Question

a) What volume (to the nearest 0.1 mL) of 6.60-M HCl must be added to 0.450 L of 0.200-M K₂HPO₄ to prepare a pH = 7.40 buffer?

b) How many grams (to the nearest 0.01 g) of NH₄Cl (M_m = 53.49 g/mol) must be added to 400. mL of 0.964-M solution of NH₃ in order to prepare a pH = 8.75 buffer? pKb = 4.75

Answer 1

a) pH = pka2 + log(K2HPO4/KH2PO4)

NO of mol of K2HPO4 = 0.45*0.2 = 0.09 mol

NO of mol of KH2PO4 formed = HCl = v*6.6 mol

pka2 = 7.2

pH = 7.4

7.4 = 7.2+log((0.09-(v*6.6))/(v*6.6))

v = 0.00527 L

volume of HCl must be taken = 5.27 ml

b)
    pH = 14-(pb+log(NH4Cl/NH3))

NO Of mol of NH3 = 0.4*0.964 = 0.3856 mol

no of mol of NH4Cl must add = x

8.75 = 14 - (4.75+log(x/0.3856))

x = 1.22

no of mol of NH4Cl must add = x = 1.22 mol

mass of NH4Cl must add = 1.22*53.49 = 65.26 g