Chapter 7.18 - Ionization Constants of Weak Bases

In the previous section, we saw ionization constant (K_a) of weak acids. In this section, we will see ionization constant (K_b) of weak bases. Later in this section, we will also see the relation between K_a and K_b

• In an earlier section, we saw that:
    ♦ Strong bases like NaOH undergo complete ionization when added to water
    ♦ Weak bases do not undergo complete ionization when added to water
• We can write the main features of the ionization constant of weak bases in 7 steps:
1. Let XOH be a weak base. It's ionization process can be represented by the following equation:
XOH(aq) ⇌ X⁺(aq) + OH^-(aq)
2. Let c be the initial concentration of XOH
    ♦ That is., at t = 0, [XOH] = 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 XOH undergoing ionization = c𝛼
4. When c𝛼 moles of XOH undergoes ionization, the number of moles of XOH remaining will be equal to (c - c𝛼)
• Also, from the stoitiometric coefficients, we can write:
    ♦ When c𝛼 moles of XOH undergoes dissociation,
        ✰ c𝛼 moles of X⁺ will be formed
        ✰ c𝛼 moles of OH^- 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 base XOH
    ♦ It is denoted as K_b
• So we can write $\mathbf\small{\rm{K_b=\frac{(c\alpha)^2}{c-c\alpha}}}$
• Thus we get Eq.7.10: $\mathbf\small{\rm{K_b=\frac{c^2\alpha^2}{c(1-\alpha)}}}$
(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.7.11: $\mathbf\small{\rm{K_b=\frac{[X^+][OH^-]}{[XOH]}}}$
7. It is clear that, if [X⁺] and [OH^-] are larger, K_b will be larger
• [X⁺] and [OH^-] will be larger for strong acids
◼ So we can write:
Strong acids will have a large value of K_b

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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). We saw similar solved examples involving K_a in the previous section also
• In our present case, we have K_b in place of K_a
    ♦ If this K_b is known, we can calculate the concentrations at equilibrium
    ♦ Once the concentrations are known, we can calculate pH and 𝛼
         ✰ (Recall that, basic solutions have pH greater than 7)
    ♦ The following solved examples demonstrate the procedure

Solved example 7.68
The pH of 0.004 M hydrazine solution is 9.7. Calculate it's ionization constant K_b and pK_b
Solution:
1. The balanced equation for the dissociation of hydrazine is:
NH₂NH₂(aq) + H₂O(l) ⇌ NH₂NH₃⁺(aq) + OH^-(aq)
2. Given that, pH is 9.7
• We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 9.7 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -9.7
• So [H⁺] will be equal to the antilog of -9.7
• Thus we get: [H⁺] = antilog (-9.7) = 1.9952 × 10^-10 M
3. We have: [H⁺][OH^-] = 10^-14
• So [OH^-] = 5.012 × 10^-5 M
4. From the stoitiometric coefficients, it is clear that:
[NH₂NH₃⁺] will be same as [OH^-]
5. So we get: [NH₂NH₃⁺] = 5.012 × 10^-5
• So [NH₂NH₂] = (0.004 - 5.012 × 10^-5) = 0.00394 ≃ 0.004
6. Next we calculate K_b. We have:
$\mathbf\small{\rm{K_b=\frac{[NH_2NH_3^+][OH^-]}{[NH_2NH_2]}=\frac{(5.012 \times 10^{-5})^2}{0.004}}}$ = 6.36 × 10^-7
7. So pK_b = -log₁₀(K_b) = -log₁₀(6.36 × 10^-7) = 6.2

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Relation between K_a and K_b

Relation between K_a and K_b can be explained with the help of an example. It can be written in 9 steps:
1. Consider the reaction between NH₃ and H₂O. The balanced equation is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
• Here NH₃ acts as the base because, it accepts a proton and also produces OH^- ions in the aqueous solution
• So we can write the base dissociation constant as: $\mathbf\small{\rm{K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}}}$
2. In this reaction, we can identify the conjugate acid and conjugate base
(See step 13 in section 7.12)
• We can write them as:
    ♦ NH₃ is the base, H₂O is the acid
    ♦ NH₄⁺ is the conjugate acid, OH^- is the conjugate base
3. We see that, NH₄⁺ is the conjugate acid
• Let us see it's reaction with water. The balanced equation is:
NH₄⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NH₃(aq)
• Here NH₄⁺ indeed acts as the acid because, it donates a proton and also produces H₃O⁺ ions in the aqueous solution
• So we can write the acid dissociation constant as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][NH_3]}{[NH_4^+]}}}$
4. Let us add the reactions in (1) and (3)
• It can be written in 4 steps:
(i) On the left side, we will be having four items:
NH₃(aq) + H₂O(l) + NH₄⁺(aq) + H₂O(l)
(ii) On the right side, we will be having four items:
NH₄⁺(aq) + OH^-(aq) + H₃O⁺(aq) + NH₃(aq)
(iii) So after addition, we will get:
NH₃(aq) + H₂O(l) + NH₄⁺(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) + H₃O⁺(aq) + NH₃(aq)
(iv) NH₃(aq) and NH₄⁺(aq) are common on both sides. So the net reaction is:
H₂O(l) + H₂O(l) ⇌ OH^-(aq) + H₃O⁺(aq)
5. The above net reaction is familiar to us
• We know the ionization constant of that reaction. It is: K_w = 10^-14
6. Now let us find the product of two items:
    ♦ K_b that we wrote in (1)
    ♦ K_a that we wrote in (3)
• We get: K_aK_b = $\mathbf\small{\rm{\frac{[H_3O^+][NH_3]}{[NH_4^+]}\times \frac{[NH_4^+][OH^-]}{[NH_3]}}}$
⇒ K_aK_b = [H₃O⁺][OH^-]
• We have seen this product before. It is equal to K_w
    ♦ Also, K_w is the ionisation constant of the net reaction obtained in (4)
• So we can write: K_aK_b = K_w
7. Let us write a summary of the above steps:
• We added the two reactions in (1) and (3) and obtained the net reaction
• We obtained the product of the ionization constants of the reactions in (1) and (3)
• We found that:
    ♦ The product
    ♦ is equal to
    ♦ The ionization constant of the net reaction
• The reactions in (1) and (3) are related to conjugate acid-base pair
◼ So we can write:
For a conjugate acid-base pair, K_aK_b = K_w
8. We can write this information as a general rule. It can be written in 3 steps:
(i) We have a few reactions
    ♦ We write the equilibrium constants of those reactions : K₁, K₂, K₃ . . .
    ♦ We write the product of those constants: K₁ × K₂ × K₃ . . .
(ii) We add the reactions and write the net reaction
• We denote the equilibrium constant of this net reaction as K_NET
(iii) Then K_NET will be equal to the product in (i)
• Thus we get Eq.7.10: K_NET = K₁ × K₂ × K₃ . . .
9. Consider the result that we wrote in 6 : K_aK_b = K_w
• We know that, the K values involve negative powers. In order to avoid those negative powers and thus make it more presentable, we can use logarithms
• We get:
log₁₀(K_a) + log₁₀(K_b) = log₁₀(K_w) = log₁₀(10^-14)
• Multiplying throughout by -1, we get:
-log₁₀(K_a) + -log₁₀(K_b) = -log₁₀(K_w) = -log₁₀(10^-14)
• But we have:
    ♦ -log₁₀(K_a)= pK_a
    ♦ -log₁₀(K_b)= pK_b
    ♦ -log₁₀(K_w)= pK_w
    ♦ -log₁₀(10^-14)= -(-14) = 14
• Thus we get Eq.7.11: pK_a + pK_b = pK_w = 14

Solved example 7.69
Determine the degree of ionization and pH of a 0.05 M ammonia solution. The ionization constant of ammonia is 1.77 × 10^-5. Also calculate the ionization constant of the conjugate acid of ammonia
Solution:
1. The balanced equation is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
• Let at equilibrium, x moles each of NH₄⁺ and OH^- be present
• From the balanced equation, it is clear that, if x mol each of NH₄⁺ and OH^- are formed, the same x mol of NH₃ would be consumed
• So the concentration of NH₃ at equilibrium would be (0.05 - x) mol
2. For this reaction, K_b can be obtained as: $\mathbf\small{\rm{K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}}}$
• Substituting the values from (1), we get:
1.77 × 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.05-x}}}$
⇒ x² + (1.77 × 10^-5)x - 0.0885 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.000932 or -0.000949
• negative value is not acceptable. So we take x = 0.000932
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [NH₄⁺] = [OH^-] = x = 0.000932 M
   ♦ [NH₃] = (0.05 - x) = 0.049
5. Once we know [OH^-], we can calculate the pH
• We have: pH = (14 - pOH)= [14 - -log₁₀([OH^-])]
= [14 - -log₁₀(0.00932)] = [14 - 3.031] = 10.97
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.000932
• c = 0.05 M
• So 𝛼 = ^0.00932⁄_0.05 = 0.018
7. We can find K_a of the conjugate acid of ammonia as follows:
• We have seen that, for a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}= \frac{10^{-14}}{1.77 \times 10^{-5}}}}$ = 5.64 × 10^-10

Solved example 7.70
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constant of the corresponding conjugate base
Solution:
• For a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• In this problem, K_a values are given
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}}}$
(i) K_b of the conjugate base of HF = $\mathbf\small{\rm{\frac{10^{-14}}{6.8 \times 10^{-4}}}}$ = 1.47 × 10^-11
(i) K_b of the conjugate base of HCOOH = $\mathbf\small{\rm{\frac{10^{-14}}{1.8 \times 10^{-4}}}}$ = 5.55 × 10^-11
(i) K_b of the conjugate base of HCN = $\mathbf\small{\rm{\frac{10^{-14}}{4.8 \times 10^{-9}}}}$ = 2.08 × 10^-6

Solved example 7.71
The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b
Solution:
1. Given that pH = 9.95
• This is greater than 14. So it is a basic solution
• pOH will be (14 - 9.95) = 4.05
2. From pOH, we can calculate [OH^-]
[OH^-] = antilog of -4.05 = 8.913 × 10^-5
3. We are asked to find the ionization constant K_b
• We have: $\mathbf\small{\rm{K_b=\frac{[positive \; ion][OH^-]}{[C_{18}H_{21}NO_3]}}}$
• [positive ion] will be same as [OH^-]
• Thus we get [positive ion] = 8.913 × 10^-5 M
4. Next we want [C₁₈H₂₁NO₃] at equilibrium
• Assume that, only a very small portion of the solute undergoes dissociation
    ♦ That is., 𝛼 is very small
• Then [C₁₈H₂₁NO₃] will be same as the initial concentration, which is 0.005
5. Thus we get: $\mathbf\small{\rm{K_b=\frac{(8.913 \times 10^{-5})^2}{0.005}}}$ = 1.588 × 10^-6
6. pK_b = -log(K_b) = -log(1.588 × 10^-6) = 5.79

Solved example 7.72
What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline
Solution:
1. The balanced equation is:
C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₄⁺(aq) + OH^-(aq) 
• Let at equilibrium, x moles each of C₆H₅NH₄⁺ and OH^- be present
• From the balanced equation, it is clear that, if x mol each of C₆H₅NH₄⁺ and OH^- are formed, the same x mol of C₆H₅NH₂ would be consumed
• So the concentration of C₆H₅NH₂ at equilibrium would be (0.001 - x) mol
2. For this reaction, K_b can be obtained as: $\mathbf\small{\rm{K_b=\frac{[C_6H_5NH_4^+][OH^-]}{[C_6H_5NH_2]}}}$
• Substituting the values from (1), we get:
4.27 × 10^-10 = $\mathbf\small{\rm{\frac{x^2}{0.001-x}}}$
⇒ x² + (4.27 × 10^-10)x - 4.27 × 10^-13 = 0
3. Solving this quadratic equation, we get: x = 6.532 × 10^-7 or -6.536 × 10^-7
• negative value is not acceptable. So we take x = 6.532 × 10^-7
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [C₆H₅NH₄⁺] = [OH^-] = x = 6.532 × 10^-7 M
   ♦ [NH₃] = (0.001 - x) = 0.00099
5. Once we know [OH^-], we can calculate the pH
• We have: pH = (14 - pOH)= [14 - -log₁₀([OH^-])]
= [14 - -log₁₀(6.532 × 10^-7)] = [14 - 6.18] = 7.82
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 6.532 × 10^-7
• c = 0.001 M
• So 𝛼 = 6.532 × 10^-7/0.001 = 0.000653
7. We can find K_a of the conjugate acid of ammonia as follows:
• We have seen that, for a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}= \frac{10^{-14}}{4.27 \times 10^{-10}}}}$ = 2.34 × 10^-5

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In the next section, we will see polybasic acids and polyacidic bases


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