Question

How many grams of dry NH4Cl need to be added to 2.50 L of a 0.700 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.56? Kb for ammonia is 1.8×10−5.

Express your answer numerically in grams to three significant figures.

Answer 1

Dissociation equation is given below

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 - 8.56 = 5.44
4.74 = 5.44 + log[NH3]/[NH₄⁺]
log[NH3]/[NH₄]⁺ = 4.74 - 5.44 = -0.7
[NH3]/[NH₄]⁺ = 0.1995
[NH₄]⁺ = 0.700/0.1995 = 3.51M
Let us calculate
Kb = [NH₄]⁺[OH^-]/[NH3] = 1.8×10^-5
pH = 8.56 pOH = 14 - 8.56 = 5.44
[OH^-] = 3.63×10^-6
1.8×10^-5 = [NH₄]⁺[3.63×10^-6]/[0.700]
[NH₄]⁺ = [0.700]×(1.8×10^-5)/[3.63×10^-6] = 3.47 M (so we are agreed!)
In 2.50 L we have 2.5×3.47 =8.67 mol
Mass of NH4Cl needed = 5.71×53.4917 = 464.08 g NH4Cl