Question

A buffer solution with a volume of 0.0255 L consists of 0.61 M hypochlorous acid (HClO), a weak acid, plus 0.61 M potassium hypochlorite (KClO). The acid dissociation constant of hypochlorous acid, Ka, is 4.0 x 10^−8.

1. Determine the pH of the buffer solution after the addition of 0.0039 mol rubidium hydroxide (RbOH), a strong base. (Assume no change in solution volume.)

Answer 1

7.62

Step-by-step explanation:

According to the Henderson-Hasselbach equation, the pH value o a buffer can be calculated by the following equation:

`pH = pK_a + log(([A^-])/([HA]))`

Let's identify the variables:

`pK_a = -log(K_a) = -log(4.0\cdot 10^(-8)) = 7.40`

`[A^-] = [ClO^-]`

`[HA] = [HClO]`

Identify the moles of each component:

`n_(ClO^-) = 0.61~M\cdot 0.0255~L = 0.015555~mol`

`n_(HClO^) = 0.61~M\cdot 0.0255~L = 0.015555~mol`

The strong base reacts with acid to produce more of the basic component:

`OH^- (aq) + HClO (aq)\rightarrow ClO^- (aq) + H_2O (l)`

Hydroxide is the limiting reactant. The new amounts of the weak acid and base are:

`n_(HClO) = 0.015555~mol - 0.0039~mol = 0.011655~mol`

`n_(ClO^-) = 0.015555~mol + 0.0039~mol = 0.019455~mol`

Let's use the molar amounts in the equation, as we have exactly same volume for each component in the buffer:

`pH = 7.40 + log((0.019455~mol)/(0.011665~mol)) = 7.62`