Exercice 6 | Tampons et titrage des acides et des bases.

HCl → H⁺ + Cl^-
NaOH → Na⁺ + HO^-

Réaction : H⁺ + HO^- → H₂O

n_H+,0 = [H⁺] x V_H+ = 0.20 x 20.0 x 10^-3 = 4.0 x 10^-3 mol

n_HO-,0 = [HO^-] x V_HO- = 0.15 x 20.0 x 10^-3 = 3.0 x 10^-3 mol

HO^- est le réactif limitant et à la fin n_H+ = n_H+,0 - n_HO-,0 = 1.0 x 10^-3 mol

[H⁺] = $\frac{{\mathrm{n}}_{\mathrm{H}+}}{\mathrm{V}}$ = $\frac{1.0 \times {10}^{-3}}{40.0 \times {10}^{-3}}$ = 2.5 x 10^-2 M

pH = - log [H⁺] = 1.60