Question

If you needed to perform a reaction in a controlled pH environment that was fairly basic, you might choose to use the ammonium and ammonia buffer system. How many grams of solid ammonium chloride would you have to add to 1.250 liters of 0.165M NH3 to obtain a buffered solution of pH = 8.85? (Kb of NH3 = 1.8 x 10-5)

Answer 1

Kb = 1.8 x10^-5

-log(Kb) = -log(1.8 x10^-5)

Pkb = 4.74

NH3 = 1.250L of 0.165M

number of moles of NH3 = 0.165M x 1.250L = 0.20625 moles = 0.2063 moles

number of moles of NH3 = 0.2063 moles

PH = 8.85

PH + POH = 14

POH = 14- PH

POH = 14 - 8.85

POH = 5.15

let be number of moles of NH4Cl = n moles

POH = PKb + log[salt]/[base]

5.15 = 4.74 + log( n/0.2063)

5.15 - 4.74 = log(n/0.2063)

0.41 = log(n/0.2063)

log(n/0.2063) = 0.41

n/0.2063 = 10^0.41

n/0.2063 = 2.57

n = 2.57 x0.2063 = 0.5302

number of moles of NH4Cl = 0.5302 moles

molar mass of NH4Cl = 53.49 g/mole

mass of one mole of NH4Cl = 53.49 grams

mass of 0.5302 moles of NH4Cl = 0.5302 x 53.49 = 28.36 grams

mass of NH4Cl required = 28.36 grams.