What mass of NH4Cl must be added to 0.750 L of a 0.1M solution of NH3, to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.) Kb of NH3 = 1.8 x10-5 %3D Kw= 1 x 10-14

Answer 1

Kb for NH3 = 1.8E-5 from the problem. You need Ka. You know KaKb = Kw; therefore, Ka = Kw/Kb = 1E-14/1.8E-5 = 5.55E-10 and pKa = 9.26 . Now use the Henderson-Hasselbalch equation which is pH = pKa + log [(base)/(acid)].
9.26 = 9.26 + log B/A where B is base and A is acid.
0 = log (0.1/A)
1 = (0.1/A) or A = 0.1 M
So (NH4Cl) = 0.1 M
mols NH4Cl = M x L = 0.1 x 0.750 L = 0.075
Then grams NH4Cl = mols x molar mass = ?

Answer 2

What mass of NH4Cl must be added to 0.750 L of a 0.1M solution of NH3, to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.) Kb of NH3 = 1.8 x10-5 %3D Kw= 1 x 10-14