Question

The buffer capacity of a solution may be defined as the number of moles of H+ that will change the pH of 1.00 L of the buffer by 1.00 pH units. What is the buffer capacity of a solution which is 0.10 M in acetic acid (Ka = 1.8 × 10-5) and 0.30 M in sodium acetate, in units of mol (H+) per liter? Please show full work. (Answer key says 0.21 mol).

Answer 1

Given K_a, find pK_a as pK_a = -log (K_a) = -log (1.8*10^-5) = 4.74.

Find the pH of the buffer by using the Henderson-Hasslebach equation as

pH = pK_a + log [A^-]/[HA] (HA denotes acetic acid and A^- denotes acetate)

====> pH = 4.74 + log (0.30 M)/(0.10 M)

====> pH = 4.74 + log (3.00) = 4.74 + 0.477 = 5.217 ≈ 5.22.

The pH of the buffer changes by 1.0 unit due to the addition of H⁺. We know that the addition of H⁺ lowers the capacity of the pH of the buffer; hence, the new pH is (5.22 – 1.0) = 4.22.

Again, find the ratio of A^- and HA using the Henderson-Hasslebach equation as

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

====> 4.22 = 4.74 + log [A^-]/[HA]

====> log [A^-]/[HA] = -0.52

====> [A^-]/[HA] = antilog (-0.52) = 0.30199 ≈ 0.302

====> [A^-] = 0.302*[HA]

We had 1.0 L of 0.10 M acetic acid and 0.30 M sodium acetate; hence, moles HA present in the buffer initially = (1.0 L)*(0.10 M) = 0.10 mole; moles A^- present in the buffer initially = (1.0 L)*(0.30 M) = 0.30 mole.

H⁺ reacts with A^- to form to HA as below.

A^- + H⁺ ------> HA

As per the stoichiometric equation,

1 mole H⁺ = 1 mole A^- = 1 mole HA

Let x moles H⁺ be added to the buffer; therefore, x moles H⁺ will react with x moles A^- to form x moles HA. The number of moles of HA and A^- at equilibrium are (0.10 + x) mole and (0.30 – x) mole respectively.

Since the volume of the solution stays same, we have, after the addition of H⁺,

[HA] = (0.10 + x)mole/(1.0 L) = (0.10 + x) M

[A^-] = (0.30 – x)mole/(1.0 L) = (0.30 – x) M

As per the problem,

(0.30 – x) = 0.302*(0.10 + x)

====> 0.30 – x = 0.0302 + 0.302x

====> 0.2698 = 1.302x

====> x = (0.2698)/(1.302) = 0.2072 ≈ 0.21

The number of moles of H⁺ added is 0.21 mole (ans).