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pH of a mixture strong acid / weak base

Strict method calculation

Calculate for example the $pH$ of $100\;mL$ $HCl\; 0.20\; M$ mixed with $100\;mL$ $CH_3COONa\; 0.10\; M$ $(pK_a=4.75)$

Equations

$c_{[HCl]}$ $=$ $\frac{0.10\cdot 0.20 }{0.10+0.10}$ $=$ $0.10\; M$ $c_{[CH_3COONa]}$ $=$ $\frac{0.10\cdot 0.10}{0.10+0.10}$ $=$ $0.050\; M$ (1) Electroneutrality: $[Cl^-]$ $+$ $[OH^-]$ $+$ $[CH_3COO^-]$ $=$ $[H_3O^+]$ $+$ $[Na^+]$ (2) Conservation of matter: $[Cl^-]=0.10$ (3) Conservation of matter: $[Na^+]=0.05$ (4) Conservation of matter: $[CH_3COOH]$ $+$ $[CH_3COO^-]$ $=$ $0.050$ (5) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$ (6) Acidity constant: $10^{-4.75}$ $=$ $\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$

Resolution

Elimination of all variables as a function of $[H_3O^+]$ $10^{4.75}[H_3O^+]^3$ $+$ $(1-0,10\cdot10^{4.75})[H_3O^+]^2$ $+$ $(-0.10-0.050-10^{-9.25})[H_3O^+]$ $-$ $10^{-14}$ $=$ $0$ Solving this →   3rd degree equation, we find: $[H_3O^+]$ $=$ $5.000\cdot 10^{-2} \frac{mol}{L}$ $pH=1.301$