Fraction Remaining During Selective Precipitation

Introduction

Knowing which ion precipitates first is only half the story of a separation. The real test is how completely the first ion has been removed by the time the second one starts to come out. If a lot of the first ion is still dissolved at that moment, the two cannot be cleanly separated by simply stopping the addition. This page shows how to calculate that residue and how to judge whether it is small enough.

Core explanation

The key moment. As the shared counter-ion concentration rises, the first salt precipitates and its ion is steadily removed, always keeping Q = Ksp for that salt. The second salt begins to form when the counter-ion reaches the second threshold. At that instant, the counter-ion concentration is known exactly, so the residual concentration of the first ion follows from the first salt's Ksp.

The calculation.

1. Find the second threshold: [X]₂ = the counter-ion concentration at which salt 2 reaches Q = Ksp₂, using the second ion's initial concentration (it has not yet been consumed). 2. Insert [X]₂ into the first salt's Ksp expression and solve for the first ion. 3. Divide by the first ion's initial concentration to get the fraction remaining.

Worked illustration: halides with silver. A solution is 0.010 mol dm⁻³ in both I⁻ and Cl⁻, and Ag⁺ is added slowly. Using Ksp(AgI) = 8.3 × 10⁻¹⁷ and Ksp(AgCl) = 1.8 × 10⁻¹⁰:

- AgI starts at [Ag⁺] = 8.3 × 10⁻¹⁵ mol dm⁻³. - AgCl starts at [Ag⁺] = 1.8 × 10⁻⁸ mol dm⁻³. - At that point [I⁻] = 8.3 × 10⁻¹⁷ ÷ 1.8 × 10⁻⁸ = 4.6 × 10⁻⁹ mol dm⁻³. - Fraction remaining = 4.6 × 10⁻⁹ ÷ 0.010 = 4.6 × 10⁻⁷, or about 0.00005%.

The iodide is essentially completely removed before chloride begins: an excellent separation.

A less favourable pair. Repeat for Br⁻ and Cl⁻ (Ksp(AgBr) = 5.0 × 10⁻¹³). At the AgCl onset, [Br⁻] = 5.0 × 10⁻¹³ ÷ 1.8 × 10⁻⁸ = 2.8 × 10⁻⁵ mol dm⁻³, a fraction of 2.8 × 10⁻³, or 0.28%. This fails the common 0.1% criterion, so bromide and chloride cannot be separated cleanly this way.

The ratio rule. For two 1:1 salts sharing a counter-ion, dividing the two Ksp expressions at the moment both are saturated gives [ion 1] ÷ [ion 2] = Ksp₁ ÷ Ksp₂. With equal starting concentrations, the fraction remaining of ion 1 equals the Ksp ratio directly. For salts of different types, no such shortcut exists and the full two-step calculation is required.

Dilution. If a large volume of reagent is added, all concentrations fall. For careful work, use the volume at the moment the second salt begins, or work in moles and convert. In most textbook problems the reagent is concentrated and dilution is neglected, which should be stated as an assumption.

Formulae

At the onset of salt 2: [X]₂ = (Ksp₂ ÷ [ion 2]ᵃ)^(1/b). Residual ion 1 from Ksp₁ with [X]₂. Fraction remaining = [ion 1]residual ÷ [ion 1]initial. For two MX salts: [ion 1] ÷ [ion 2] = Ksp₁ ÷ Ksp₂.

Step-by-step reasoning

1. Decide which salt precipitates first by comparing thresholds. 2. Calculate the threshold counter-ion concentration for the second salt. 3. Substitute it into the first salt's Ksp to find the residual first ion. 4. Express the residue as a fraction or percentage of the original. 5. Compare with the chosen criterion, such as 0.1%, and give a verdict.

Visual explanation

Plot log[ion 1] against log[counter-ion] as reagent is added. The line stays flat until the first threshold, then slopes steeply downwards along the Ksp line. Mark the second threshold on the horizontal axis; read the height of the sloping line there. That height is the residual log concentration — the lower it is, the cleaner the separation.

Real-world analogy

Two groups leave a cinema through a single door. The first group files out while the second waits for its turn. The quality of the separation is judged by how many of the first group are still inside when the second group starts moving. A handful is acceptable; a crowd means the groups mix at the door.

Real-world example

Potentiometric titrations of a halide mixture with silver nitrate show two end-points, one for iodide and one for chloride. The iodide end-point is sharp because almost no iodide remains when chloride starts to react. For bromide–chloride mixtures the first end-point is distorted and slightly early, exactly as the 0.28% residue predicts, and co-precipitation of mixed crystals adds further error.

Why?

Why is the residue calculated using the second threshold rather than the end of the addition? Stopping the addition at the second threshold is the best one can do without contaminating the first solid. The residue at that point is therefore the minimum achievable loss of the first ion into the filtrate.

Common misconception

"Once the first salt has started to precipitate, all of its ion is removed." Precipitation lowers the dissolved ion only until Q = Ksp at the current counter-ion level. A measurable amount remains until the counter-ion concentration is raised much further.

Worked example

Question: A solution contains 0.010 mol dm⁻³ Ba²⁺ and 0.010 mol dm⁻³ Sr²⁺. Sulfate is added slowly. What percentage of barium remains dissolved when SrSO₄ starts to form? Ksp(BaSO₄) = 1.1 × 10⁻¹⁰; Ksp(SrSO₄) = 3.4 × 10⁻⁷.

Reasoning: SrSO₄ threshold: [SO₄²⁻] = 3.4 × 10⁻⁷ ÷ 0.010 = 3.4 × 10⁻⁵ mol dm⁻³. Residual [Ba²⁺] = 1.1 × 10⁻¹⁰ ÷ 3.4 × 10⁻⁵ = 3.2 × 10⁻⁶ mol dm⁻³. Fraction = 3.2 × 10⁻⁶ ÷ 0.010 = 3.2 × 10⁻⁴.

Answer: About 0.032% of the barium remains, so the separation meets the 0.1% criterion.

Quick check

1. Using the ratio rule, what fraction of ion A remains when BX starts, if AX and BX are 1:1 salts with Ksp 1 × 10⁻¹⁵ and 1 × 10⁻¹⁰ and equal starting concentrations? Answer: Fraction = 1 × 10⁻¹⁵ ÷ 1 × 10⁻¹⁰ = 1 × 10⁻⁵, or 0.001%.

Exam focus

Examiners usually ask for "the concentration of X remaining when Y just begins to precipitate". Use the second salt's threshold, then the first salt's Ksp, and finish with a percentage and a clear judgement. State any assumptions about dilution and about the second ion's concentration being unchanged.

Advanced insight

The fraction remaining is an ideal lower limit. Activity corrections at high ionic strength, complexation of the first ion by excess counter-ion (for example AgCl₂⁻ forming at high chloride) and solid-solution formation between isomorphous salts such as AgBr and AgCl all make real separations worse than the simple calculation predicts.

Summary

To judge a selective precipitation, find the counter-ion concentration at which the second salt begins, then use the first salt's Ksp to calculate how much of the first ion remains. Express this as a fraction of the starting amount and compare with a criterion such as 0.1%. For two 1:1 salts at equal concentration, the fraction remaining equals the ratio of Ksp values.

Practice questions

1. Show that bromide cannot be quantitatively separated from chloride (both 0.010 mol dm⁻³) by adding silver ions, given Ksp(AgBr) = 5.0 × 10⁻¹³ and Ksp(AgCl) = 1.8 × 10⁻¹⁰. Answer: At AgCl onset [Ag⁺] = 1.8 × 10⁻⁸, so [Br⁻] = 2.8 × 10⁻⁵ mol dm⁻³, which is 0.28% remaining, above the 0.1% limit. 2. A solution is 0.10 mol dm⁻³ in Cl⁻ and 0.0010 mol dm⁻³ in I⁻. Calculate the fraction of iodide remaining when AgCl begins. Answer: AgCl onset [Ag⁺] = 1.8 × 10⁻⁹; [I⁻] = 8.3 × 10⁻¹⁷ ÷ 1.8 × 10⁻⁹ = 4.6 × 10⁻⁸ mol dm⁻³; fraction = 4.6 × 10⁻⁵ (0.0046%). 3. Why does increasing the concentration of the second ion make the separation worse? Answer: A higher second-ion concentration lowers its threshold, so the second salt starts earlier, when more of the first ion is still dissolved. 4. State two reasons why real residues may be larger than the calculated value. Answer: Co-precipitation or solid-solution formation, and complexation or activity effects that change the effective ion concentrations.