Answer to Question #96886 in General Chemistry for Boye Zephaniah

Question

A 0.065 M solution of the weak acid HA has a pH of 3.6 at 25°C .
(a) what is the ionization constant of the acid at 25°C, the percentage ionization of HA, and the ratio of A/AH in the solution.
(b) what is the pH of a 0.05 M HCL solution?
(c) How much stronger in acidity is the HCL solution than the HA solution?
(d) what is the Kb and pKb of the conjugate base, A?

Answer 1

(a)

$$[H^+] = [A^-]$$

$$[HA] = c(HA) - [H^+]$$

$$[H^+] = 10^{-pH} = 10^{-3.6} = 0.00025 M$$

$$k_a = {[H^+][A^-] \over [HA]} = {0.00025*0.00025 \over 0.065 - 0.00025} = 9.7*10^{-7}$$

Ratio $A^-/HA$ in the solution:

$${[A^-] \over [HA]} = {0.00025 \over 0.065 - 0.00025} = 0.003861$$

(b)

Chloric acid is a strong acid, the degree of dissociation is taken equal to 1, then:

$$c(H^+) = c(HCl)$$

$$pH = -lg(c(H^+)) = -lg(0.05) = 1.3$$

(c)

$${[H^+]_{HCl} \over [H^+]_{HA} } = {10^{-1.3} \over 10^{-3.6}} = 10^{2.3} = 199.5 ≈ 200$$

(d)

The stronger the conjugate base, the weaker the acid.

$$k_b = {k_w(H_2O) \over K_a(HA)} = {10^{-14} \over 9.7*10^{-7}} ≈ 10^6$$

$$pK_b = -lgK_b = -6$$