Question

Consider how to prepare a buffer solution with pH = 7.33 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.418-M solution of weak acid with 0.359 M sodium hydroxide.

Weak Acid	Conjugate Base	Ka	pKa
HNO2	NO2-	4.5 x 10-4	3.35
HClO	ClO-	3.5 x 10-8	7.46
HCN	CN-	4.0 x 10-10	9.40

How many L of the sodium hydroxide solution would have to be added to the acid solution of your choice?

L

Answer 1

First, the best buffer will be the one with the pKa value closer to the pH desired. In this case for a pH of 7.33, the pKa of 7.46 would be the best and the ideal. So the system to use will be HClO and ClO-.

We need to know the concentration of ClO^- using an ice chart:

r: HClO --------> ClO^- + H⁺   Ka = 10^-7.46 = 3.47x10^-8
i: 0.418 0 0
e: 0.418-x x x

3.48x10^-8 = x² / 0.418-x
3.48x10^-8*0.418 = x²
x = [ClO^-] = 1.2x10^-4 M

As this is in 1 L solution the concentration is the same of the number of moles. When we add the base, we know that this will increase the ClO^- concentration and decrease the HClO. so using the HH, we can calculate the volume needed:

[ClO^-] = 1.2x10^-4 moles / Vt
[OH] = 0.358 * Vb / Vt
[HClO] = 0.418 / Vt

Replacing this in the HH equation we have:

pH = pKa + log [(1.2x10^-4+(0.358*Vb) / Vt) / (0.418-(0.358*Vb) / Vt)]
pH = pKa + log [1.2x10^-4+(0.358*Vb) / 0.418-(0.358*Vb)]

From here we solve for Vb:
7.33 - 7.46 = log [1.2x10^-4+(0.358*Vb) / 0.418-(0.358*Vb)]
10^-0.13 =  [1.2x10^-4+(0.358*Vb) / 0.418-(0.358*Vb)]
0.74 =  [1.2x10^-4+(0.358*Vb) / 0.418-(0.358*Vb)]
0.74(0.418 - 0.358Vb) = 1.2x10^-4+0.358Vb
0.3093 - 0.2649Vb = 1.2x10^-4 + 0.358Vb
0.3093 - 1.2x10^-4 = (0.358+0.2649)Vb
Vb = 0.496 L

Hope this helps