b) employer, employee
c) Owner, Servant
d) Principle, Agent
Answer:
b) employer, employee

Question 5.
Labour turnover is the rate at which employees ………….. the organisation.
a) enter
b)leave
c) salary
d)None of the above
Answer:
b) leave

II. Very Short Answer Questions.

Question 1.
Give the meaning of Human Resource.
Answer:
In an organisation, the human resource is the employees who are inevitable for the survival and success of the enterprise.

Question 2.
What is Human Resource Management?
Answer:
The Branch of Management that deals with managing resource is known as “Human Resource Management (HRM)

Question 3.
State two features of HRM.
Answer:
Features of Human Resource Management.
Universally relevant: Human Resource Management has universal relevance.
Goal-oriented: The accomplishment of organisational goals is made possible through the best utilisation of human resources in an organisation.

Question 4.
Mention two characteristics of Human Resource. [OWE]
Answer:

• HR is Only factor of production that lives.
• HR can be Work as a team.
• HR – Exhibits innovations and creativity.

Question 5.
What are the managerial functions of HRM.
Answer:
The functions of human resource management may be classified as under: I Managerial function – Planning, Organising, Directing, Controlling.
II Operative function – Procurement, Development, Compensation, Retention, Integration, Maintenance.

III. Short Answer Questions.

Question 1.
Define the term Human Resource Management.
Answer:
HRM – Definition:
“HRM as that part of management process which is primarily concerned with the human constituents of an organisation”. E.F.L. BRECH

Question 2.
What are the characteristics of Human resources? [OWE] [MOM]
Answer:

1. Human resource exhibits innovation and creativity.
2. Human resources alone can think, act, analyse and interpret.
3. Human resources are emotional beings.
4. Human resources can be motivated either financially or non-financially.
5. The behaviour of human resources is unpredictable.
6. Over the years human resources gain value and appreciation.
7. Human resources are movable.
8. Human resources can work as a team.

Question 3.
What is the significance of Human resource?
Answer:

• HR is possible only through human relations.
• HR managers all other factors of production.
• HR can be utilised at all levels of management.
• HR is helpful to utilise all other resources.
• HR helps in industrial relation.
• HR well protected by legal frame works.

Question 4.
State the functions of Human Resource Management.
Answer:
The functions of human resource management may be classified as under: I Managerial function – Planning, Organising, Directing, Controlling.
II Operative function – Procurement, Development, Compensation, Retention, Integration, Maintenance.

1. Planning: Planning is deciding in advance what to do, how to do and who is to do it. It bridges the gap between where we are and where we want to go.
2. Organising: It includes division of work among employees by assigning each employee their duties, delegation of authority as required.
3. Procurement: Acquisition deals with job analysis, human resource planning, recruitment, selection, placement, transfer and promotion.
4. Development: Development includes performance appraisal, training, executive development, career planning and development, organisational development.
5. Compensation: It deals with job evaluation, wage and salary administration, incentives, bonuses, fringe benefits and social security schemes.

IV. Long Answer Questions.

Question 1.
Explain the characteristics of Human Resource. [OWE] [MOM] [LACE] (any 5) Characteristics
Answer:

•  HR is the Only factor of production that lives.
•  HR can Work as a team.
•  HR Exhibits innovation and creativity.
•  HR is Movable.
•  HR Over years – gain values.
•  HR can Motivated either in cash or in kind.
•  HR the Labour of employees that is hired and not the employee himself.
•  HR alone Act, Analyse, Think and interpret.
•  HR Create all other resources.
•  HR are Emotional beings.

Question 2.
Describe the significance of Human Resource Management. (MEERA) (JCE) (any 5)
Answer:
The role of human resource management is the process of acquiring, training, appraising, and compensating employees. The significance of human resource management is given below:

1. To identify manpower needs: Determination of manpower needs in an organisation is very important as it is a form of investment.
2. To ensure the correct requirement of manpower: At any time the organisation should
   not suffer from shortage or surplus manpower which is made possible through human resource management
3. To select the right man for the right job: Human resource management ensures the right talent to select the right employee for the right job.
4. To update the skill and knowledge: Human resource management enables employees to remain up-to-date through training and development programmes.
5. To appraise the performance of employees: A periodical appraisal of the performance of employees through human resource management activities boosts up good performers and motivates slow performers.

Question 3.
Elaborate on the Managerial functions of Human Resource Management.
Answer:
Managerial Functions : [COP] [D]
Controlling”

• It is comparing the actuals with the standards and to check whether activities are going on as per plan and rectify deviations.
• The control process includes fixing the standards, measuring actual performance, comparing the actual with standard laid down, measuring deviations and taking corrective decisions.
• This is made possible through observation, supervision, reports, records and audit.

Organising:

• It includes division of work among employees assigning duties to each employee.
• Delegation of authority as required, creation of accountability to make employees responsible.

Planning:

• Planning is thinking before doing.
• Planning is deciding in advance What to do, How to do, Who is to do it.
• It bridges the gap between where we are and where we want to go.
• It involves determination of objectives, policies, procedures, rules, strategies, programmes and budgets.

Directing:

• It involves issue of orders and instructions along with supervision, guidance and motivation to get the best out of employees.
• This reduces waste of time, energy and money and early attainment of organisational objectives.

Question 4.
Discuss the Operative functions HRM.
Answer:
Operating functions of HRM:

1. Procurement: Acquisition deals with job analysis, human resource planning, recruitment, selection, placement and promotion.
2. Development: It includes performance appraisal, training, executive development, and organizational development.
3. Compensation: It deals with job evaluation, wage and salary administration, incentives, bonus schemes.
4. Retention: This is made possible through health and safety, social security, job satisfaction and quality of work life.
5. Maintenance: This encourages employees to work with job satisfaction, reducing labour turnover, for human resource.

12th Commerce Guide Human Resource Management Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
create all other resources.
a) Skill
b) workforce
c) Natural resources
d) Human resource
Answer:
d) Human resources

Question 2.
The function of HRM concerned with …………………………
a) Motivating
b) Maintaining
c) Hiring
d) All of these
Answer:
d) All of those

Question 3.
How to encourage the best performer in an organisation?
a) Bonus
b) incentives
c) Both (a) and (b)
d) NOTA
Answer:
c) Both (a) and (b)

Question 4.
Human Resource Exbhits ……………………
a) Talent
b) Creativity
c) Innovation
d) All of those
Answer:
d) All of those

Question 5.
Who are inevitable for the survival and success of the enterprise?
a) Employee
b) Proprietors
c) Debtors
d) Creditors
Answer:
a) Employee

Question 6.
HR is ………………
a) Im-morable
b) In-tangble
c) Movable
d) NOTA
Answer:
c) Movable

II. Match the following

Question 1.
Match List I with List – II

Answer:
a) (i) 4,(ii) 1,(iii) 2,(iv) 3

III. Very short answer questions.

Question 1.
Define the term Human Resource.
Answer:
Man of all sources available to him, can grow and develop”.

Question 2.
Give two points of difference between HR and HRM .[MO]
Answer:

IV. Long Answer Questions.

Question 1.
Explain the features of HRM? [CUGAI]
Answer:
Continuous Process:
As long as there is HR in the running of an organisation, the activities relating to managing , HR exists.

Universal Relevent:

• HRM has universal relevance.
• The approach and style varies depending the nature of organisation structure and is applicable at all levels.

Goal Oriented :
The accomplishment of organisational goals is made possible through best utilisation of HR is an organisation.

A Systematic Approach:
HRM lavs emphasis on a systematic approaching in managing the tasks performed by HR of an organisation.

Intangible:
HRM is an intangible function which can be measured only by results.

Question 2.
Differentiate HR from HRM.[MOPET]
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.3

Question 1.
The probability density function of X is given by

Find the value of k.
Solution:
Since f(x) is a probability density function

Question 2.
The probability density function of X is

Find
(i) P(0.2 ≤ X < 0.6)
(ii) P(1.2 ≤ X < 1.8)
(iii) P(0.5 ≤ X < 1.5)
Solution:

Question 3.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function

Find
(i) the value of k
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Solution:

(i) Since f(x) is a probability density function

Question 4.
The probability density function of X is given by

Find
(i) the value of k
(ii) the distribution function
(iii) P(X < 3)
(iv) P(5 ≤ X)
(v) P(X ≤ 4)
Solution:
(i) Since f is a probability density function

(ii) The distribution function F(x) = \(\int_{-\infty}^{x}\) f(u) du

Question 5.
If X is the random variable with probability density function f(x) given by,

then find
(i) the distribution function F(x)
(ii) P(-0.5 ≤ x ≤ 0.5)
Solution:
Distribution function F(x) = \(\int_{-\infty}^{x}\) f(u) du
case 1.
x < -1
F(x) = \(\int_{-\infty}^{x}\) f(u) du = 0

Question 6.
If X is the random variable with distribution function F(x) given by,

Find
(i) the probability density function f(x)
(ii) P(0.3 ≤ X ≤ 0.6)
Solution:
(i) Probability density function

(ii) P(0.3 ≤ X ≤ 0.6)
P(a ≤ X ≤ b) = F(b) – F(a)
P(0.3 ≤ X ≤ 0.6) = F(0.6) – F(0.3)
= \(\frac { 1 }{ 2 }\) [(0.6)² + (0.6)] – \(\frac { 1 }{ 2 }\) [(0.3)² + (0.3)]
= \(\frac { 1 }{ 2 }\) (0.96 – 0.39)
= 0.285

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2

Question 1.
Three fair coins are tossed simultaneously. Find the probability mass function for a number of heads that occurred.
Solution:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable that denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Probability mass function

Question 2.
A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces and ‘5’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) P(4 ≤ X < 10)
(iv) P(X ≥ 6)
Solution:
Let X be the random variable denotes the total score in two thrown of a die.
Sample space S

n (S) = 36
X = {2, 4, 6, 8, 10}

(i) Probability mass function

(ii) Cumulative distribution function

= \(\frac { 36 }{ 36 }\)
= 1

Question 3.
Find the probability mass function and cumulative distribution function of a number of girl children in families with 4 children, assuming equal probabilities for boys and girls.
Solution:
Let X be the random variable denotes the number of girl child among 4 children
X = {0, 1, 2, 3, 4}

(i) Probability mass function

(ii) Cumulative distribution

Question 4.
Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by

Find (i) the value of k.
(ii) cumulative distribution function
(iii) P(X ≥ 1)
Solution:

(i) Given f is a probability mass function
\(\sum_{x}\) f(x) = 1
Probability mass function is

(ii) Cumulative distribution function
F(0) = P(x ≤ 0)
= P(x = 0)
= \(\frac { 1 }{ 8 }\)
F(1) = P(x ≤ 1)
= P(x = 0) + P(x = 1)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
F(2) = P(x ≤ 2)
= P(x = 0) + P(x = 1) + P(x = 2)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) = 1

(iii) P(X ≥ 1)
= P(X = 1) + P(X = 2)
= \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) + \(\frac { 7 }{ 8 }\)

Question 5.
The cumulative distribution function of a discrete random variable is given by

Find
(i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Solution:

f(-1) = P(X = -1) = F(-1) – F(-1) = 0.15 – 0 = 0.15
f(0) = P(X = 0) = F(0) – F(-1) = 0.35 – 0.15 = 0.20
f(1) = P(X = 1) = F(1) – F(0) = 0.60 – 0.35 = 0.25
f(2) = P(X = 2) = F(2) – F(1) = 0.85 – 0.60 = 0.25
f(3) = P(X = 3) = F(3) – F(2) = 1 – 0.85 = 0.15

(ii) P(X < 1)
P(X < 1) = P(X = -1) + P(X = 0)
= 0.15 + 0.20
= 0.35

(iii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3)
= 0.25 + 0.15
= 0.40

Question 6.
A random variable X has the following probability mass function.

Find
(i) the value of k
(ii) P(2 ≤ X < 5)
(iii) P(3 < X)
Solution:
(i) Given f(x) in a probability mass function
\(\sum_{x}\) f(x) = 1
k² + 2k² + 3k² + 2k + 3k = 1
6k² + 5k = 1
6k² + 5k – 1 = 0
(k + 1) (6k – 1) = 0
k = \(\frac { 1 }{ 6 }\)
(k ≠ -1 neglecting negative terms)
Probability mass function

(ii) P(2 ≤ X < 5)
P(2 ≤ X < 5) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 2 }{ 36 }\) + \(\frac { 3 }{ 36 }\) + \(\frac { 2 }{ 6 }\)
= \(\frac { 2+3+12 }{ 36 }\) + \(\frac { 17 }{ 36 }\)

(iii) P(X > 3) = P(X = 4) + P(X = 5)
= \(\frac { 2 }{ 6 }\) + \(\frac { 3 }{ 6 }\)
= \(\frac { 5 }{ 6 }\)

Question 7.
The cumulative distribution function of a discrete random variable is given by

Find
(i) the probability mass function
(ii) P(X < 3) and (iii) P(X ≥ 2)
Solution:
(i) Probability mass function

(ii) P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 5 }\)
= \(\frac { 4 }{ 5 }\)

(ii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 10 }\)
= \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.1

Question 1.
Suppose X is the number of tails that occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in it. inverse – images.
Solution:
Let X is the random variable that denotes the number of tails when three coins are tossed simultaneously.
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ ‘X’ takes the values 0,1, 2, 3
i.e., X (HHH) = 0 ; X (HHT, HTH, THH) = 1 ; X (HTT, THT, TTH) = 2 ; X (TTT) = 3

Question 2.
In a pack of 52 playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and number of points in its inverse images.
Solution:
Let X be the random variable of number of black cards occur.
∴ X = {0, 1, 2}
The sample space consists of 52C₂ = 1326
X = 0, X (both are white cards) = 26C₂ = 325
X = 1, X (one black and one white cards)
= 26C₁ × 26C₁
= 676
X = 2, X (both are black cards) = 26C₂ = 325

Question 3.
An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
Solution:
Number of mangoes = 5
Number of Apples = 4
Total number of fruits = 9
Let ‘X’ be the random variable that denotes the number of apples taken, then it takes the values 0, 1, 2, 3
X (MMM) = 0
X (AMM (or) MAM (or) MMA) = 1
X (AAM (or) AMA (or) MAA) = 2
X (AAA) = 3

Question 4.
Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win Rs 15 for each red ball selected and we lose Rs 10 for each black ball selected. If X denotes the winning amount, then find the values of X and the number of points in its inverse images.
Solution:
Let X be the random variable that denotes the j winning amount.
X (Both are black balls) = Rs 2 (-10) = -Rs 20
X (one red and one black ball) = Rs 15 – Rs 10 = Rs 5
X (both are red ball) = Rs 2 (15) = Rs 30
∴ X = {-20, 5, 30}
The sample space consists of ¹⁴C₂ = 91
X = – 20, Both black balls = ⁸C₂ = 28
X = 5, One black, one red ball = ⁸C₁ × 6C₁
= 8 × 6 = 48
X = 30, Both are white balls = ⁶C₂ = 15

Question 5.
A six-sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on the remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Solution:
Let X be the random variable denotes the total score is two throws of a die.
Sample Space S

n (S) = 36
X = {4, 5, 6, 7, 8}
From the sample space

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf 8 Securities Exchange Board of India Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 8 Securities Exchange Board of India

I. Choose The Correct Answer.

Question 1.
Securities Exchange Board of India was first established in the year…………………
a) 1988
b) 1992
c) 1995
d) 1998
Answer:
a) 1988

Question 2.
The headquarters of SEBI is…………..
a) Calcutta
b) Bombay
c) Chennai
d) Delhi
Answer:
b) Bombay

Question 3.
In which year SEBI was constituted as the regulator of capital markets in India?
a) 1988
b) 1992
c) 2014
d) 2013
Answer:
a) 1988

Question 4.
Registering and controlling the functioning of collective investment schemes as………….
a) Mutual Funds
b) Listing
c) Rematerialisation
d) Dematerialization
Answer:
d) Dematerialization

Question 5.
SEBI is empowered by the Finance ministry to nominate ………………….. members on the Governing body of every stock exchange.
a) 5
b) 3
c) 6
d) 7
Answer:
b) 3

Question 6.
The process of converting physical shares into electronic form is called ………………
a) Dematerialisation
b) Delisting
c) Materialisation
d) Debarring
Answer:
a) Dematerialisation

Question 7.
Trading is dematerialized shares commenced on the NSE is ………………………
a) January 1996
b) June 1998
c) December 1996
d) December 1998
Answer:
c) December 1996

Question 8.
…………..was the first company to trade its shares in Demat form.
a) Tata Industries
b) Reliance Industries
c) Infosys
d) Birla Industries
Answer :
b) Reliance Industries

Question 9.
…………….. enables small investors to participate in the investment on the share capital of large companies.
a) Mutual Funds
b) Shares
c) Debentures
d) Fixed deposits
Answer:
a) Mutual Funds

Question 10.
PAN stands for …………………
a) Permanent Amount Number
b) Primary Account Number
c) Permanent Account Number
d)Permanent Account Nominee
Answer:
c) Permanent Account Number

II. Very Short Answer Questions.

Question 1.
Write short notes on SEBI.
Answer:
Securities and Exchange Board of India (SEBI) was first established in the year 1988 as a non-statutory body for regulating the securities market.

Question 2.
Write any two objectives of SEBI.
Answer:
Control Over Brokers:
The important object is to supervise or check the activities of the Brokers, and other intermediaries in order to control the capital market.

Regulation of Stock Exchange :

• The first objective of SEBI is to regulate the Stock Exchange.
• So that efficient services may be provided to all the parties operating there.

Question 3.
What is a Demat Account?
Answer:
A Demat account holds all the shares that are purchased in electronic or dematerialized form.

Question 4.
Mention the headquarters of SEBI.
Answer:
Head Quarters – Mumbai [Bandra Kurla – Complex]
.
Question 5.
What are the various ID proofs?
Answer:
Proof of identity: PAN card, voter’s ID, passport, driver’s license, bank attestation, IT returns, electricity bill, telephone bill, ID cards with applicant’s photo issued by the central or state government are the ID proofs.

III. Short Answer Questions.

Question 1.
What is meant by Dematerialization?
Answer:

•  Dematerialization [DEMAT] is the process by which physical share certificates of an investor are taken back by the Company or Register and destroyed.
• Then an equivalent number of securities in the Electronic form is credited to the investor’s account with his Depository Participant. [DP]
•  DEMAT is done at the request of the Investor.
•  He has to open an account with a DP.
•  In DEMAT A/c – Purchase – Credited his account.
•  In DEMAT A/ c – Sales – Debited his account

Question 2.
What are the documents required for a Demat account?
Answer:

• For opening a Demat account, we submit proof of identity and address along with a passport size photo and the account opening form.
• Documents for Identity: Pan card, Voters ID, Passport, Driver’s License, IT Returns, Electricity and Telephone Bills are the Identity documents.
• Documents for Address: Ration card, Passport, Voter’s ID card, Driving License, Telephone Bills, Electricity bills are the address documents.

Question 3.
What is the power of SEBI under the Securities Contract Act?
Answer:

•  Power to grant License to the new stock exchange.
•  Power to direct any stock exchange to amend the rules.
•  Power to supersede governing body of any stock exchange.
•  Power to ask for information and accounts of the stock exchange.
•  Power to suspend the business of the stock exchange.
•  Power to prohibit contracts of the stock exchange.

Question 4.
What is meant by Insider trading?
Answer:
Insider trading means the buying and selling of securities by directors, promoters, etc., who have access to some confidential information about the company. This affects the interests of the general investors and is essential to check this tendency.

Question 5.
Draw the organizational structure of SEBI.
Answer:
Organization Structure of SEBI

IV. Long Answer Questions.

Question 1.
What are the functions of SEBI?
Answer:
functions:

•  SEBI is the NODAL agency which safeguards the interests of an investor in the Indian Financial Market.
  SEBI performs:
  
• Safeguarding the interest of the investors by means of adequate education and guidance.
• SEBI makes Rules and Regulations that must be strictly followed by the participants of the financial market.
• Regulating and Controlling the business on Stock Exchange.
• Registration of Brokers and sub-brokers is made mandatory.
• Conduct inspection and inquiries of stock exchanges, intermediaries, and self-regulating organizations.
• Barring insider trading in securities.
• Prohibiting deceptive and unfair methods by intermediaries.
• Registering and Controlling share agents, bankers, trustees, registrars, merchant bankers, underwriters, managers, etc.
• SEBI regulates mergers and amalgamation as a way to protect the interest of the investors. Registering and Controlling the function of collective investment schemes such as mutual funds.
• Promoting self-regulatory organizations intermediaries.
  Carrying out steps in order to develop the capital markets by having an accommodating approach.

Question 2.
Explain the powers of SEBI.
Answer:

The various powers of a SEBI are explained below:

1. Powers Relating to Stock Exchanges and Intermediaries: SEBI has wide powers to get the information from the stock exchange and intermediaries regarding their business transactions for inspection.
2. Power to Impose Monetary Penalties: SEBI has the power to impose monetary penalties on capital market intermediaries for violations.
3. Power to Initiate Actions in Functions Assigned
4. Power to Regulate Insider Trading: SEBI has the power to regulate insider trading or can regulate the functions of merchant bankers.
5. Powers Under Securities Contracts Act: For the regulation of the stock exchange, the Ministry of Finance issued a notification, for delegating several of its powers under the securities contract Act.

Question 3.
What are the benefits of Dematerialisation? (advantages)
Answer:
Benefits of Dematerialisation:

1. The risks relating to physical certificates like loss, theft, forgery are eliminated completely with a Demat Account.
2. The risk of paperwork enables quicker transactions and higher efficiency in trading.
3. The shares which are created through mergers and consolidation of companies are credited automatically in the Demat account.
4. There is no stamp duty for the transfer of securities.
5. Certain banks also permit holding of both equity and debt securities in a single account
6. A Demat account holder can buy or sell any amount of shares.

12th Commerce Guide Securities Exchange Board of India Additional Important Questions and Answers

I.Choose the correct answer.

Question 1.
In which year is SEBI Act being passed by the Indian Parliament?
a) 1990
b) 1991
c) 1992
d) 1993
Answer:
c) 1992

Question 2.
…………….. is the foremost objective of SEBI.
a) Security
b) Regulate
c) Control
d) NOTA
Answer:
a) Security

Question 1.
SEBI got the statutory powers in the year _____
(a) 1988
(b) 1992
(c) 1969
(d) 1980
Answer:
(b) 1992

Question 4.
Which is an Apex body that maintains and Regulates the capital market?
a) Stock Exchange
b) SEBI
c) OTCEI
d) NSE
Answer:
b) SEBI

Question 5.
Name the three key functions of SEBI.
Answer:
SEBI performs three key functions.
They are:

1. quasi-legislative
2. quasi-judicial
3. quasi-executive

Question 6.
SEBI has the following number of members including the chairman.
a) 4
b) 5
c) 6
d) 7
Answer:
c) 6

Question 7.
Pick the odd one out:
a) Voter ID
b) PAN
c) ID card
d) PostCard
Answer:
d) Post Card

Question 8.
which one of the following is not correctly matched?
a) DEMAT – Dematerialisation
b) PAN – Proof
c) SEBI -12 members
d) HQ – Bendrakurla
Answer :
c) SEBI -12 members

Question 9.
Choose the correct statement.
(i) SEBI Act was passed in the year 1992 in Indian Parliament.
(ii) It protects the interest of the Investors.
(iii) It Regulates and controls the stock exchange.
a) (i) is correct
b) (ii) is correct
c) (iii) is correct
d) All (i), (ii) and (iii) are correct
Answer :
d) All (i), (ii) and (iii) are correct

II. Match the following

Question 1.

Answer :
a) (i) – 2, (ii) – 1, (iii) – 4, (iv) – 3

III. Assertion and Reason :

Question 1. Assertion (A): DEMAT is done at the request of the investor.
Reason (R): A DEMAT A/c holder can buy or sell any amount of shares.
a) (A) is True (R) is True
b) (A) and (R) are False
c) (A) is True (R) is False
d) (A) is False (R) is True
Answer :
a) (A) is True (R) is True

Question 2.
Assertion (A): SEBI is a supervisory (Apex) Body.
Reason (R): So, It cannot regulates and control the stock exchange.
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are True
d) Both (A) and (R) are False
Answer :
a) (A) is True (R) is False|

IV. Very Short answer questions.

Question 1.
Write a note on PAN.
Answer:
PAN, or permanent account number, is a unique 10-digit alphanumeric identity allotted to each taxpayer by the Income Tax Department. It also serves as identity proof.