Question

A 0.16 M solution of a weak acid has a pH of 4.0. Calculate the Ka. You may assume that the amount of H ion is much greater than 1 x 10-7 M and much less than 0.16 M.

Answer 1

Answer:

The dissociation constant of weak acid is $6.254\times 10^{-8}$.

Explanation:

The pH of the solution = 4.0

$pH=-\log[H^+]$

$4=-\log[H^+}$

$[H^+]=10^{-4} M$

$HA\rightleftharpoons A^-+H^+$

initially

  c

At equilibrium

 c-x    x    x

Concentration of acid = c = [HA] = 0.16 M

Concentration of $H^+$ ions = x = $10^{-4} M$

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(c-x)}$

$K_a=\frac{10^{-4}\times 10^{-4}M}{(0.16 M-10^{-4} M)}$

$=6.254\times 10^{-8}$

The dissociation constant of weak acid is $6.254\times 10^{-8}$.