Question

The $\left[H^{+}\right]$ of a resulting solution that is $0.01M$ acetic acid $(K_a=1.8\times {10}^{-5})$ and $0.01M$ in benzoic acid $(K_a=6.3\times {10}^{-5})$:

• A. $9\times {10}^{-4}$
• B. $81\times {10}^{-4}$
• C. $9\times {10}^{-5}$
• D. $2.8\times {10}^{-3}$

Answer 1

The correct option is A $9\times {10}^{-4}$

GIVEN

$K_{a1}$ = $1.8\times {10}^{-5}$, $C_1=0.01$

$K_{a2}$ = $6.3\times {10}^{-5}$, $C_2=0.01$

SOLUTION
The dissociation constant for both the acids is $K_a$=${10}^{-5}$

Here $\alpha <0.05$ So, neglect $\alpha$ with respect to 1 for both the acids

$\left[H^{+}\right]$=$\sqrt{K_{a1}{C}_1+K_{a2}{C}_2}$

where,

$\left[H^{+}\right]$ = $\sqrt{1.8\times {10}^{-5}\left(0.01\right)+6.3\times {10}^{-5}\left(0.01\right)}$

$\left[H^{+}\right]$ = $\sqrt{1.8\times {10}^{-7}+6.3\times {10}^{-7}}$

$\left[H^{+}\right]$ = $\sqrt{8.1\times {10}^{-7}}$ =$\sqrt{81\times {10}^{-8}}$

$\left[H^{+}\right]$ = $9\times {10}^{-4}$

The correct option: A.