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pH of a buffer

Strict method calculation

Calculate for example the $pH$ of $100\;mL$ $CH_3COOH\; 0.20\; M$ mixed with $300\;mL$ $CH_3COONa\; 0.10\; M$ $(pK_a\;=\;4.75)$

Equations

$c_{[CH_3COOH]}$ $=$ $\frac{0.10\cdot 0,20 }{0.10+0.30}$ $=$ $0.050\; M$ $c_{[CH_3COONa]}$ $=$ $\frac{0.30\cdot 0,10}{0.10+0.30}$ $=$ $0.075\; M$ (1)Electroneutrality: $[CH_3COO^-]$ $+$ $[OH^-]$ $=$ $[H_3O^+]$ $+$ $[Na^+]$ (2)Conservation of matter: $[CH_3COO^-]$ $+$ $[CH_3COOH]$ $=$ $0.075$ $+$ $0.050$ (3)Conservation of matter: $[Na^+]=0.075$ (4)Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$ (5)Acidity constant: $\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ $=$ $10^{-4.75}$

Resolution

Elimination of all variables as a function of $[H_3O^+]$ $10^{4.75}[H_3O^+]^3$ $+$ $(1$ $+$ $0.075\cdot 10^{-14})[H_3O^+]^2$ $+$ $(-0.05$ $-$ $10^{-9.25})[H_3O^+]$ $-$ $10^{-14}$ $=0$ Solving the →   3rd degree equation, we find $[H_3O^+]$ $=$ $1.117\cdot 10^{-5} \frac{mol}{L} $ $pH=4.93$