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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Chapter 7.21 - The Buffer solution

In the previous section, we saw common ion effect and the hydrolysis of salts. In this section, we will see buffers

• A buffer solution (or simply buffer) is a specially prepared aqueous solution
   ♦ It is prepared in such a way as to have a certain pH value
         ✰ Even when some acid is added, it's pH will not decrease much
         ✰ Even when some base is added, it's pH will not increase much
• A buffer can be prepared by two methods:
Method 1:
By mixing a weak acid and it’s conjugate base
Method 2:
By mixing a weak base and it’s conjugate acid

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• We will first see how the buffer is prepared. After that we will see how buffer works
Preparation by method 1:
This can be written in 3 steps:
1. We need a weak acid
• Consider the weak acid CH₃COOH
   ♦ Take an aqueous solution of this CH₃COOH
   ♦ Let us call it solution 1
• We know that CH₃COOH will dissociate in aqueous solution as:
   ♦ CH₃COOH ⇌ CH₃COO^- + H⁺
   ♦ The solution 1 will be at equilibrium according to this equation
2. Next we want the conjugate base of CH₃COOH
   ♦ The conjugate base of CH₃COOH is CH₃COO^-
• We prepare a second solution
   ♦ Let us call it solution 2
• The solution 2 is an aqueous solution of CH₃COONa (sodium acetate)
• The CH₃COONa dissociates completely
   ♦ CH₃COONa  → CH₃COO^- + Na⁺
   ♦ So solution 2 will contain only CH₃COO^-  and Na⁺
         ✰ Thus solution 2 will contain the required conjugate base
3. Next we add solution 2 to solution 1
• That means, we are adding CH₃COO^- ions and Na⁺ ions to solution 1
   ♦ Na⁺ ions are stable. They do not take part in the reaction
• So the net effect is that, [CH₃COO^-] in the resulting solution increases
• Due to this increase in [CH₃COO^-], the equilibrium mentioned in (1) will shift towards the left
• Thus a new equilibrium will be reached. The buffer is ready

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Next we will see how this buffer works. It can be written in 5 steps:
1. Assume that, a strong acid is added to the buffer
• The strong acid will supply a large number of H⁺ ions
   ♦ So we would expect the pH of the buffer to fall
2. But the very purpose of the buffer is to keep the pH at the same value
• Indeed the buffer can do this because, it has a lot of excess CH₃COO^- ions
• Those ions will consume the incoming H⁺ ions. The equation is:
CH₃COO^- + H⁺ ⇌ CH₃COOH
3. Thus most of the incoming H⁺ ions are neutralized, keeping the pH from falling too much
4. At the beginning of this discussion, we said that:
A buffer (by method 1) is a mixture of a weak acid and it’s conjugate base
• For our present case. we chose CH₃COOH as the weak acid
• The pure solution of CH₃COOH is a mixture of CH₃COOH and it’s conjugate base CH₃COO^-
• Then why do we add CH₃COONa ?
• The answer is that:
The pure mixture will not have enough CH₃COO^- ions to neutralize the incoming H⁺ ions
5. To this prepared buffer, instead of adding a strong acid, we can add a strong base. Even then, the pH will not change much
• This is because, the incoming OH^- will react with CH₃COOH:
CH₃COOH + OH^- ⇌ CH₃COO^- + H₂O
• Thus the incoming OH^- ions are neutralized

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• Now we will see how the buffer is prepared by the second method. After that we will see how that buffer works
Preparation by method 2:
This can be written in 3 steps:
1. We need a weak base
• Consider the weak base NH₄OH
   ♦ Take an aqueous solution of this NH₄OH
   ♦ Let us call it solution 1
• We know that NH₄OH will dissociate in aqueous solution as:
   ♦ NH₄OH ⇌ NH₄⁺ + OH^-
   ♦ The solution 1 will be at equilibrium according to this equation
2. Next we want the conjugate acid of NH₄OH
   ♦ The conjugate acid of NH₄OH is NH₄⁺
• We prepare a second solution
   ♦ Let us call it solution 2
• The solution 2 is an aqueous solution of NH₄Cl (ammonium chloride)
• The NH₄Cl dissociates completely
   ♦ NH₄Cl  → NH₄⁺ + Cl^-
   ♦ So solution 2 will contain only NH₄⁺  and Cl^-
         ✰ Thus solution 2 will contain the required conjugate acid
3. Next we add solution 2 to solution 1
• That means, we are adding NH₄⁺ ions and Cl^- ions to solution 1
   ♦ Cl^- ions are stable. They do not take part in the reaction
• So the net effect is that, [NH₄⁺] in the resulting solution increases
• Due to this increase in [NH₄⁺], the equilibrium mentioned in (1) will shift towards the left
• Thus a new equilibrium will be reached. The buffer is ready

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Next we will see how this buffer works. It can be written in 5 steps:
1. Assume that, a strong base is added to the buffer
• The strong base will supply a large number of OH^- ions
   ♦ So we would expect the pH of the buffer to rise
2. But the very purpose of the buffer is to keep the pH at the same value
• Indeed the buffer can do this because, it has a lot of excess NH₄⁺ ions
• Those ions will consume the incoming OH- ions. The equation is:
NH₄⁺ + OH^- ⇌ NH₄OH
3. Thus most of the incoming OH^- ions are neutralized, keeping the pH from rising too much
4. At the beginning of this discussion, we said that:
A buffer (by method 2) is a mixture of a weak base and it’s conjugate acid
• For our present case. we chose NH₄OH as the weak base
• The pure solution of NH₄OH is a mixture of NH₄OH and it’s conjugate acid NH₄⁺
• Then why do we add NH₄Cl ?
• The answer is that:
The pure solution will not have enough NH₄⁺ ions to neutralize the incoming OH^- ions
5. To this prepared buffer, instead of adding a strong base, we can add a strong acid. Even then, the pH will not change much
• This is because, the incoming H⁺ will react with NH₄OH:
NH₄OH + H⁺ ⇌ NH₄⁺ + H₂O
• Thus the incoming H⁺ ions are neutralized

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A summary of the discussion so far, can be given in a flowchart form. It is shown in the fig.7.21 below:

Fig.7.21

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• Next we will see two solved examples which will demonstrate the buffer action
Solved example 7.79
A buffer is prepared by adding 0.050 M CH₃COONa to 0.05 M CH₃COOH
(a) What is it’s pH ?
(b) What will be the new pH if 0.001 moles of HCl is added to 1 liter of the buffer. Assume that, volume remains at 1 liter even after adding the HCl
(c) What will be the pH of a solution obtained by adding 0.001 moles of HCl to 1 liter of pure water?
Solution:
Part (a):
1. Consider the original solution which is: 0.050 M CH₃COOH
• It’s dissociation can be written as:
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
2. Let at equilibrium, x moles of CH₃COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COOH] = 0.050-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = x
3. To this equilibrium, we are adding 0.050 M CH₃COONa
• This CH₃COONa will undergo complete dissociation according to the equation:
CH₃COONa ⇌ CH₃COO^- + Na⁺
• Since there is complete dissociation, 0.050 moles of CH₃COO^- will be produced
4. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
5. Based on (2) and (3), we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.050-x
         ✰ [CH₃COO^-] = 0.050+x
         ✰ [H⁺] = x
6. Even when a new equilibrium is attained, the equilibrium constant will not change
• The equilibrium constant for this reaction can be obtained from the data book: K_a = 1.76  × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.76 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.050+x)x}{(0.050-x)}}}$
7. x will be very small when compared to 0.050
• This is because, CH₃COOH is a weak acid. It will not give much CH₃COO^- ions
   ♦ So (0.050 + x) can be taken as 0.050
   ♦ For example, (0.050 + 0.0000001) can be taken as 0.050 for practical purposes
• Similarly, (0.050 – x) can be taken as 0.050
8. Thus the result in (6) becomes:
$\mathbf\small{\rm{1.76 \times 10^{-5}=\frac{(0.050)x}{(0.050)}=x}}$
(The reader can opt not to approximate (0.050+x) and (0.050-x) as 0.050. 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 1.76  × 10^-5)
9. Thus we get: [H⁺] = x = 1.76  × 10^-5
• So pH = -log₁₀([H⁺]) = -log₁₀(1.76  × 10^-5) = 4.75 

Part (b):
1. We take 1 liter of the buffer prepared in part (a). To that 1 liter, we add 0.001 moles of HCl
• Given that, the volume does not change. So even after adding HCl, the volume is 1 liter
   ♦ Thus [HCl] = 0.001 M
2. HCl is a strong acid. It dissociates completely
• So we get: [H⁺] = [Cl^-] = 0.001
3. The 1 liter solution will contain the following species:
CH₃COOH, CH₃COO^-, H⁺, Na⁺ and Cl^-
• Na⁺ and Cl^- are stable. They will not take part in the reaction. We can ignore them
   ♦ The concentrations of the remaining species are:
         ✰ [CH₃COOH] = (0.050-x) = 0.050
         ✰ [CH₃COO^-] = (0.050+x) = 0.050
         ✰ [H⁺] = (x + 0.001) = 0.001
• We make the above approximations because x (calculated as 1.76  × 10^-5 in part a), is very small when compared to 0.050 and 0.001
4. The H⁺ will react with CH₃COO^- according to the equation:
CH₃COO^- + H⁺ ⇌ CH₃COOH (aq)
• Nearly all the 0.001 moles of H⁺ will be used up in this way
   ♦ So [CH₃COO^-] will decrease by 0.001
   ♦ Also [CH₃COOH] will increase by 0.001
   ♦ Let y be the final concentration of [H⁺]
5. Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COO^-] = (0.050 - 0.001) = 0.049
         ✰ [CH₃COOH] = (0.050 + 0.001) = 0.051
         ✰ [H⁺] = y
6. The reaction in (4) is the reverse of
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
• So for the reaction in (4), we have to take the reciprocal of K_a
• We get: $\mathbf\small{\rm{\frac{1}{K_a}= \frac{1}{1.76 \times 10^{-5}}=\frac{[CH_3COOH]}{[CH_3COO^-][H^+]}=\frac{0.051}{0.049y}}}$
⇒ y = 1.8318  × 10^-5
7. Thus we get: [H⁺] = y = 1.8318 × 10^-5
So pH = -log₁₀([H⁺]) = -log₁₀(1.8318  × 10^-5) = 4.74

Part (c):
1. When 0.001 moles of HCl is added to water, all those HCl molecules will dissociate into H⁺ and Cl^- ions
• So [H⁺] = 0.001
2. Then pH = -log₁₀([H⁺]) = -log₁₀(0.001) = 3.00
3. Let us compare the three pH values
• In part (a), we get:
pH of the buffer = 4.75
• In part (b) we get:
pH after adding 0.001 moles of HCl = 4.74
• In part (c) we get:
pH when 0.001 moles of HCl is added to pure water = 3.00
◼  That means, the buffer is effective in resisting pH change. If there was no buffer, the pH would have fallen from 4.75 to 3. But due to the buffer action, the pH falls from 4.75 to 4.74 only

Solved example 7.80
A buffer is prepared by adding 0.0350 M NH₄Cl to 0.0500 M NH₄OH
(a) What is it’s pH ?
(b) What will be the new pH if 0.001 moles of NaOH is added to 1 liter of the buffer. Assume that, volume remains at 1 liter even after adding the NaOH
(c) What will be the pH of a solution obtained by adding 0.001 moles of NaOH to 1 liter of pure water?
Solution:
Part (a):
1. Consider the original solution which is: 0.050 M NH₄OH
• It’s dissociation can be written as:
NH₄OH ⇌ NH₄⁺ + OH^-
2. Let at equilibrium, x moles of NH₄⁺ be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [NH₄OH] = 0.050-x
         ✰ [NH₄⁺] = x
         ✰ [OH^-] = x
3. To this equilibrium, we are adding 0.0350 M NH₄Cl
• This NH₄Cl will undergo complete dissociation according to the equation:
NH₄Cl ⇌ NH₄⁺ + Cl^-
• Since there is complete dissociation, 0.0350 moles of NH₄⁺ will be produced
4. So the various species in the resulting solution are:
NH₄OH, NH₄⁺, OH^- and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
NH₄OH ⇌ NH₄⁺ + OH^-
5. Based on (2) and (3), we can write:
   ♦ At the new equilibrium,
         ✰ [NH₄OH] = 0.0500-x
         ✰ [NH₄⁺] = 0.0350+x
         ✰ [OH^-] = x
6. Even when a new equilibrium is attained, the equilibrium constant will not change
• The equilibrium constant for this reaction can be obtained from the data book: K_b = 1.77  × 10^-5
• So we can write: $\mathbf\small{\rm{K_b=1.75 \times 10^{-5}=\frac{[NH_4^+][OH^-]}{[NH_4OH]}=\frac{(0.0350+x)x}{(0.0500-x)}}}$
7. x will be very small when compared to 0.0500 and 0.0350
• This is because, NH₄OH is a weak base. It will not give much NH₄⁺ ions
   ♦ So (0.0500 + x) can be taken as 0.0500
   ♦ For example, (0.0500 + 0.0000001) can be taken as 0.0500 for practical purposes
• Similarly, (0.0350 – x) can be taken as 0.0350
8. Thus the result in (6) becomes:
$\mathbf\small{\rm{1.77 \times 10^{-5}=\frac{(0.0350)x}{(0.0500)}=0.7x}}$
x = 2.529 × 10^-5
(The reader can opt not to approximate (0.0500+x) as 0.0500 and (0.0350+x) as 0.0350. 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 2.529  × 10^-5)
9. Thus we get: [OH^-] = x = 2.529  × 10^-5
• So pOH = -log₁₀([OH^-]) = -log₁₀(2.529  × 10^-5) = 4.597
• So pH = (14 - pOH) = (14 - 4.597) = 9.403

Part (b):
1. We take 1 liter of the buffer prepared in part (a). To that 1 liter, we add 0.001 moles of NaOH
• Given that, the volume does not change. So even after adding NaOH, the volume is 1 liter
   ♦ Thus [NaOH] = 0.001 M
2. NaOH is a strong base. It dissociates completely
• So we get: [OH^-] = [Na⁺] = 0.001
3. The 1 liter solution will contain the following species:
NH₄OH, NH₄⁺, OH^-, Na⁺ and Cl^-
• Na⁺ and Cl^- are stable. They will not take part in the reaction. We can ignore them
   ♦ The concentrations of the remaining species are:
         ✰ [NH₄OH] = (0.050-x) = 0.0500
         ✰ [NH₄⁺] = (0.0350+x) = 0.0350
         ✰ [OH^-] = (x + 0.001) = 0.001
• We make the above approximations because x (calculated as 2.529  × 10^-5 in part a), is very small when compared to 0.050, 0.0350 and 0.001
4. The OH^- will react with NH₄⁺ according to the equation:
NH₄⁺ + OH^- ⇌ NH₄OH
• Nearly all the 0.001 moles of OH^- will be used up in this way
   ♦ So [NH₄⁺] will decrease by 0.001
   ♦ Also [NH₄OH] will increase by 0.001
   ♦ Let y be the final concentration of [OH^-]
5. Then we can write:
   ♦ At equilibrium,
         ✰ [NH₄⁺] = (0.0350 - 0.001) = 0.0340
         ✰ [NH₄OH] = (0.0500 + 0.001) = 0.0510
         ✰ [OH^-] = y
6. The reaction in (4) is the reverse of
NH₄OH ⇌ NH₄⁺ + OH^-
• So for the reaction in (4), we have to take the reciprocal of K_b
• We get: $\mathbf\small{\rm{\frac{1}{K_b}= \frac{1}{1.77 \times 10^{-5}}=\frac{[NH_4OH]}{[NH_4^+][OH^-]}=\frac{0.0510}{0.034 \, y}}}$
⇒ y = 2.655  × 10^-5
7. Thus we get: [OH^-] = y = 2.655 × 10^-5
• So pOH = -log₁₀([OH^-]) = -log₁₀(2.655 × 10^-5) = 4.576
• So pH = (14 - pOH) = (14 - 4.576) = 9.424

Part (c):
1. When 0.001 moles of NaOH is added to water, all those NaOH molecules will dissociate into Na⁺ and OH^- ions
• So [OH^-] = 0.001
2. Then pOH = -log₁₀([OH^-]) = -log₁₀(0.001) = 3.00
So PH = (14 - pH) = (14 - 3) = 11
3. Let us compare the three pH values
• In part (a), we get:
pH of the buffer = 9.403
• In part (b) we get:
pH after adding 0.001 moles of NaOH = 9.424
• In part (c) we get:
pH when 0.001 moles of NaOH is added to pure water = 11
◼  That means, the buffer is effective in resisting pH change. If there was no buffer, the pH would have increased from 9.403 to 11. But due to the buffer action, the pH increased from 9.403 to 9.424 only

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• In the above two solved examples:
   ♦ Part (a) demonstrates how we can calculate the pH of buffers
   ♦ Parts (b) and (c) demonstrate how buffer helps to avoid large changes in pH
• In the next section, we will see a few more solved examples related to this category


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