140 mL de HCl 0,175 M et 350 mL de NaOH 0,0750 M sont combinés.

Calculer le pH de cette solution.

Question

140 mL de HCl 0,175 M et 350 mL de NaOH 0,0750 M sont combinés.

Calculer le pH de cette solution.

| Chimie générale 3

Accepted Answer

n_H+,0 = [H⁺]₀ x V_HCl = 140 x 10^-3 x 0.175

n_H+,0 = 2.45 x 10^-2 mol

n_HO-,0 = [HO^-]₀ x V_NaOH = 350 x 10^-3 x 0.0750

n_HO-,0 = 2.63 x 10^-2 mol

There is an excess of HO^- which will neutralize all the [H⁺].

n_HO-,f = n_HO-,0 - n_H+,0

n_HO-,f = 2.63 x 10^-2 - 2.45 x 10^-2 = 1.75 x 10^-3 mol

[HO^-]_f = $\frac{n_{{HO}^{-}, f}}{V}$

[HO^-]_f = $\frac{1.75 \times 10^{- 3}}{490 \times 10^{- 3}}$

[HO^-]_f = 3.57 x 10^-3 M

pOH = -log [HO^-]_f = 2.45

pH = 14 – pOH = 11.6

Midterm 2 - Question 5 | Chimie générale 3