Question

How many grams of solid Na2CO3 and solid NaHCO3 are needed to prepare 600 mL of 250 mM bicarbonate buffer at pH = 4.5? Carbonic acid has two pKa values at 3.77 and 10.20

Answer 1

To prepare 600 ml (0.600 L) of 250 mM (0.250 M) bicarbonate buffer with pH 4.5

using Na2CO3 (base) and NaHCO3 (acid)

pKa = 10.20

Using Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

or,

pH = pKa + log(Na2CO3/NaHCO3)

4.5 = 10.20 + log(Na2CO3/NaHCO3)

(Na2CO3) = 2 x 10^-6(NaHCO3)

we have,

(NaHCO3) + (Na2CO3) = 0.250 M x 0.600 L = 0.150 mol

substituting from above,

(NaHCO3) + 2 x 10^-6(/NaHCO3) = 0.150 mol

moles (NaHCO3) needed = 0.150/1.000002 = 0.1499997 mol

mass NaHCO3 needed = 0.1499997 mol x 84.007 g/mol = 12.6 g

moles (Na2CO3) needed = 0.15 - 0.1499997 = 3 x 10^-7 mol

mass Na2CO3 needed = 3 x 10^-7 mol x 106 g/mol = 3.18 x 10^-5 g

[formula, moles = mass/molar mass]