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Question

1(a) A buffer solution composed of chloroacetic acid, HC2H2ClO2, and sodium chloroacetate, NaC2H2ClO2, contains 0.122 mol and 0.108 mol of each component respectively in 1.00 L of solution. When H3O^+ is added to this buffer it gets consumed (neutralized) as shown below. Determine the moles of each buffer component that remain in solution after 0.014 moles of H3O^+ is added to the buffer. 

(b) Determine the buffer pH after the addition of 0.014 moles of H3O+. (Ka chloroaceticacid = 1.4 x 10^-3) 

2(a) A buffer composed of hypochlorous acid (HClO, Ka= 3.5 x 10^-8) and sodium hypochlorite, NaClO, is prepared. Determine the pH of this buffer if the [HClO] is 3 times smaller than the [NaClO]. 

(b) You need to prepare a pH 5.50 buffer for a lab experiment. You decide prepare this buffer using the weak acid HA ( Ka= 2.2 x 10^-6) and its conjugate base, A-. Determine the ratio, [HA]/[A-], in this buffer.

Answer 1

1(a) The buffer reaction can be written as:

HC₂H₂ClO₂ (aq) <==> C₂H₂ClO₂^- (aq)

When 0.014 moles of H₃O⁺ is added, it will react with C₂H₂ClO₂^- (base) to promote the reverse reaction. In this process, the concentration of C₂H₂ClO₂^- will decrease and the concentration of HC₂H₂ClO₂ (acid) will increase.

HC₂H₂ClO₂ (aq) ← C₂H₂ClO₂^- (aq) + H₃O⁺ (aq)

mol of HC₂H₂ClO₂ formed = 0.122 + 0.014 = 0.136 mol HC₂H₂ClO₂

mol of C₂H₂ClO₂^- remaining after neutralization = 0.108 - 0.014 = 0.094 mol C₂H₂ClO₂^-

1(b) Use the Hendersen-Haaselbach equation to solve for the pH.

pH = pK_a + log ([C₂H₂ClO₂^- ] / [HC₂H₂ClO₂])

K_a = 1.4 x 10^-3 ; pK_a = - log K_a = - log (1.4 x 10^-3) = 2.85

[C₂H₂ClO₂^- ] = 0.094 mol / 1.00 L = 0.094 M

[HC₂H₂ClO₂] = 0.136 mol / 1.00 L = 0.136 M

pH = 2.85 + log (0.094 / 0.136)

= 2.85 - 0.16 = 2.69

2(a) Buffer reaction is:

HClO (aq) <==> ClO^- (aq)

[HClO] is 3 times smaller than [ClO^-]. This equates to [HClO] = 1/3 [ClO^-] OR [ClO^-] = 3[HClO].

Use the Hendersen-Haaselbach equation again.

pH = pK_a + log ([ClO^- ] / [HClO])

K_a = 3.5 x 10^-8 ; pK_a = - log K_a = - log (3.5 x 10^-8) = 7.46

Plug in the pK_a value and [ClO^-] = 3[HClO] in the above equation, you get

pH = 7.46 + log (3[HClO] / [HClO]) = 7.46 + log (3/1) = 7.46 + 0.48 = 8.94

2(b) HA (aq) <==> A- (aq)

K_a = 2.2 x 10^-6 ; pK_a = - log K_a = - log (2.2 x 10^-6) = 5.66

pH = 5.50

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

5.50 = 5.66 + log ([A^- ] / [HA])

log ([A^- ] / [HA]) = 5.50 - 5.66

log ([A^- ] / [HA]) = -0.16

Take the antilog of both sides, you get

[A^- ] / [HA] = 10^-0.16 = 0.69

Ratio of [HA] / [A^- ] = 1 / 0.69 = 1.5

This means that the concentration of [HA] is 1.5 times larger than [A^-].