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$pH$ of a weak base

Strict method calculation

Calculate for example the $pH$ of $NH_3\; 0.1\; M$ with $pK_a\;=\;9.20$

Equations

(1) Electroneutrality: $[OH^-]$ $=$ $[H_3O^+]+[NH_4^+]$ (2) Conservation of matter: $[NH_3]+[NH_4^+]$ $=$ $0.10$ (3) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$ (4) Acidity constant: $\frac{[NH_3][H_3O^+]}{[NH_4^+]}$ $=$ $10^{-9.20}$

Resolution

Elimination of all variables as a function of $[H_3O^+]$ : $10^{9.20}[H_3O^+]^3$ $+$ $(1$ $+$ $10^{8.2})[H_3O^+]^2$ $+$ $(-10^{-4.8})[H_3O^+]$ $-$ $10^{-14}$ $=0$ Solving this →    3rd degree equation, we find: $[H_3O^+]$ $=$ $7.998\cdot 10^{-12} \frac{mol}{L} $ $pH=11.097$