Question

A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution

Answer 1

Answer: 2.75%

Explanation:

$pH=-log [H+]$

$3.26 = -log [H+]$

$[H+] = 5.495\times 10^{-4} M$

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

initial      0.020     0           0

eqm        0.020 -x    x      x

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

$K_a=\frac{[x][x]}{[0.020-x]}$

$x=5.495\times 10^{-4}$

$K_a=\frac{[5.495\times 10^{-4}]^2}{[0.020-5.495\times 10^{-4}]}$

$K_a =1.553\times 10^{-5}$

percent dissociation = $\frac{[H^+_eqm]}{[Acid_{initial}]}\times 100$

percent dissociation=$\frac{5.495\times 10^{-4}}{0.020}\times 100$

Thus percent dissociation= 2.75 %