Question

Calculate the $pH$ of solution obtained by mixing $10mL$ of $0.2MHCl$ and $40mL$ of $0.1M{H}_{2}S{O}_4$.

Answer 1

$H^{+}=\frac{N_1{V}_1+N_2{V}_2}{V_1+V_2}=\frac{.2\times 10+.1\times 40}{50}=\frac{6}{50}=1.2\times {10}^{-1}$

$pH=-log{H}^{+}=1-log1.2=1-.07=.93\approx 1$