Chemistry: Abuffer 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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Abuffer solution with a volume of 0.0255 l consists

xatziry1717

15.01.2022

7.62

Explanation:

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

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

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:

$pH = 7.40 + log(\frac{0.019455~mol}{0.011665~mol}) = 7.62$

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