Fixer Bath Selection: Ammonia or Thiosulfate?
A film-processing studio mixes its own fixing bath from sodium thiosulfate. A fresh 1.00 L bath is 1.00 M in thiosulfate and must dissolve at least 0.050 mol of AgCl carried over from each roll of film before it is spent. The new technician proposes switching to 1.00 M ammonia solution, which is cheaper and available in bulk; the studio also has rock salt and scrap zinc sheet. Relevant data: Ksp(AgCl) = 1.8 × 10⁻¹⁰; Kf[Ag(NH₃)₂]⁺ = 1.7 × 10⁷; Kf[Ag(S₂O₃)₂]³⁻ = 2.9 × 10¹³; E°(Ag⁺/Ag) = +0.80 V; E°(Zn²⁺/Zn) = −0.76 V. The studio is in a rented building with poor ventilation, and the landlord has just raised the issue of silver discharge into the drain. Last month a drum of ammonia was left open and the whole floor smelled of it for days.
1. Calculate how much AgCl 1.00 M ammonia can dissolve per litre and decide whether the technician's proposal meets the 0.050 mol requirement.
Read the worked answer
Combine the two equilibria: AgCl(s) ⇌ Ag⁺ + Cl⁻ (Ksp) Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ (Kf) Overall: AgCl(s) + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ + Cl⁻ K = Ksp × Kf = (1.8 × 10⁻¹⁰)(1.7 × 10⁷) = 3.1 × 10⁻³.
Set up solubility s: [Ag(NH₃)₂]⁺ = s, [Cl⁻] = s, [NH₃] = 1.00 − 2s. s² / (1.00 − 2s)² = 3.1 × 10⁻³ → s / (1.00 − 2s) = 0.0557. Solving: s = 0.0557 − 0.111s → s = 0.0557/1.111 = 0.050 M.
Judgement: the figure lands exactly on the 0.050 mol per litre requirement - less than 1% margin. That is not a workable specification: the second ammonia molecule consumed per silver, evaporation losses (the open drum emptied itself into the room air), and chloride already present in a partly used bath all push the capacity below the requirement after a single roll. Volatility also means the bath composition drifts during a shift.
Verdict: reject ammonia. It is thermodynamically borderline and operationally unstable, and the ventilation problem in the brief is a genuine safety argument on its own.
2. Do the same for the thiosulfate bath and explain why the two ligands give such different capacities.
Read the worked answer
Same treatment: AgCl(s) + 2S₂O₃²⁻ ⇌ [Ag(S₂O₃)₂]³⁻ + Cl⁻ K = Ksp × Kf = (1.8 × 10⁻¹⁰)(2.9 × 10¹³) = 5.2 × 10³.
Applying s²/(1.00 − 2s)² = 5.2 × 10³ gives an absurd s ≈ 70 M, i.e. the equilibrium lies so far to the right that it is not the controlling factor. Here the limit is reagent supply, not thermodynamics: each Ag⁺ is bound by two thiosulfate ions, so 1.00 mol of ligand can carry at most 0.50 mol of silver.
Capacity ≈ 0.50 mol AgCl per litre, ten times the 0.050 mol requirement - a tenfold margin, and one that survives a partly spent bath, chloride build-up and modest dilution.
Why the difference: both ligands are chelating, but they are not equally good. Thiosulfate binds through a soft sulfur donor to a soft Ag⁺ centre, so the complexation constant is about six orders of magnitude larger than that of the nitrogen donor in ammonia; [Ag(S₂O₃)₂]³⁻ is also a stable bis-chelate. That change in stability constant (10⁷ → 10¹³) is multiplied directly into the dissolution equilibrium, converting a borderline reagent into an overwhelming one. Verdict: keep the thiosulfate bath.
3. The landlord objects to silver going down the drain. Explain, with potentials, how the scrap zinc sheet recovers the silver, and why this works even though the free Ag⁺ concentration in the bath is tiny.
Read the worked answer
The redox couple: Anode: Zn(s) → Zn²⁺ + 2e⁻, E° = −0.76 V (oxidation) Cathode: 2Ag⁺ + 2e⁻ → 2Ag(s), E° = +0.80 V (reduction) E°cell = 0.80 − (−0.76) = 1.56 V. ΔG° = −nFE° = −2 × 96500 × 1.56 ≈ −3.0 × 10⁵ J mol⁻¹ - strongly spontaneous, so silver plates out on the zinc as a grey sludge that is filtered, dried and sold, keeping the metal out of the drain.
The tiny-free-Ag⁺ question: the potential is set by the free Ag⁺ activity, which is small because most silver is locked up as [Ag(S₂O₃)₂]³⁻. That only shifts the actual cell potential, not the direction. As each trace of free Ag⁺ is reduced, the complex dissociates to replace it (Le Chatelier), so the bath is stripped essentially completely - the complex acts as a reservoir, not a brake. The Nernst equation then explains why a real cell reads lower than 1.56 V yet still stays positive.
Reject the alternatives: rock salt would merely re-precipitate AgCl, converting a dissolved discharge problem into a solid-waste problem with no recovery; ammonia on its own is, by the previous calculation, barely able to hold the silver at all.