# pH of a strong base

## Strict method calculation

Calculate for example the $pH$ of $NaOH\; 0.1\; M$

### Equations

(1) Electroneutrality: $[OH^-]$ $=$ $[H_3O^+]+[Na^+]$ (2) Conservation of matter: $[Na^+]=0.10$ (3) Ionic product of water: $[H_3O^+][OH^-]$ $=$ $10^{-14}$

### Resolution

$[OH^-]^2$ $-$ $0.10[OH^-]$ $-$ $10^{-14}$ $=0$ so: $[OH^-]$ $=$ $\frac{0.1\pm \sqrt{0,01+4\cdot 10^{-14}}}{2}$ $\approx$ $0.10\frac{mol}{L}$ The negative solution must be rejected, so: $pH$ $=$ $14-pOH$ $\approx$ $13.00$