Calculate the pH of a solution obtained by mixing $200\mathrm{mL}$ of $0.1\mathrm{MHCl}$ with $800\mathrm{mL}$ of $0.125{\mathrm{MCH}}_3\mathrm{COOH}\text{. }$$\left(K_a\text{ for }{\mathrm{CH}}_3\mathrm{COOH}=1.8\times {10}^{-5}\right)$

Concentration of HCl after mixing (c₂) =$\frac{200\times 0.1}{200+800}$=0.02 M

Concentration of CH₃COOH after mixing (c₁)= $\frac{800\times 0.125}{800+200}=0.1\mathrm{M}$

${\left[{\mathrm{H}}_3\mathrm{O}\right]}^{+}$ of mixture of weak acid and strong acid = $\frac{{\mathrm{c}}_2+\sqrt{{\mathrm{c}}_2^2+4{\mathrm{k}}_{\mathrm{a}}{\mathrm{c}}_1}}{2}$

=$\frac{0.02+\sqrt{{\left(0.02\right)}^2+4\times 1.8\times {10}^{-5}\times 0.1}}{2}$

on solving ${\left[{\mathrm{H}}_3\mathrm{O}\right]}^{+}$ = 2.01$\times$10^-2

p^H = 2- log2.01 = 2 - 0.3032 = 1.6968