Question

Calculate the $pH$ of solution when $100\text{mL}$, $0.1\text{M}$ $C{H}_{3}COOH$ and $100\text{mL}$, $0.1\text{M}HCOOH$ are mix together (Given : $K_a\left(C{H}_{3}COOH\right)=2\times {10}^{-5},K_a\left(HCOOH\right)=6\times {10}^{-5})$ (Given: $\log 2=0.30$)

Answer 1

As the volume is doubled, concentration would be half. The total $\left[H^{+}\right]$ is obtained by addition $\left[H^{+}\right]$ from both acids.
$\left[H^{+}\right]=\sqrt{(K_{a_1}{C}_1+K_{a_2}{C}_2)}=\sqrt{(2\times {10}^{-5}\times \frac{0.1}{2}+6\times {10}^{-5}\times \frac{0.1}{2})}=2\times {10}^{-3}$
$\therefore pH=-\log \left[H^{+}\right]=-\log (2\times {10}^{-3})=2.7$