CSIR-NET Chemical Science: Ionic Equilibrium - Most Important Questions!! (Download PDF)

By Neetesh Tiwari|Updated : July 12th, 2021

Hello Gradians,

It will not be a new thing for you if we say that Gradeup is considered the synonym for success. For the past 5 years, At BYJU'S Exam Prep, we have been promoting talented candidates of the CSIR-NET category to the path of success. As a result, BYJU'S Exam Prep has been an integral part of the preparation journey of many Aspirants.

As we all know, the CSIR-NET exam is very near, and there is no time left for revision. At this point, you must be seeking some ready-made authentic short revision notes or some important questions to practice just before the exam. How will you react if you get all these study materials in one place? It will be beneficial for you at this time, right! We at BYJU'S Exam Prep comes up with Important Questions on Ionic Equilibrium for Chemical Sciences to help you score good marks, which our experienced subject-matter experts have meticulously designed. So what are you waiting for? Scroll down the article below and start study for the CSIR NET Exam; these questions will be fruitful during last-minute revision to fetch more marks in the exams. Students can also download it as a PDF file and save it for future purposes.

We have been providing short study notes for the Chemical Science subject, which can be checked here. We have experienced immense love and support from all of you through these short notes in the CSIR-NET category, and we hope you all continue to support us in the same way!

Hello Gradians,

It will not be a new thing for you if we say that Gradeup is considered the synonym for success. For the past 5 years, At BYJU'S Exam Prep, we have been promoting talented candidates of the CSIR-NET category to the path of success. As a result, BYJU'S Exam Prep has been an integral part of the preparation journey of many Aspirants.

As we all know, the CSIR-NET exam is very near, and there is no time left for revision. At this point, you must be seeking some ready-made authentic short revision notes or some important questions to practice just before the exam. How will you react if you get all these study materials in one place? It will be beneficial for you at this time, right! We at BYJU'S Exam Prep comes up with Important Questions on Ionic Equilibrium for Chemical Sciences to help you score good marks, which our experienced subject-matter experts have meticulously designed. So what are you waiting for? Scroll down the article below and start study for the CSIR NET Exam; these questions will be fruitful during last-minute revision to fetch more marks in the exams. Students can also download it as a PDF file and save it for future purposes.

We have been providing short study notes for the Chemical Science subject, which can be checked here. We have experienced immense love and support from all of you through these short notes in the CSIR-NET category, and we hope you all continue to support us in the same way!

 Ionic Equilibrium - Most Important Questions

1. At 25°C, a monobasic acid has a dissociation constant of 1.8 × 10–5. Calculate its degree of dissociation at the same temperature and concentration of 0.20 M. What will be the hydrogen ion concentration(mol dm-3) provided by it?

  1. 0.001167
  2. 0.001897
  3. 0.01297
  4. 0.00156

2. What is the pH of a 0.01 M CH3COONa aqueous solution at 25°C (Ka for CH3COOH = 1.75× 10–5). Kw= 1.008 × 10–14.

  1. 1.57
  2. 3.48
  3. 8.38
  4. 5.55

3. 20 mole of NH4OH and 0.25 mole of NH4Cl per litre make up a buffer solution. Determine the solution's pH level. At normal temperature, the dissociation constant of NH4OH is 1.8 × 10–5.

  1. 1.121
  2. 3.452
  3. 9.161
  4. 10.20

4. Aniline as a base has a dissociation constant of 5.93 ×10–10 at 25°C. At 25°C, the ionic product of water is 1.008 × 10–14 . Calculate the percentage hydrolysis of aniline hydrochloride in a 0.1 M solution.

  1. 1.01
  2. 3.67
  3. 2.59
  4. 1.30

5. Aniline, acetic acid, and water all have dissociation constants of 3.83× 10–10, 1.75× 10–5, and 1.008 ×10–14 at 25°C, respectively. In a decimolar solution, calculate the % hydrolysis of aniline acetate.

  1. 3.212
  2. 1.219
  3. 4.523
  4. 1.890

6. Determine the pH of an aqueous solution made up of 25 mL of 0.2 M HCl and 50 mL of 0.25 M NaOH. Kw= 10–14 mol2 dm–6 at 25°C

  1. 13
  2. 5.60
  3. 4.56
  4. 11

7. In a litre of buffer solution, there are 0.2 moles of acetic acid and 0.25 moles of potassium acetate. Calculate the pH change in the solution after adding 0.5 mL of 1 M HCl. At normal temperature, acetic acid has a dissociation constant of 1.75 ×  10–5. (The volume shift due to HCl additions may be ignored.)

  1. 0.002
  2. 0.0045
  3. 0.089
  4. 0.0074

8. Calculate the pH and concentrations of hydrogen and hydroxyl ions in a 3.2 × 10–3M Ba(OH)2 solution in water at 25°C.

  1. 12.56, 2.37 ×10–11
  2. 11.81, 1.57 ×10–12
  3. 14.56, 3.12 ×10–10
  4. 10.21, 4.23 ×10–10

9. Determine the degree of hydrolysis of a 0.10 M sodium acetate solution at 25°C. Kw = 1.008 ×  10–14 and Ka = 1.75×  10–5.

  1. 7.589 × 10–5
  2.  8.956 × 10–5
  3.  6.542 × 10–5
  4. 5.674 × 10–5

10. Determine the pH of a 1 ×10–7 M HCl aqueous solution at 25°C. Take Kw = 10–14 mol2 dm–6

  1. 7.05
  2. 7.0
  3. 6.79
  4. 5.99

Answer Keys

1. B

2. D

3. C

4. D

5. B

6. A

7. A

8. B

9. A

10. C

Solutions

Solution 1: 

 

Solution 2: 

pKw = –log Kw = – log (1.008 × 10–14) = 13.997 = 14

pKw = – log Ka = – log (1.75 × 10–5) = 4.76

log c = log (102) = –2

From the given pH value ,the solution is alkaline

Solution 3: 

pOH = pKb + log ([Salt]/[Base])

pKb = – log Kb = – log(1.81 × 10–5) = 4.7423

pOH = 4.7423 + og  0.25/0.20 = 4,839

pH = 14 – 4.839 = 9.161

Solution 4: 

 

Solution 5: 

 

 

Solution 6: 

Number of milli moles of the acid in the solution = 25 × 0.2 = 5

Number of milli moles of the alkali in the solution = 50 × 0.25 = 12.5

Number of milli moles of the alkali left in the solution after the addition of the acid = 12.5 – 5 = 7.5

Total volume of the solution = 50 + 25 = 75 ml

[H+][OH] = 10–14 mol2 dm–6 at 25°C

[OH] = 0.10 mol dm–3

[H+] = 10–14 mol2 dm6/0.10 mol dm–3 = 10–13 mol dm–3

pH=13

Solution 7: 

pH = pKa + log ([Salt]/[Acid])

pKa = – log Ka = – log 1.75 × 10–5 = 4.76

pH of the solution before adding HCl will be given by

pH = pKq + log (0.25/0.20) = 4.76 + 0.0969 = 4.8569

Amount of HCl added = 0.5 ml of 1M HCl = 0.5 millimole = 0.0005 mole

Assuming HCl to be completely dissociated, the amount of H+ ions added will be = 0.0005 mole

The addition of H+ ion will result in the reaction

CH3COO + H+  CH3COOH

i.e., the amount of acetic acid will increase and that of the salt will correspondingly decrease by 0.005 mole.

 After adding HCl.

[CH3COOH] = 0.20 + 0.0005 = 0.2005 mol dm–3

[CH3COOK] = 0.25 – 0.005 = 0.2495 mol dm–3

(The change of volume resulting from the addition of HCl is negligible)

pH = pKa + log ([Salt]/[Acid]) = 4.76 + log (0.2495/0.2005)= 4.8549

Change of pH = 4.8569 – 4.8549 = 0.002

Solution 8: 

Ba(OH)2 is a strong base which dissociates to furnish two moles of OH ions for one mole of the base:

Ba(OH)2 --> Ba2+ + 2OH

Hence, [OH] = 2 × 3.2 × 10–3 M = 6.4 × 10–3 mol dm–3

pOH = – log [OH] = – log (6.4 × 10–3) = – log 6.4 – log 10–3

=> – 0.8062 – (–3) = – 0.8062 + 3 = 2.19

Since pH + pOH = 14 at 25°C

pH = 14 – 2.19 = 11.81

Solution 9: 

 

Solution 10: 

 

Most Important Questions on Ionic Equilibrium - Download PDF Now

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Neetesh TiwariNeetesh TiwariMember since Aug 2020
Working Experience of more than 2 years in Education Industry. Currently associated to CSIR-NET at Gradeup. Guiding students to crack the exam and fulfil their dreams.
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