[13%] A chemist prepares an ammonia/ammonium chloride buffer from solid ammonium chloride, 0.80 M NH — Acid-Base Equilibrium and Solubility Chemistry Question
Problem Context
[13%] A chemist prepares an ammonia/ammonium chloride buffer from solid ammonium chloride, 0.80 M NH3, and distilled water.
What is the pH of the 0.80 M NH3 solution? (The Ka of NH4+ is 5.6 × 10-10.)
Model Answer
The major reaction is: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Keq = (Kw/Ka of NH4+)
Keq = (1 × 10-14)/(5.6 × 10-10)
Keq = 1.8 × 10-5
Since this is the major reaction, [NH4+] ≈ [OH–], and since it is not very favorable, [NH3] ≈ 0.80 M.
[NH4+][OH−] / [NH3] = [OH−]2 / [0.80] = 1.8 × 10-5
[OH–]2 = 1.44 × 10-5
[OH–] = 3.8 × 10-3 M
pH = 14 + log[OH–] = 11.60
What mole ratio of ammonia to ammonium chloride is required for the buffer to have pH = 9.20?
Model Answer
pH = pKa of NH4+ + log([NH3]/[NH4+])
9.20 = 9.25 + log([NH3]/[NH4+])
log([NH3]/[NH4+]) = –0.05
[NH3]/[NH4+] = 0.89
Calculate the volume of ammonia solution and mass of ammonium chloride needed to make up 500 mL of the pH = 9.20 buffer with a total concentration of nitrogen-containing species (NH3 or NH4+) of 0.50 mol L-1.
Model Answer
There are two constraints on the number of moles of NH3 and of NH4+. From the fact that the total concentration of nitrogen species is 0.500 M in 0.500 L,
(mol NH3) + (mol NH4+) = 0.250 mol
From the calculation in (b), (mol NH3) = 0.89(mol NH4+)
Solving the two simultaneous equations gives:
0.89(mol NH4+) + (mol NH4+) = 0.250 mol
(mol NH4+) = 0.132 mol
0.132 mol NH4+ × (53.49 g NH4Cl/mol) = 7.06 g NH4Cl
(mol NH3) = 0.89(mol NH4+) = 0.89(0.132 mol) = 0.117 mol NH3
0.117 mol NH3/(0.80 M NH3) = 0.147 L of the 0.80 M NH3 solution
To the solution described in (c) is added 10.0 mL of 0.10 M AgNO3.
i. Some AgCl precipitates from this mixture. Justify that this is the case. The Ksp of AgCl is 1.8 × 10-10 and the Kf of Ag(NH3)2+ is 1.6 × 107.
Model Answer
mol Ag+ = (0.0100 L)×(0.10 mol/L) = 0.0010 mol Ag+. Since the amounts of both chloride and ammonia are over 100 times this amount, we can neglect the changes of these latter two species as AgCl(s) and Ag(NH3)2+ form. So [Cl–] = 0.132 mol/(0.510 L) = 0.259 M and [NH3] = 0.117 mol/(0.510 L) = 0.229 M.
There are a number of ways to approach the analysis of whether a precipitate forms. Let us assume that no precipitate forms, so all of the Ag+ (initial concentration = 0.0010 mol/0.51 L = 1.96 × 10-3 M) remains in solution. Of course, much of it will be in the form of the complex ion:
[Ag(NH3)2+] / ([Ag+][NH3]2) = [Ag(NH3)2+] / ([Ag+][0.229]2) = Kf = 1.6 × 107
[Ag(NH3)2+] / [Ag+] = 8.4 × 105
This means that almost all the silver ion in solution is in the form of the complex ion, so [Ag+] = [Ag(NH3)2+]/(8.4 × 105) = (1.96 × 10-3)/(8.4 × 105) = 2.3 × 10-9 M.
Qsp = [Ag+][Cl–] = [2.3 × 10-9][0.259] = 6.0 × 10-10 > Ksp. This means that there will be a precipitate of AgCl.
ii. What is the final concentration of free Ag+ in solution after this mixture achieves equilibrium?
Model Answer
Since there is a precipitate, [Ag+][Cl–] = Ksp
[Ag+][0.259] = 1.8 × 10-10
[Ag+] = 6.95 × 10-10 M
iii. What is the mass of AgCl that precipitates from solution?
Model Answer
From the complexation equilibrium,
[Ag(NH3)2+] / ([Ag+][NH3]2) = [Ag(NH3)2+] / ([6.95×10−10][0.229]2) = Kf = 1.6 × 107
[Ag(NH3)2+] = 5.8 × 10-4 M
This accounts for (5.8 × 10-4 mol/L)×(0.510 L) = 3.0 × 10-4 mol of silver. The amount of free Ag+ is negligible, so the amount of silver that must have precipitated as AgCl is (1.00 × 10-3 mol total Ag) – (3.0 × 10-4 mol complexed Ag) = 7.0 × 10-4 mol AgCl.
(7.0 × 10-4 mol AgCl)×(143.4 g mol-1) = 0.100 g AgCl