Question

if a 0.12M solution of an acid is 4.2% ionized, calculate the pH of the solution and the dissociation constant of the acid

Answer 1

Solution.

$Ka = \frac{\alpha^2 \times C}{1-\alpha}$

$Ka = 2.21 \times 10^{-4}$

$pH = -lg(\alpha \times C) = 2.30$

Answer:

$pH = 2.30$

$Ka = 2.21 \times 10^{-4}$