Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is [tex]3.59\times 10^{-4}\%[/tex]
Explanation:
Dissociation of weak acid is represented as:
[tex]HA\rightleftharpoons H^+A^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.188 M and [tex]\alpha[/tex] = ?
[tex]K_a=2.43\times 10^{-12}[/tex]
Putting in the values we get:
[tex]2.43\times 10^{-12}=\frac{(0.188\times \alpha)^2}{(0.188-0.188\times \alpha)}[/tex]
[tex](\alpha)=3.59\times 10^{-6}=3.59\times 10^{-4}\%[/tex]
Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is [tex]3.59\times 10^{-4}\%[/tex]
