Question

A 100.0 mL sample of 0.18 M HCl is titrated with 0.27 M KOH at 25°C. Determine the pH of the solution after the addition of 66.67 mL of the KOH titrant.

Answer 1

pH = 7.0

Step-by-step explanation:

The HCl reaction with KOH as follows:

HCl + KOH → H₂O + KCl

To find pH first we need to know how many moles of each reactant reacts as follows:

HCl = 0.1000L * (0.18 mol / L) = 0.018 moles HCl

KOH = 0.06667L * (0.27mol / L) = 0.018 moles KOH

That means all HCl reacts with all KOH and you will have in solution just water. The equilibrium of water is:

H₂O(l) ⇆ H⁺(aq) + OH⁻(aq)

K = 1x10⁻¹⁴ = [H⁺] [OH⁻]

As the amount of H⁺ = OH⁻:

1x10⁻¹⁴ = [H⁺]²

[H⁺] = 1x10⁻⁷M

as pH = -log [H⁺]

pH = 7.0