Solubility of Silver Chloride in Ammonia
Solid AgCl is shaken with 0.100 M aqueous NH3 until equilibrium is reached at 25 °C. For AgCl, Ksp = 1.8×10⁻¹⁰. For the complex ion Ag(NH3)2⁺, the overall formation constant is β2 = 1.7×10⁷. The base ionization of NH3 has Kb = 1.8×10⁻⁵; you may neglect the small amount of NH3 consumed by protonation when estimating the free NH3 concentration. Assume volumes are unchanged and that no other silver complexes form. In a separate experiment, the same mass of AgCl is shaken with pure water under otherwise identical conditions. The solubility of AgCl in pure water is often quoted as about 1.3×10⁻⁵ M. The temperature is held constant, and the solid, if present, has unit activity.
1. Estimate the solubility of AgCl in 0.100 M NH3 and compare it with the solubility in pure water. State clearly which approximations you use.
Read the worked answer
Let s be the molar solubility of AgCl. Then [Cl⁻] = s and total dissolved silver is s. Let [Ag⁺] be the free ion. Ksp = [Ag⁺][Cl⁻] gives [Ag⁺] = Ksp/s.
The complex formation is Ag⁺ + 2NH3 ⇌ Ag(NH3)2⁺, with β2 = [complex]/([Ag⁺][NH3]²). Thus [complex] = β2(Ksp/s)[NH3]².
Total silver: s = [Ag⁺] + [complex] = Ksp/s + β2Ksp[NH3]²/s.
Multiply by s: s² = Ksp(1 + β2[NH3]²).
With [NH3] ≈ 0.100 M, β2[NH3]² = 1.7×10⁷×1.0×10⁻² = 1.7×10⁵, which dominates 1. Hence s² ≈ (1.8×10⁻¹⁰)(1.7×10⁵) = 3.06×10⁻⁵, so s ≈ 5.5×10⁻³ M.
In pure water s = √Ksp = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M. The ratio is about 5.5×10⁻³/1.34×10⁻⁵ ≈ 410. Ammonia dramatically increases solubility by tying up Ag⁺ as a complex.
2. Check whether the assumption that [NH3] remains 0.100 M is quantitatively acceptable. If it is not, indicate whether the true solubility is larger or smaller than your estimate and why.
Read the worked answer
The estimate used [NH3] = 0.100 M. Forming Ag(NH3)2⁺ consumes about 2s ≈ 0.011 M of NH3, because each dissolved silver binds two ammonia molecules. That is about 11% of the initial 0.100 M, so the approximation is only fair.
Using [NH3] = 0.100 − 2s and neglecting free Ag⁺ in the total-silver balance gives s² ≈ Kspβ2(0.100 − 2s)², or s ≈ 5.53×10⁻³(1 − 20s).
Solving s = 5.53×10⁻³ − 0.1106s gives 1.1106s = 5.53×10⁻³, so s ≈ 4.98×10⁻³ M.
Thus the true solubility is about 5.0×10⁻³ M, roughly 10% lower than the simple estimate. The direction is clear: depleting free NH3 reduces the complexation term, so the estimate is an overestimate.