Question

The ionisation 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 ph

Answer 1

Answer: Degree = 0.124, [H+] = [F-] = 0.00247 M, [HF] = 0.0175 M

Explanation:

$K_a = \frac{[H^+][F^-]}{[HF]} = 3.04 \times 10^{-4}$

At equilibrium,

[HF] = 0.02 - x and since x is very small and negligible, [HF] approx = 0.02

[H+] = [F-] = x

Substitute these values into the Ka expression

$\frac{x^2}{0.02} = 3.04 \times 10^{-4}\\x = \sqrt{3.04 \times 10^{-4} \times 0.02} = 0.00247 M$

x = [H+] = [F] = 0.00247 M

[HF] = 0.02 - 0.00247 = 0.01753 M

pH = - lg [H+] = 2.6

Degree of dissociation = 0.00247/0.02 = 12.4%