Question

What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 2.43 × 10 − 12 .

Answer 1

Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is $3.59\times 10^{-4}\%$

Explanation:

Dissociation of weak acid is represented as:

$HA\rightleftharpoons H^+A^-$

 cM              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.188 M and $\alpha$ = ?

$K_a=2.43\times 10^{-12}$

Putting in the values we get:

$2.43\times 10^{-12}=\frac{(0.188\times \alpha)^2}{(0.188-0.188\times \alpha)}$

$(\alpha)=3.59\times 10^{-6}=3.59\times 10^{-4}\%$

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is $3.59\times 10^{-4}\%$