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.
