# $pH$ of a strong acid

## Strict method calculation

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

### Equations

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

### Resolution

(2) and (3) in (1): $[H_3O^+]^2$ $-$ $0.10[H_3O^+]$ $-$ $10^{-14}$ $=0$ so: $[H_3O^+]$ $=$ $\frac{0,1\pm \sqrt{0,01+4\cdot 10^{-14}}}{2}$ $\approx$ $0.10\frac{mol}{L}$ The negative solution is to be rejected, so we have: $pH\approx 1.00$