Question

A 45.0 mL buffer solution is 0.430 M Na2HPO4 (aq) and 0.330 M KH2PO4 (aq). The ka for H2PO4^- is 6.2*10^-8

a. what is the pH of the buffer solution

b. what are the pH and the pH change resulting from the addition of 9.0*10^-4 mol of NaOH to the buffer solution

c. what are the pH and the pH change resulting from the addition of 9.5*10^-4 mol of nitric acid to the initial buffer system

d. how many mL of 0.400 M sodium hydroxide would you have to add to the buffer in order to exhaust its buffering capacity

Answer 1

a)

mmoles of H2PO4- = 0.330 x 45 = 14.85

mmoles of HPO42- = 0.430 x 45 = 19.35

pKa = 7.21

pH = pKa + log [salt / acid]

    = 7.21 + log [19.35 / 14.85]

pH = 7.32

b)

mmoles of H2PO4- = 0.330 x 45 = 14.85

mmoles of HPO42- = 0.430 x 45 = 19.35

mmoles of NaOH = 0.9

pH = pKa + log [salt + C / acid - C]

     = 7.21 + log [19.35 + 0.9 / 14.85 - 0.9]

pH = 7.37

pH change = 7.37 - 7.32 = 0.05

pH change = 0.05

c)

mmoles of HNO3 = 0.95

pH = pKa + log [salt - C / acid + C]

     = 7.21 + log [19.35 - 0.95 / 14.85 + 0.95]

pH = 7.28

pH change = - 0.04

d)

mmoles of H2PO4- = 14.85

mmoles of H2PO4 = mmoles of NaOH . then buffer fails.

14.85 = 0.4 x V

V = 37.125

volume of NaOH = 37.12 mL