Question

Calculate h+ , ch3coo- and c7h5o2- in a solution that is 0.02 m acetic acid and 0.01 m in benzoic acid. Ka(acetic)=1.8

Answer 1

$\bold{[H^{+}]=10^{-3}}$

$\bold{[CH_{3}COO^{-}]=3.6\times10^{-4}}$

$\bold{[C_{6}H_{5}COO^{-}]=6.4\times 10^{-4}}$

Step by step explanation:

The dissociation reaction of acetic acid is as follows.

$\bold{CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}}$

$\bold{K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}}$

The dissociation reaction of acetic acid is as follows

$\bold{C_{5}H_{5}COOH\rightleftharpoons C_{6}H_{5}COO^{-}+H^{+}}$

$\bold{K_{a}=\frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}}$

$\bold{[H^{+}]=\sqrt{K_{1}C_{1}+K_{2}C_{2}}}$

$=\sqrt{1.8\times 10^{-5}\times 0.02 +6.4\times10^{-5}\alpha 0.01}$

$=\sqrt{0.36\times 10^{-6}+0.64\times 10^{-6}}$

$=10^{-3}$

Hence, $\bold{[H^{+}]=10^{-3}}$

Concentration of $\bold{CH_{3}COO^{-}}$:

$1.8\times 10^{-5}=\frac{[CH_{3}COO^{-}]\times 10^{-3}}{0.02}$

$[CH_{3}COO^{-}]=\frac{3.6\times 10^{-7}}{10^{-3}}=3.6\times10^{-4}$

Hence, $\bold{[CH_{3}COO^{-}]=3.6\times10^{-4}}$

Concentration of $\bold{C_{6}H_{5}COO^{-}}$

$6.4\times 10^{-5}=\frac{[C_{6}H{5}COO^{-}]\times 10^{-3}}{0.01}$

$[C_{6}H_{5}COO^{-}]=\frac{6.4\times 10^{-7}}{10^{-3}}=6.4\times 10^{-4}$

Hence, $\bold{[C_{6}H_{5}COO^{-}]=6.4\times 10^{-4}}$