Calculate for example the $pH$ of $CH_3COOH\; 0.1\; M$ with $pK_a\;=\;4.75$
(1) Electroneutrality: $[CH_3COO^-]$ $+$ $[OH^-]$ $=$ $[H_3O^+]$ (2) Conservation of matter: $[CH_3COO^-]$ $+$ $[CH_3COOH]$ $=$ $0.10$ (3) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$ (4) Acidity constant: $\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ $=$ $10^{-4.75}$
Elimination of all variables in function of $[H_3O^+]$ $10^{4.75}[H_3O^+]^3$ $+$ $[H_3O^+]^2$ $+$ $(-0.1-10^{-9.25}[H_3O^+]$ $-$ $10^{-14}=0$ Solving this → 3rd degree equation , we find: $[H_3O^+]$ $=$ $1.3243\cdot 10^{-3} \frac{mol}{L} $ $pH=2.878$