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Page No 229: - Chapter 7 Equilibrium class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 7.46:

The ionization constant of acetic acid is 1.74 × 10^–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer:

Method 1

Since Ka >> Kw, :

Method 2

Degree of dissociation,

c = 0.05 M

Ka = 1.74 × 10^–5

Thus, concentration of CH₃COO– = c.α

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

Question 7.47:

It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.

Answer:

Let the organic acid be HA.

Concentration of HA = 0.01 M

pH = 4.15

Now,

Then,

Question 7.48:

Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Answer:

(i) 0.003MHCl:

Since HCl is completely ionized,

Now,

Hence, the pH of the solution is 2.52.

(ii) 0.005MNaOH:

Hence, the pH of the solution is 11.70.

(iii) 0.002 HBr:

Hence, the pH of the solution is 2.69.

(iv) 0.002 M KOH:

Hence, the pH of the solution is 11.31.

Question 7.49:

Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution.

b) 0.3 g of Ca(OH)₂dissolved in water to give 500 mL of solution.

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Answer:

(a) For 2g of TlOH dissolved in water to give 2 L of solution:

(b) For 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution:

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

13.6 × 1 mL = M₂ × 1000 mL

(Before dilution) (After dilution)

13.6 × 10^–3 = M₂ × 1L

M₂ = 1.36 × 10^–2

[H⁺] = 1.36 × 10^–2

pH = – log (1.36 × 10^–2)

= (– 0.1335 + 2)

= 1.866 ∼ 1.87

Question 7.50:

The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.

Answer:

Degree of ionization, α = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H₃O⁺ = c.α

= 0.1 × 0.132

= 0.0132

Now,

Question 7.51:

The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

Answer:

c = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)

Question 7.52:

What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer:

Kb = 4.27 × 10^–10

c = 0.001M

pH =?

α =?

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.

Question 7.53:

Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74.

How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

Answer:

When HCl is added to the solution, the concentration of H⁺ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

Case II: When 0.1 M HCl is taken.

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

Question 7.54:

The ionization constant of dimethylamine is 5.4 × 10^–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer:

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

And,

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Question 7.55:

Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a) Human muscle-fluid, 6.83

(b) Human stomach fluid, 1.2

(d) Human saliva, 6.4.

Answer:

(a) Human muscle fluid 6.83:

pH = 6.83

pH = – log [H⁺]

∴6.83 = – log [H⁺]

[H⁺] =1.48 × 10^–7 M

(b) Human stomach fluid, 1.2:

pH =1.2

1.2 = – log [H⁺]

∴[H⁺] = 0.063

(c) Human blood, 7.38:

pH = 7.38 = – log [H⁺]

∴ [H⁺] = 4.17 × 10^–8 M

(d) Human saliva, 6.4:

pH = 6.4

6.4 = – log [H⁺]

[H⁺] = 3.98 × 10^–7

Question 7.56:

The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Answer:

The hydrogen ion concentration in the given substances can be calculated by using the given relation:

pH = –log [H⁺]

(i) pH of milk = 6.8

Since, pH = –log [H⁺]

6.8 = –log [H⁺]

log [H⁺] = –6.8

[H⁺] = anitlog(–6.8)

(ii) pH of black coffee = 5.0

Since, pH = –log [H⁺]

5.0 = –log [H⁺]

log [H⁺] = –5.0

[H⁺] = anitlog(–5.0)

(iii) pH of tomato juice = 4.2

Since, pH = –log [H⁺]

4.2 = –log [H⁺]

log [H⁺] = –4.2

[H⁺] = anitlog(–4.2)

(iv) pH of lemon juice = 2.2

Since, pH = –log [H⁺]

2.2 = –log [H⁺]

log [H⁺] = –2.2

[H⁺] = anitlog(–2.2)

(v) pH of egg white = 7.8

Since, pH = –log [H⁺]

7.8 = –log [H⁺]

log [H⁺] = –7.8

[H⁺] = anitlog(–7.8)

Question 7.57:

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Answer:

Question 7.58:

The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Answer:

Solubility of Sr(OH)₂ = 19.23 g/L

Then, concentration of Sr(OH)₂

Question 7.59:

The ionization constant of propanoic acid is 1.32 × 10^–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answer:

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have: