Amount of grams to make buffer solution?

Question

What is the amount of ammonium chloride that has to be dissolved in 141 mL of 0.205 M NH3 (Kb= 1.8*10-5) to make it a buffer with a pH of 9.01, in g?

Answer 1

We assume that there is no change in volume when we add the ammonium chloride. For simplicity we use the Henderson-Hasselbalch equation (and the assumptions that go with it)

pH= pKa+log[conj.base]/[acid]

here conj.base is NH3 and acid is NH4+
NH4Cl dissociates into NH4+ and Cl- 100%, thus [NH4Cl]= [NH4+]
but [NH4Cl]= mole/V= mass/(MW*V)

Also, since Ka=Kw/Ka, we have pKa=pKw-pKb =14-pKb

so

pH= 14-pKb + log [NH3]/[NH4+]=
= 14-pKb + log [NH3]/[NH4Cl] =
= 14-pKb + log [NH3]/ (x/MW*V) =>
pH= 14-pKb + log [NH3]*MW*V/x =>

9.01= 14-(-log(1.8*10^-5)) +log(0.205*53.5*0.141/x) =>
log 1.55/x = -0.25 =>
1.55/x= 10^-0.25 =>
x=1.55*10^0.25 = 2.756 g

check for numerical errors