Question

How many grams of dry NH4Cl need to be added to 2.10 L of a 0.700 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.57? Kb for ammonia is 1.8×10−5.

Answer 1

first convert the Kb in to pKa

Kb = 1.8×10^−5
pKb = -log(Kb) = -log(1.8×10^−5) = 4.74

pKa + pKb = 14
pKa = 14 - pKb = 14 - 4.74 = 9.26

now use the handerson equation

pH = pKa + log([A^-]/[HA])

here A^- = NH3

HA = NH4Cl

moles of NH3 = 0.7 M x 2.10 L = 1.47 moles

lets NH2Cl concentration = x

put all these values in the above eqation

8.57 = 9.26 + log(1.47/x)

log(1.47/x) = 8.57-9.26 = -0.69

1.47/x = 10^-0.69

1.47/x = 0.2042

x = 1.47 / 0.2042

x =7.2 moles

mass of NH4Cl = molar mass x moles

= 53.5 g /mol x 7.2 moles

= 385.2 grams