Question

How many grams of dry NH4Cl need to be added to 1.60 L of a 0.100 M solution of ammonia, NH3 , to prepare a buffer solution that has a pH of 8.55? Kb for ammonia is 1.8×10−5 . May you please show step by step how the answer was achieved?

Answer 1

NH4Cl ----------> NH4(+) + Cl(-)
NH3 + H2O <----------> NH4(+) + OH(-)
The Kb is small so the it is possible to say that all the NH4(+) is released from NH4Cl and the concentration of NH3 will not change .

Conce. of NH3 = 0.1 M

pOH = 14 - 8.55 = 6.45
Cconcentration of OH(-) = 10 ^{- pOH} = 3.54 * 10^-7

1.8 * 10^-5 = [ Conce.(OH) * Conce.(NH4) ] / Conce.(NH3)

1.8 * 10^-5 * 0.1 M= [ 3.54 * 10^-7 * conce (NH4) ]

conce NH4 = 5.04 M

grams of NH4Cl = [ 1.6 liters ] * [ Conce.(NH4) ] * [ molecular waigh of NH4Cl ]

grams of NH4Cl = 1.6 * 5.04 M * 53.5 g/mol = 431.42 g