$pH$ of a mixture weak acid / strong base

Strict method calculation

Calculate for example the $pH$ of $100\;mL$ $CH_3COOH\; 0.20\; M$ $(pK_a=4.75)$ mixed with $100\;mL$ $NaOH\; 0.10\; M$

Equations

$c_{[CH_3COOH]}$ $=$ $\frac{0.10\cdot 0.20 }{0.10+0.10}$ $=$ $0.10\; M$ $c_{[NaOH]}$ $=$ $\frac{0.10\cdot 0.10}{0.10+0.10}$ $=$ $0.050\; M$ (1) Electroneutrality: $[CH_3COO^-]$ $+$ $[OH^-]$ $=$ $[H_3O^+]$ $+$ $[Na^+]$ (2) Conservation of matter: $[Na^+]=0.050$ (3) Conservation of matter: $[CH_3COOH]+[CH_3COO^-]$ $=$ $0.10$ (4) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$ (5) Acidity constant: $10^{-4.75}$ $=$ $\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$

Resolution

Elimination of all variables as a function of $[H_3O^+]$ : $10^{4.75}[H_3O^+]^3$ $+$ $(1$ $+$ $0.050\cdot10^{4,75})[H_3O^+]^2$ $+$ $(0.050$ $-$ $0,10-10^{p-14})[H_3O^+]$ $-$ $10^{-14}$ $=0$ Solving this →   3rd degree equation, we find: $[H_3O^+]$ $=$ $1.778\cdot 10^{-5} \frac{mol}{L}$ $pH=4.75$