Question

1. How many grams (to the nearest 0.01 g) of NH₄Cl (M_m = 53.49 g/mol) must be added to 650. mL of 0.951-M solution of NH₃ in order to prepare a pH = 10.50 buffer?

2. What volume (to the nearest 0.1 mL) of 4.40-M HCl must be added to 0.450 L of 0.150-M K₂HPO₄ to prepare a pH = 6.60 buffer?

Answer 1

1. pH = pKa + log(base/acid)

let x amount of NH4Cl is added

10.50 = 9.25 + log((0.951 x 650 - x)/x)

17.783x = 618.15 - x

x = 32.91 mmol

grams of NH4Cl to be added = 32.91 mmol x 53.49/1000 = 1.76 g

2. moles of each species in buffer

6.60 = 7.20 + log([K2HPO4]/[KH2PO4])

0.251[KH2PO4] = [K2HPO4]

[KH2PO4] + [K2HPO4] = 0.150 x 0.450 = 0.0675 mols

[KH2PO4] + 0.251[KH2PO4] = 0.0675

[KH2PO4] = 0.054 mols

[K2HPO4] = 0.0675 - 0.054 = 0.0135 mols

So we have to add HCl = 0.054 mols/4.40 M = 0.012 L = 12.3 ml