Chapter 7.22 - Solved Examples Related to Buffer Solutions and pH

In the previous section, we saw two solved examples which showed us:
   ♦ How we can calculate the pH of buffers
   ♦ How buffer helps to avoid large changes in pH
• In this section, we will see a few more solved examples related to this category

Solved example 7.81
The ionization constant of phenol is 1 × 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be the degree of ionization if the solution is also 0.01 M in sodium phenolate?
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M C₆H₆O (phenol)
• It’s dissociation can be written as:
C₆H₆O ⇌ C₆H₅O^- + H⁺
2. Let at equilibrium, x moles of C₆H₅O^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [C₆H₆O] = 0.050-x
         ✰ [C₆H₅O^-] = x
         ✰ [H⁺] = x
3. The equilibrium constant for this reaction is given as: K_a = 1 × 10^-10
• So we can write: $\mathbf\small{\rm{K_a=1 \times 10^{-10}=\frac{[C_6H_5O^-][H^+]}{[C_6H_6O]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.050
• This is because, C₆H₆O is a weak acid. It will not give much C₆H₅O^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1 \times 10^{-10}=\frac{x^2}{(0.050)}}}$
⇒ x = 2.236  × 10^-6
(The reader can opt not to approximate (0.050-x) as 0.050. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 2.236  × 10^-6)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [C₆H₅O^-] = x = 2.236  × 10^-6
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 2.236 × 10^-6 = 0.05 × 𝛼
⇒ 𝛼 = 4.472 × 10^-5

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M C₆H₅ONa (sodium phenolate)
• This C₆H₅ONa will undergo complete dissociation according to the equation:
C₆H₅ONa ⇌ C₆H₅O^- + Na⁺
• Since there is complete dissociation, 0.01 moles of C₆H₅O^- will be produced
2. So the various species in the resulting solution are:
C₆H₆O, C₆H₅O^-, H⁺ and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
C₆H₆O ⇌ C₆H₅O^- + H⁺
3. Let at equilibrium, x moles of C₆H₅O^- be produced from C₆H₆O
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [C₆H₆O] = 0.05-x
         ✰ [C₆H₅O^-] = 0.01+x
         ✰ [H⁺] = x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1 × 10^-10
• So we can write: $\mathbf\small{\rm{K_a=1 \times 10^{-10}=\frac{[C_6H_5O^-][H^+]}{[C_6H_6O]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1 \times 10^{-10}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 5 × 10^-10
(The reader can opt not to approximate (0.050-x) and (0.01+x) as 0.05 and 0.01 respectively. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 5  × 10^-10)
7. Thus we get: x = 5  × 10^-10
• We know that, x = c 𝛼
• So we get: 5 × 10^-10 = 0.05 × 𝛼
⇒ 𝛼 = 10^-8
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 4.472 × 10^-5
    ♦ In part (b), we have : 𝛼 = 10^-8
• It is clear that, when sodium phenolate is added, dissociation of phenol is suppressed
• The mixture of phenol and sodium phenolate can act as a buffer
• Note that, this example demonstrates common ion effect also

Solved example 7.82
Calculate the degree of ionization of 0.05 M acetic acid if it's pK_a value is 4.74. How is the degree of dissociation affected when it's solution also contains (a) 0.01 M (b) 0.1 M in HCl
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M CH₃COOH (acetic acid)
• It’s dissociation can be written as:
CH₃COOH ⇌ CH₃COO^- + H⁺
2. Let at equilibrium, x moles of CH₃COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = x
3. The pK_a of this reaction is given as 4.74
• We know that, pK_a = -log₁₀(K_a)
   ♦ So K_a = antilog (-pK_a) = antilog (-4.74) = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.05
• This is because, CH₃COOH is a weak acid. It will not give much CH₃COO^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{x^2}{(0.05)}}}$
⇒ x = 9.53  × 10^-4
(The reader can opt not to approximate (0.05-x) as 0.05. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 9.53 × 10^-4)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [CH₃COO^-] = x = 9.53  × 10^-4
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 9.53 × 10^-4 = 0.05 × 𝛼
⇒ 𝛼 = 0.01907

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.01 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH ⇌ CH₃COO^- + H⁺
3. Let at equilibrium, x moles of CH₃COO^- be produced from CH₃COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = 0.01+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 9.0985 × 10^-5
(The reader can opt not to approximate (0.050-x) and (0.01+x) as 0.05 and 0.01 respectively. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 9.0985 × 10^-5)
7. Thus we get: x = 9.0985  × 10^-5
• We know that, x = c 𝛼
• So we get: 9.0985 × 10^-5 = 0.05 × 𝛼
⇒ 𝛼 = 0.0018
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.01907
    ♦ In part (b), we have : 𝛼 = 0.0018
• It is clear that, when 0.01 M HCl is added, dissociation of acetic acid is suppressed

Part (c):
1. To the equilibrium in part (a), we are adding 0.1 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.1 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH ⇌ CH₃COO^- + H⁺
3. Let at equilibrium, x moles of CH₃COO^- be produced from CH₃COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = 0.1+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.1 + x) can be taken as 0.1
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{(0.1)x}{(0.05)}}}$
⇒ x = 9.0985 × 10^-6
(The reader can opt not to approximate (0.05-x) and (0.1+x) as 0.05 and 0.1 respectively. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 9.0985 × 10^-6)
7. Thus we get: x = 9.0985  × 10^-6
• We know that, x = c 𝛼
• So we get: 9.0985 × 10^-6 = 0.05 × 𝛼
⇒ 𝛼 = 0.00018
◼ Let us compare the three results:
    ♦ In part (a), we have : 𝛼 = 0.01907
    ♦ In part (b), we have : 𝛼 = 0.0018
    ♦ In part (c), we have : 𝛼 = 0.00018
• It is clear that:
   ♦ When 0.01 M HCl is added, dissociation of acetic acid is suppressed
   ♦ When 0.1 M HCl is added, dissociation of acetic acid is even more suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.83
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate it's degree of ionization in it's 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M NaOH
Solution:
Part (a):
1. Consider the original solution which is: 0.02 M NH₄OH (dimethylamine)
• It’s dissociation can be written as:
(CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH^-
2. Let at equilibrium, x moles of (CH₃)₂NH₂⁺ be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [(CH₃)₂NH] = 0.02-x
         ✰ [(CH₃)₂NH₂⁺] = x
         ✰ [OH^-] = x
3. The equilibrium constant of this reaction is given as: K_b = 5.4 × 10^-4
• So we can write: $\mathbf\small{\rm{K_b=1 \times 5.4^{-4}=\frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}=\frac{(x)(x)}{(0.02-x)}}}$
4. x will be very small when compared to 0.02
• This is because, (CH₃)₂NH is a weak base. It will not give much (CH₃)₂NH₂⁺ ions
   ♦ So (0.02 - x) can be taken as 0.02
   ♦ For example, (0.02 - 0.0000001) can be taken as 0.02 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{5.4 \times 10^{-4}=\frac{x^2}{(0.02)}}}$
⇒ x = 3.286 × 10^-3
(The reader can opt not to approximate (0.02-x) as 0.02. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 3.286  × 10^-3)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [OH^-] = [(CH₃)₂NH₂⁺] = x = 3.286 × 10^-3
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 3.286 × 10^-3 = 0.02 × 𝛼
⇒ 𝛼 = 0.1643

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.1 M NaOH (sodium hydroxide)
• This NaOH will undergo complete dissociation according to the equation:
NaOH ⇌ Na⁺ + OH^-
• Since there is complete dissociation, 0.1 moles of OH^- will be produced
2. So the various species in the resulting solution are:
(CH₃)₂NH, (CH₃)₂NH₂⁺, OH^- and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
(CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH^-
3. Let at equilibrium, x moles of (CH₃)₂NH₂⁺  be produced from (CH3)(CH₃)₂NH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [(CH₃)₂NH] = 0.02-x
         ✰ [(CH₃)₂NH₂⁺] = x
         ✰ [OH^-] = 0.1+x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_b = 5.4 × 10^-4
• So we can write: $\mathbf\small{\rm{K_b=5.4 \times 10^{-4}=\frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}=\frac{(0.1+x)x}{(0.02-x)}}}$
5. As before, x will be very small when compared to 0.02
   ♦ So (0.02 - x) can be taken as 0.02
   ♦ Also, (0.1 + x) can be taken as 0.1
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{5.4 \times 10^{-4}=\frac{(0.1)x}{(0.02)}}}$
⇒ x = 1.08 × 10^-4
(The reader can opt not to approximate (0.02-x) and (0.1+x) as 0.02 and 0.1 respectively. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 1.08 × 10^-4)
7. Thus we get: x = 1.08 × 10^-4
• We know that, x = c 𝛼
• So we get: 1.08 × 10^-4 = 0.02 × 𝛼
⇒ 𝛼 = 0.0054
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.1643
    ♦ In part (b), we have : 𝛼 = 0.0054
• It is clear that, when sodium hydroxide is added, dissociation of dimethylamine is suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.84
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?
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M CH₃CH₂COOH (acetic acid)
• It’s dissociation can be written as:
CH₃CH₂COOH ⇌ CH₃CH₂COO^- + H⁺
2. Let at equilibrium, x moles of CH₃CH₂COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃CH₂COOH] = 0.05-x
         ✰ [CH₃CH₂COO^-] = x
         ✰ [H⁺] = x
3. The K_a of this reaction is given as 1.32 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.32 \times 10^{-5}=\frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2COOH]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.05
• This is because, CH₃CH₂COOH is a weak acid. It will not give much CH₃CH₂COO^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1.32 \times 10^{-5}=\frac{x^2}{(0.05)}}}$
⇒ x = 0.000812 = 8.12  × 10^-4
(The reader can opt not to approximate (0.05-x) as 0.05. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 8.12 × 10^-4)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [CH₃CH₂COO^-] = x = 8.12  × 10^-4
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 8.12 × 10^-4 = 0.05 × 𝛼
⇒ 𝛼 = 0.0162
7. pH = -log₁₀([H⁺]) = -log₁₀([8.12 × 10^-4]) = 3.09

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.01 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃CH₂COOH, CH₃CH₂COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃CH₂COOH ⇌ CH₃CH₂COO^- + H⁺
3. Let at equilibrium, x moles of CH₃CH₂COO^- be produced from CH₃CH₂COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃CH₂COOH] = 0.05-x
         ✰ [CH₃CH₂COO^-] = x
         ✰ [H⁺] = 0.01+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.32 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.32 \times 10^{-5}=\frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2OOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.32 \times 10^{-5}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 6.6 × 10^-5
(The reader can opt not to approximate (0.05-x) and (0.01+x) as 0.05 and 0.01 respectively. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 6.6 × 10^-5)
7. Thus we get: x = 6.6  × 10^-5
• We know that, x = c 𝛼
• So we get: 6.6 × 10^-5 = 0.05 × 𝛼
⇒ 𝛼 = 0.00132
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.0162
    ♦ In part (b), we have : 𝛼 = 0.00132
• It is clear that, when 0.01 M HCl is added, dissociation of propanoic acid is suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.85
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization
constant of the acid and its degree of ionization in the solution
Solution:
1. Consider the solution which is: 0.01 M HCNO (cyanic acid)
• It’s dissociation can be written as:
HCNO ⇌ OCN^- + H⁺
2. Let at equilibrium, x moles of CH₃CH₂COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [HCNO] = 0.1-x
         ✰ [OCN^-] = x
         ✰ [H⁺] = x
3. To find x, we can make use of the given pH value
We have: x = [H⁺] = antilog(-pH) = antilog(-2.34) = 0.00457
4. So we can write: $\mathbf\small{\rm{K_a=\frac{[OCN^-][H^+]}{[HCNO]}=\frac{(x)(x)}{(0.01-x)}}}$
5. x will be very small when compared to 0.01
• This is because, HCNO is a weak acid. It will not give much OCN^- ions
   ♦ So (0.1 - x) can be taken as 0.1
   ♦ For example, (0.1 - 0.0000001) can be taken as 0.1 for practical purposes
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{K_a=\frac{(0.00457)^2}{(0.1)}=0.000209}}$ = 2.09 × 10^-4
(The reader can opt not to approximate (0.1-x) as 0.1. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 2.09 × 10^-4)
7. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [OCN^-] = x = 0.00457
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 0.00457 = 0.1 × 𝛼
⇒ 𝛼 = 0.0457

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In the next section, we will see solubility equilibria


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