Question

Suppose you have a solution that is 0.20M of HF (weak acid, Ka = 6.8x10^-4) and a 0.10M of HCl (strong acid).

What is the acid dissociation equation of the weak acid in the solution?

What is the concentration of F- in the solution?

What is the pH of this solution?

Answer 1

HF is the weak acid and ionizes as

HF (aq) --------> H⁺ (aq) + F^- (aq)

The acid ionization constant is given as

K_a = [H⁺][F^-]/[HF] = 6.8*10^-4 …….(1)

pK_a = -log (K_a) = -log (6.8*10^-4)

= 3.167

≈ 3.17 (ans).

The concentration of F^- in the solution is obtained by letting [F^-] = x M. Due to the 1:1 nature of ionization, [H⁺] = [F^-] = x M.

Therefore,

6.8*10^-4 = (x)(x)/(0.20 – x)

where x M is the equilibrium concentration of HF.

Since K_a is small, we can assume x << 0.20 M and write

6.8*10^-4 = x²/(0.20) (ignore units)

=====> x² = 1.36*10^-4

=====> x = 0.01166 ≈ 0.012

The concentration of F^- in the solution is 0.012 M (ans).

HCl is a strong acid and ionizes completely; therefore,

HCl (aq) --------> H⁺ (aq) + Cl^- (aq)

As per the stoichiometric equation,

[H⁺] = [HCl] = 0.1 M.

The concentration of H⁺ from the ionization of HF is 0.012 M.

Therefore,

[H⁺]_tot = (0.10 + 0.012) M = 0.112 M.

pH = -log [H⁺]

= -log (0.112 M)

= 0.9508

≈ 0.95 (ans).