The only common acids with pKa≈12 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+H2O⇌H3O++PO3-4; pKa3=12.35
or
HA+H2O⇌H3O++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.
