How do you buffer a solution with a pH of 12?

1 Answer
Apr 6, 2014

You find an acid with a pKa close to 12. Then you make a solution of the acid and its conjugate base in the correct proportions.

Explanation:

The only common acids with pKa12 are H3PO4 with pKa3=12.35 and ascorbic acid, H2C6H6O6 with pKa2=11.79.

Let’s use H3PO4.

How would you prepare 1.000 L of 0.1000 mol/L phosphate buffer?

We will need to use Na2HPO4⋅12H2O as the acid and Na3PO4⋅12H2O as the conjugate base.

The chemical equation is

HPO2-4+H2OH3O++PO3-4; pKa3=12.35

or

HA+H2OH3O++A-; pKa=12.35

The Henderson-Hasselbalch equation is

pH=pKa+log([A-][HA])

12.00=12.35+log([A-][HA])

log([A-][HA])=12.00 – 12.35=-0.35

[A-][HA]=10-0.35=0.447

[A-]=0.447[HA]

Also,

[A-]+[HA]=0.1000 mol/L

0.447[HA]+[HA]=1.447[HA]=0.1000 mol/L

[HA]=0.1000 mol/L1.447=0.069 13 mol/L

[A-]=0.447[HA]=0.447 × 0.069 12 mol/L=0.0309 mol/L

So, you wash 0.069 13 mol (23.24 g) Na2HPO4⋅12H2O and 0.0309 mol (11.74 g) Na3PO4⋅12H2O into a 1 L volumetric flask.

Then you add enough water to make 1.000 L of solution.