Question

Calculate the pH of the following mixture given
$K_a=1.8\times {10}^{-5}$ and $K_b=1.8\times {10}^{-5}(p{K}_a=p{K}_a=4.7447)$
$50mL$ $0.05M$ $NaOH+50mL$ of $0.1M$ ${CH}_{3}COOH$

Answer 1

$K_a=K_b=1.80\times {10}^{-5}$

$p{K}_a=p{K}_b=4.7447$

$50ml$ of $0.005M$ $NaOH+50ml$ of $0.1M$ $C{H}_{3}COOH$

$HAC+NaOH\leftrightharpoons H_{2}O+C{H}_{3}COONa$

Moles of $NaOH=0.050\times 0.05=0.0025$ $mol$

$\therefore \left[C{H}_{3}COONa\right]$ in $100ml$ $\left(0.1L\right)$ solution$=\frac{0.0025}{0.1}$ $=0.025\frac{mol}{L}$

$pH=\frac{1}{2}[p{K}_a+p{K}_w+logC]$

$=\frac{1}{2}[14+4.7447+log(0.025\left)\right]$

$pH=8.57$

$pH$ of mixture is $8.57$