Chapter 7.17 - Ionization Constants of Weak Acids

In the previous section, we completed a discussion on pH scale. In this section, we will see ionization constants of weak acids

• In an earlier section, we saw that:
    ♦ Strong acids like HCl undergo complete ionization when added to water
    ♦ Weak acids do not undergo complete ionization when added to water
• We can write the main features of the ionization constant of weak acids in 7 steps:
1. Let HX be a weak acid. It's ionization process can be represented by the following equation:
HX(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X^-(aq)
2. Let c be the initial concentration of HX
    ♦ That is., at t = 0, [HX] = c moles L^-1
3. Let 𝛼 be the extent of ionization
    ♦ That is., 𝛼 is the 'fraction of c' which undergoes dissociation
    ♦ 𝛼 is a fraction like ¹⁄₄, ²⁄₃ etc.,
        ✰ (Remember that, fractions can be expressed as decimals also)  
• Then number of moles of HX undergoing ionization = c𝛼
4. When c𝛼 moles of HX undergoes ionization, the number of moles of HX remaining will be equal to (c - c𝛼)
• Also, from the stoitiometric coefficients, we can write:
    ♦ When c𝛼 moles of HX undergoes dissociation,
        ✰ c𝛼 moles of H₃O⁺ will be formed
        ✰ c𝛼 moles of X^- will be formed
5. So we obtained the concentrations at equilibrium
• Using those concentrations, we can write the expression for the equilibrium constant
    ♦ It is called the ionization constant of the acid HX
    ♦ It is denoted as K_a
• So we can write $\mathbf\small{\rm{K_a=\frac{(c\alpha)^2}{c-c\alpha}}}$
• Thus we get Eq.7.8: $\mathbf\small{\rm{K_a=\frac{c^2\alpha^2}{c(1-\alpha)}}}$
(Note that, we do not consider the concentration of H₂O because, it is a pure liquid)
6. To write K_a, we can use the earlier method also
• We get Eq.7.9: $\mathbf\small{\rm{K_a=\frac{[H^+][X^-]}{[HX]}}}$
7. It is clear that, if [H⁺] and [X^-] are larger, K_a will be larger
• [H⁺] and [X^-] will be larger for strong acids
◼ So we can write:
Strong acids will have a large value of K_a

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• In previous sections, we have seen that:
If the equilibrium constant K is known, the concentrations at equilibrium can be calculated (see solved example 7.6 in section 7.5)
• In our present case, we have K_a in place of K
    ♦ If this K_a is known, we can calculate the concentrations at equilibrium
    ♦ Once the concentrations are known, we can calculate pH and 𝛼
    ♦ The following solved examples demonstrate the procedure

Solved example 7.62
The ionization constant of HF is 3.2 × 10^-4 . Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H₃O⁺, F^- and HF) in the solution and its pH.
Solution:
1. The balanced equation for the dissociation of HF is:
HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and F^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and F^- are formed, the same x mol of HF would be consumed
• So the concentration of HF at equilibrium would be (0.02 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][F^-]}{[HF]}}}$
• Substituting the values from (1), we get:
3.2 × 10^-4 = $\mathbf\small{\rm{\frac{x^2}{0.02-x}}}$
3. Solving this quadratic equation, we get: x = 0.002375 or -0.00269
• negative value is not acceptable. So we take x = 0.002375
4. So we can write:
• The concentrations at equilibrium are:
[H₃O⁺] = [F^-] = x = 0.0024 M
[HF] = (0.02 - x) = 0.01763
5. Once we know [H₃O⁺], we can calculate the pH
We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.0024) = 2.62
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.0024
• c = 0.002 M
• So 𝛼 = ^0.0024⁄_0.02 = 0.12

Solved example 7.63
The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H⁺, A^- and HA at equilibrium. Also, determine the value of K_a and pK_a of the monobasic acid
Solution:
1. Monobasic acid will have only one H atom
• We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 4.50 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -4.50
• So [H⁺] will be equal to the antilog of -4.50
3. Thus we get: [H⁺] = antilog (-4.50) = 0.00003162 = 3.16 × 10^-5 M
4. The balanced equation is: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A^-(aq)
• From the stoitiometric coefficients, it is clear that:
[A^-] will be equal to [H⁺]
• So we get: [A^-] = 3.16 × 10^-5 M
5. If 3.16 × 10^-5 M of [H⁺] and [A^-] are produced, the [HA] will be equal to:
(0.1 - 3.16 × 10^-5) = 0.0999684 ≃ 0.1 M
6. Once we know the concentrations, we can calculate K_a:
$\mathbf\small{\rm{K_a=\frac{[H_3O^+][A^-]}{[HA]}=\frac{(3.16 \times 10^{-5})^2}{0.1}}}$ = 1.0 × 10^-8
7. We have: pH = -log₁₀([H⁺])
• In a similar ways, we can write: pK_a = -log₁₀([H⁺])
• Thus we get: pK_a = -log₁₀(10⁸) = 8
8. Significance of pK_a can be written in 4 steps:
(i) We see that, if the 'negative power' of K_a is large, pK_a will be large
(ii) If the K_a has a large negative power, it indicates that, K_a is small
(iii) If K_a is small, it indicates that, the acid is weak
(iv) So we can write:
If pK_a is large, it will be a weak acid
9. Alternatively, we can use 𝛼 (expressed as a percentage) also to express the strength of weak acid. It can be written in 3 steps:
(i) We have: $\mathbf\small{\rm{\alpha=\frac{c \alpha}{c}=\frac{[HA]_{dissociated}}{[HA]_{initial}}}}$
(ii) So we get: $\mathbf\small{\rm{\alpha \, (percentage)=\frac{[HA]_{dissociated}}{[HA]_{initial}}\times 100}}$
(iii) When 𝛼 is expressed as a percentage, it is called percent dissociation

Solved example 7.64
Calculate the pH of 0.08 M solution of hypochlorous acid HOCl. The ionization constant of the acid is 2.5 × 10^-5. Determine percentage dissociation of HOCl
Solution:
1. The balanced equation for the dissociation of HOCl is:
HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + ClO^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and ClO^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and ClO^- are formed, the same x mol of HOCl would be consumed
• So the concentration of HOCl at equilibrium would be (0.08 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][ClO^-]}{[HOCl]}}}$
• Substituting the values from (1), we get:
2.5× 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.08-x}}}$
⇒ x² + (2.5 × 10^-5)x - 0.2 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.001402 or -0.001427
• negative value is not acceptable. So we take x = 0.001402
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [H₃O⁺] = [ClO-] = x = 0.001402 M
   ♦ [HOCl] = (0.08 - x) = 0.0786
5. Once we know [H₃O⁺], we can calculate the pH
• We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.001402) = 2.85
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.001402
• c = 0.08 M
• So 𝛼 = ^0.001402⁄_0.08 = 0.0175
7. Thus we get:
Percent dissociation = 𝛼 expressed as percentage = (0.0175 × 100) = 1.75%

Solved example 7.65
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.
Solution:
1. The balanced equation for the dissociation of acetic acid is:
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and CH₃COO^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and CH₃COO^- are formed, the same x mol of CH₃COOH would be consumed
• So the concentration of CH₃COOH at equilibrium would be (0.05 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][CH3COO^-]}{[CH3COOH]}}}$
• Substituting the values from (1), we get:
1.74 × 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.05-x}}}$
⇒ x² + (1.74 × 10^-5)x - 0.087 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.000924 or -0.000941
• negative value is not acceptable. So we take x = 0.000924
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [H₃O⁺] = [CH3COO-] = x = 0.000924 M
   ♦ [CH3COOH] = (0.05 - x) = 0.0491 ≃ 0.05 M
5. Once we know [H₃O⁺], we can calculate the pH
We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.000924) = 3.03

Solved example 7.66
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 .
Solution:
1. We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 4.15 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -4.15
• So [H⁺] will be equal to the antilog of -4.15
3. Thus we get: [H⁺] = antilog (-4.15) = 7.079 × 10^-5 M
4. The balanced equation is: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A^-(aq)
• From the stoitiometric coefficients, it is clear that:
[A^-] will be equal to [H⁺]
• So we get: [A^-] = 7.079 × 10^-5 M
5. If 7.079 × 10^-5 M of [H⁺] and [A^-] are produced, the [HA] will be equal to:
(0.01 - 7.079 × 10^-5) = 0.0099 ≃ 0.01 M
6. Once we know the concentrations, we can calculate K_a:
$\mathbf\small{\rm{K_a=\frac{[H_3O^+][A^-]}{[HA]}=\frac{(7.079 \times 10^{-5})^2}{0.01}}}$ = 5.01 × 10^-7
7. We have: pH = -log₁₀([H⁺])
• In a similar ways, we can write: pK_a = -log₁₀([K_a])
• Thus we get: pK_a = -log₁₀(5.01 × 10^-7) = 6.3

Solved example 7.67
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.
Solution:
1. Given that c = 0.1 and 𝛼 = 0.132
So [H⁺] = c𝛼 = (0.1 0.132) = 0.0132
2. pH = -log₁₀([H⁺]) = -log₁₀(0.0132) = 1.88
3. We have: $\mathbf\small{\rm{K_a=\frac{c^2\alpha^2}{c-c \alpha)}}}$
• Substituting the values, we get: $\mathbf\small{\rm{K_a=\frac{(0.0132)^2}{0.1-0.0132}}}$ = 0.002
4. pK_a = -log₁₀([H⁺]) = -log₁₀(0.002) = 2.7

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In the next section, we will see ionization constants of weak bases


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