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

1 Answer
Apr 6, 2014

Answer:

You find an acid with a #"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 #"H"_3"PO"_4# with #"p"K_text(a3) = 12.35# and ascorbic acid, #"H"_2"C"_6"H"_6"O"_6# with #"p"K_text(a2) = 11.79#.

Let’s use #"H"_3"PO"_4#.

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

We will need to use #"Na"_2"HPO"_4"·12H"_2"O"# as the acid and #"Na"_3"PO"_4"·12H"_2"O"# as the conjugate base.

The chemical equation is

#"HPO"_4^"2-" + "H"_2"O" ⇌ "H"_3"O"^+ + "PO"_4^"3-"#; #"p"K_text(a3) = 12.35#

or

#"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-"#; #"p"K_text(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"#

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

#"[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 #"0.069 13 mol (23.24 g) Na"_2"HPO"_4"·12H"_2"O"# and #"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.