How many grams of dry NH4Cl need to be added to 2.50 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.67? Kb for ammonia is 1.8×10−5.
Use the Henderson-Hasselbalch equation:
pH = pKa + log([A^-]/[HA])
Here, [A^-] = [NH3] (base)
[HA] = [NH4^+] (acid)
Rather than use Kb, it is easier to use the pKa of NH4^+ (the
conjugate acid).
Kb = 1.8 x 10^-5
pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.74
pKa + pKb = 14
pKa = 14 - pKb = 14 - 4.74 = 9.26
pH = pKa + log([A^-]/[HA])
8.67 = 9.26 + log((0.400 M NH3)/(x M NH4^+))
-0.59 = log((0.400 M NH3)/(x M NH4^+))
(0.400 M NH3)/(x M NH4^+) = 10^-0.59 = 0.2570
x M NH4^+ = (0.400 M NH3 )/(0.2570)
x M NH4^+ = 1.5564 M NH4^+
So we need the NH4CL concentration to be 1.5564 M. We have 2.5 L of
solution, so the number of moles of NH4Cl we need are
(1.5564 M NH4Cl)*(2.5 L) = 3.89 moles of NH4Cl
The mass of this number of moles is then
(3.89 moles of NH4Cl)*(53.491 g/mol) = 208.13 g of NH4Cl
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