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.
