Calculate for example the $pH$ of $100\;mL$ $CHCl_2COOH\; 0.1\; M$ $(pK_a\;=\;1.30$, $K_a\;=\;0.005)$ mixed with $100\;mL$ $HCl\; 0.01\; M$
(1) Electroneutrality: $[CHCl_2COO^-]$ $+$ $[OH^-]$ $+$ $[Cl^-]$ $=$ $[H_3O^+]$ (2) Conservation of matter: $[Cl^-]=0.0050$ (!dilution) (3) Conservation of matter: $[CHCl_2COO^-]$ $+$ $[CHCl_2COOH]$ $=$ $0.050$ (!dilution) (4) Ionic product of water: $[H_3O^+][OH^-]=10^{-14}$ (5) Acidity constant: $\frac{[CHCl_2COO^-][H_3O^+]}{[CHCl_2COOH]}$ $=$ $10^{-1.30}$
Elimination of all variables as a function of $[H_3O^+]$ $10^{1.3}[H_3O^+]^3$ $+$ $(1$ $-$ $0.005\cdot10^{1.3})[H_3O^+]^2$ $+$ $(-0.05$ $-$ $10^{-12,7}$ $-$ $0.005)[H_3O^+]$ $-$ $10^{-14}$ $=$ $0$ Resolving this → 3rd degree equation, we have: $[H_3O^+]$ $=$ $3.47\cdot 10^{-2} \frac{mol}{L} $ $pH=1.46$