You need to determine the chloride concentration in a strongly acidic industrial waste sample. Which argentometric method is most appropriate, and why?
AMohr method, because chromate indicator gives the clearest color change for chloride
BFajans method, because adsorption indicators are completely unaffected by solution pH
CVolhard method, because it uses a back-titration in acidic solution, where the Mohr method's chromate indicator would decompose or fail to precipitate correctly
DAll three methods are equally applicable; pH does not affect argentometric titrations
The Mohr method requires neutral pH (approximately 6.5–10). In acidic solution, chromate exists as dichromate (CrO₄²⁻ is protonated to give HCrO₄⁻/Cr₂O₇²⁻), which has a different Ksp relationship with Ag⁺ — making the endpoint unreliable. In basic conditions, Ag⁺ precipitates as Ag₂O before the endpoint. The Volhard method avoids this: it is performed in acidic solution (0.1–0.5 M HNO₃), where the ferric thiocyanate indicator is stable and the excess silver can be properly back-titrated. This pH constraint is one of the most important practical considerations in method selection.
Question 2 Multiple Choice
Two argentometric titrations are performed — one determining iodide (AgI, Ksp ≈ 10⁻¹⁶) and one determining chloride (AgCl, Ksp ≈ 10⁻¹⁰). At the equivalence point, what difference should be expected?
AThe AgCl titration will have a sharper equivalence point because chloride is more commonly determined by this method
BBoth titrations will have identical equivalence point sharpness because both reactions have 1:1 stoichiometry with Ag⁺
CThe AgI titration will have a sharper equivalence point because the smaller Ksp produces a steeper change in Ag⁺ concentration at the equivalence point
DThe AgI titration will have a broader equivalence point because iodide binds more tightly and slows the precipitation kinetics
Ksp directly determines endpoint sharpness. A smaller Ksp means the precipitation equilibrium lies further toward complete reaction — as the last analyte ions are consumed, the free Ag⁺ concentration spikes steeply over a tiny volume range. Plotting pAg versus volume of titrant, the inflection at the equivalence point is much steeper for AgI than for AgCl. This is why iodide determinations can achieve better precision than chloride determinations by argentometry, all else equal. The 1:1 stoichiometry is the same in both cases — the difference is entirely in the thermodynamics of precipitation.
Question 3 True / False
In the Mohr method, the chromate indicator precipitates as Ag₂CrO₄ only after virtually all the chloride has been consumed, because AgCl has a lower Ksp than Ag₂CrO₄.
TTrue
FFalse
Answer: True
This sequential precipitation is the entire mechanism of the Mohr endpoint. AgCl (Ksp ≈ 10⁻¹⁰) precipitates preferentially over Ag₂CrO₄ (Ksp ≈ 10⁻¹²·⁵ for the 2:1 stoichiometry, but with the concentration of chromate used in practice, the effective Ksp comparison favors AgCl first). Throughout the titration, Ag⁺ is consumed by the more soluble halide. Only after the halide is essentially exhausted does [Ag⁺] rise high enough to exceed the ion product for Ag₂CrO₄, producing the brick-red precipitate that signals the endpoint. The relative Ksp values are what guarantee correct sequencing.
Question 4 True / False
In the Volhard back-titration, after adding excess AgNO₃ to a chloride sample, you can immediately back-titrate the excess with thiocyanate without any additional treatment.
TTrue
FFalse
Answer: False
This is the classic Volhard method mistake. AgCl must be filtered off (or the precipitate must be coated with nitrobenzene) before the back-titration with thiocyanate. If AgCl is present when thiocyanate is added, the following reaction occurs: AgCl(s) + SCN⁻ → AgSCN(s) + Cl⁻. Since AgSCN (Ksp ≈ 10⁻¹²) is less soluble than AgCl (Ksp ≈ 10⁻¹⁰), the thiocyanate converts AgCl into AgSCN, consuming additional titrant beyond the excess Ag⁺. This produces a falsely high result for the excess silver and therefore a falsely low result for the original chloride concentration.
Question 5 Short Answer
Explain how the Ksp of a silver salt determines the sharpness of the equivalence point in a precipitation titration, and what practical consequence this has for analytical precision.
Think about your answer, then reveal below.
Model answer: Ksp sets how completely the precipitation reaction proceeds at the equivalence point. A smaller Ksp means the reaction goes more completely to completion as the analyte is consumed — at the equivalence point, the small remaining analyte concentration forces a steep, large change in the silver ion concentration over a tiny added volume. This steep inflection in the pAg titration curve makes the endpoint sharp and easy to detect reliably. A larger Ksp means the transition is gradual, spread over a larger volume range, making precise endpoint detection more difficult. Consequently, analytes forming less soluble silver salts (e.g., iodide vs. chloride) yield inherently more precise argentometric determinations.
This question connects the equilibrium chemistry (Ksp) to the practical analytical outcome (endpoint sharpness and precision). The same principle applies throughout titrimetric analysis: the completeness of the reaction at the equivalence point determines how sharp the endpoint is, which determines how precisely the analyst can identify it. Understanding Ksp not just as a solubility concept but as a predictor of analytical performance is the key insight that makes precipitation titrations a coherent topic rather than a collection of disconnected methods.