Question

How many grams of dry NH4Cl need to be added to 2.00 L of a 0.300 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.64? Kb for ammonia is 1.8×10−5.

Answer 1

NH₃(aq) + H₂O ⇋ [NH₄]⁺ + OH^-
Kb = 1.8×10^-5
Kb = [NH₄⁺][OH^-]/[NH₃] = 1.8×10^-5
-log both sides
pKb = pOH + log[NH₃]/[NH₄⁺]
pKb = -log(1.8×10−5) = 4.74 pOH = 14.0 - 8.64 = 5.36
4.74 = 5.36 + log[NH₃]/[NH₄⁺]
log[NH₃]/[NH₄⁺] = 4.74 - 5.36 = -0.62
[NH₃]/[NH₄⁺] = 0.24
[NH₄⁺] = 0.300/0.24 = 1.25M
Now let me check doing it my way:
Kb = [NH₄⁺][OH^-]/[NH₃] = 1.8×10^-5
pH = 8.64 pOH = 14 - 8.64 = 5.36
[OH^-] = 4.365×10^-6
1.8×10^-5 = [NH₄⁺][4.365×10^-6]/[0.300]
[NH₄⁺] = [0.300]×(1.8×10^-5)/[4.365×10^-6] = 1.2371 M (so we are agreed!)
In 2.00 L we have 2.0×1.2371 = 2.5462 mol
X MWt NH₄Cl = 2.5462 × 53.4917 = 136.2005 g NH₄Cl

136.2005 grams of NH₄Cl needed