What is the pH of a buffered system made by dissolving 17.42 g of KH2PO4 and 20.41 g of K2HPO4 in water...?

Question

What is the pH of a buffered system made by dissolving 17.42 g of KH2PO4 and 20.41 g of K2HPO4 in water to give a volume of 200.0 mL? The Ka2 for dihydrogen phosphate is 6.2x10^-8 and the equilibrium reaction of interest is

K2PO4- + H2O <--> H3O+ + HPO4-

Answer Choices

a. 7.38

b. 7.58

c. 7.03

d. 7.17

Answer 1

H2PO41-  ----&gt;  H+ + HPO42-

Ka2 = [H+][HPO42-] / [H2PO41-]

The molar mass of KH2PO4 is 136 g/mol.

17.42 g KH2PO4 x (1 mol / 136 g) = 0.128 mol

M = mol / L = 0.128 mol / 0.200 L = 0.640 M

The molar mass of K2HPO4 is 174 g/mol.

20.41 g K2HPO4 x (1 mol / 174 g) = 0.117 mol

M = mol / L = 0.117 mol / 0.200 L = 0.586 M

Ka2 = 6.2 x 10^-8

pKa2 = - log (6.2 x 10^-8) = 7.21

Henderson-Hasselbalch equation:

pH = pKa + log (base/acid) , where the acid and base make up a conjugate pair.

pH = 7.21 + log (0.586 / 0.640) = 7.21 - 0.04 = 7.17.

d. 7.17 is the correct answer.