Question

Calculate the volume of 0.25mol/L sodium dihydrogen phosphate that must be added to 150mL of 0.50mol/L sodium hydrogen phosphate to prepare a buffer solution of pH 7.40

Answer 1

sodium dihydrogen phosphate = NaH2PO4

sodium hydrogen phosphate = Na2HPO4

We know that pKa of NaH2PO4 = 7.2

pH = pKa + log [Na2HPO4] / [NaH2PO4]

[Na2HPO4] = molarity x volume = 0.50 mol/L x 0.15 L = 0.075 mol

[NaH2PO4] = molarity x volume = 0.25 mol/L x V L = 0.25 V mol

Given that pH = 7.40

pH = pKa + log [Na2HPO4] / [NaH2PO4]

7.40 = 7.2 + log (0.075)/( 0.25 V)

(0.075)/( 0.25 V) = 1.58

V = 0.189 L

V = 189 mL

Therefore,

189 mL sodium dihydrogen phosphate must be added to 150mL of 0.50mol/L sodium hydrogen phosphate to prepare a buffer solution of pH 7.40 .