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^âˆ’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.)

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Kerouac Kerouac · Answer 1 · 2020-01-25T22:22:54+01:00

Answer:

7.62

Explanation:

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

[tex]pH = pK_a + log(\frac{[A^-]}{[HA]})[/tex]

Let's identify the variables:

[tex]pK_a = -log(K_a) = -log(4.0\cdot 10^{-8}) = 7.40[/tex]

[tex][A^-] = [ClO^-][/tex]

[tex][HA] = [HClO][/tex]

Identify the moles of each component:

[tex]n_{ClO^-} = 0.61~M\cdot 0.0255~L = 0.015555~mol[/tex]

[tex]n_{HClO^} = 0.61~M\cdot 0.0255~L = 0.015555~mol[/tex]

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

[tex]OH^- (aq) + HClO (aq)\rightarrow ClO^- (aq) + H_2O (l)[/tex]

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

[tex]n_{HClO} = 0.015555~mol - 0.0039~mol = 0.011655~mol[/tex]

[tex]n_{ClO^-} = 0.015555~mol + 0.0039~mol = 0.019455~mol[/tex]

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

[tex]pH = 7.40 + log(\frac{0.019455~mol}{0.011665~mol}) = 7.62[/tex]