Question

(b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq) ? Justify your answer including the calculation of percent ionization for each solution.

Answer 1

Answer: The percent ionization is higher for 0.010 M solution of $HC_2H_3O_2$

Explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

 cM           0M           0M

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

So dissociation constant will be:

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

a) Given c= 0.10 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.10\times \alpha)^2}{(0.10-0.10\times \alpha)}$

$(\alpha)=0.013$

$\%(\alpha)=0.013\times 100=1.3\%$

b) Given c= 0.010 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.010\times \alpha)^2}{(0.010-0.010\times \alpha)}$

$(\alpha)=0.041$

$\%(\alpha)=0.041\times 100=4.1\%$

The percent ionization is higher for 0.010 M solution of $HC_2H_3O_2$