What is the percent dissociation of a 0.015M solution of hydrofluoric acid given the Ka of HF is 6.3x10-4?

1) Chemical equilibrium:

HF + H2O ⇄ H3O (+) + F (-)

0.15 M - x x + x

2) Ka = [x]^2 / [0.15 - x]

3) Solution:

6.3x10^-4 = [x^2] / (0.15 - x)

=> x^2 = (0.15 - x) * 6.3 * 10^ - 4

x^2 = 0.0000945 - 0.00063x

x^2 + 0.00063x - 0.0000945 = 0

Use the quadratic formula to find the solution: x = 0.00941

4) percent dissociation:

% = [0.00941 M / 0.15 M] * 100 = 6.27%

Answer: 6.27%