A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution before the addition of any LiOH. A)1.74 B)1.05 C)0.74 D)0.57 E)1.57

Question

Quizplus · Accepted Answer

HClO is a weak acid, so we need to use the equation for the ionization of a weak acid:

HClO + H2O ↔ H3O+ + ClO-

The equilibrium constant for this reaction is called the acid dissociation constant (Ka) and is given by:

Ka = [H3O+][ClO-]/[HClO]

We can use the Ka value to determine the pH of the solution before the addition of any LiOH:

Ka = 4.0 × 10^-8

[H3O+][ClO-]/[HClO] = 4.0 × 10^-8

Let x be the concentration of H3O+ in mol/L.

At equilibrium, [ClO-] = x mol/L and [HClO] = (0.18 - x) mol/L.

Substituting these values into the equation for Ka, we get:

x^2/(0.18 - x) = 4.0 × 10^-8

Solving for x, we get:

x = 1.9 × 10^-4 mol/L

The pH of the solution is given by:

pH = -log[H3O+]

pH = -log(1.9 × 10^-4)

pH = 3.74

Therefore, the pH of the solution before the addition of any LiOH is 0.74 (choice C).

A 1000 ML Sample of 0