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

 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). K_w= 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. K_w= 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: 

pK_w = –log K_w = – log (1.008 × 10^–14) = 13.997 = 14

pK_w = – log K_a = – log (1.75 × 10^–5) = 4.76

log c = log (10²) = –2

From the given pH value ,the solution is alkaline

Solution 3: 

pOH = pK_b + log ([Salt]/[Base])

pK_b = – log K_b = – 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 = pK_a + log ([Salt]/[Acid])

pK_a = – log K_a = – log 1.75 × 10^–5 = 4.76

pH of the solution before adding HCl will be given by

pH = pK_q + 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 = pK_a + 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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