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

Apr 6, 2014

You find an acid with a $\text{p"K_"a}$ 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 "p"K_a" ≈ 12 are ${\text{H"_3"PO}}_{4}$ with $\text{p} {K}_{\textrm{a 3}} = 12.35$ and ascorbic acid, ${\text{H"_2"C"_6"H"_6"O}}_{6}$ with $\text{p} {K}_{\textrm{a 2}} = 11.79$.

Let’s use ${\text{H"_3"PO}}_{4}$.

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

We will need to use $\text{Na"_2"HPO"_4"·12H"_2"O}$ as the acid and $\text{Na"_3"PO"_4"·12H"_2"O}$ as the conjugate base.

The chemical equation is

$\text{HPO"_4^"2-" + "H"_2"O" ⇌ "H"_3"O"^+ + "PO"_4^"3-}$; $\text{p} {K}_{\textrm{a 3}} = 12.35$

or

$\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-}$; $\text{p} {K}_{\textrm{a}} = 12.35$

The Henderson-Hasselbalch equation is

"pH" = "p"K_"a" + 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"

$\text{0.447[HA]" + "[HA]" = "1.447[HA]" = "0.1000 mol/L}$

$\text{[HA]" = ("0.1000 mol/L")/1.447 = "0.069 13 mol/L}$

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

So, you wash $\text{0.069 13 mol (23.24 g) Na"_2"HPO"_4"·12H"_2"O}$ and $\text{0.0309 mol (11.74 g) Na"_3"PO"_4"·12H"_2"O}$ into a 1 L volumetric flask.

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