Question

Describe, stepwise, how you would prepare 0.5 L of a 0.15 M phosphate buffer, pH 12.5, using K3PO4·H2O (f.w. = 230.28) and K2HPO4 (f.w. = 174.18). Assume that after dissolving the salts your pH-meter reading is 11.9.

Answer 1

pKa = 12.35

We have,

pH = pKa + log([PO4^3-]/[HPO4^2-])

12.5 = 12.35 + log([PO4^3-]/[HPO4^2-])

[PO4^3-] = 1.41[HPO4^2-]

and,

[HPO4^2-] + [PO4^3-] = 0.15 M x 0.5 L = 0.075 mol

[HPO4^2-] + 1.41[HPO4^2-] = 0.075

[HPO4^2-] = 0.031 mol

[PO4^3-] = 0.075 - 0.031 = 0.044 mol

grams of K3PO4.H2O required = 0.044 mol x 230.28 g/mol = 10.13 g

grams of K2HPO4 required = 0.031 mol x 174.18 g/mol = 5.40 g

to prepare a buffer of pH 12.5 from buffer of pH 11.9

pH = 12.5 - 11.9 = 0.6

pH from lower to higher would need addition of base to buffer

pOH = 14 - 0.6 = 13.4

Amount of base to be added [OH-] = 3.98 x 10^-14 M