Ion Product and Precipitation Direction

Introduction

When two clear solutions are mixed, will a precipitate appear? The answer comes from comparing the actual state of the mixture with the equilibrium state. Just as the reaction quotient Q tells us which way a gas-phase equilibrium will shift, the ion product Qsp tells us whether ions will leave solution as a solid or whether more solid can dissolve. The only complications are getting the concentrations right after mixing and remembering the powers.

Core explanation

The ion product. For a salt MₓAᵧ, the ion product is written exactly like Ksp but uses the concentrations actually present at a given moment:

Qsp = [Mᵐ⁺]ˣ [Aⁿ⁻]ʸ (current values)

Three cases.

Comparison State of solution What happens --- --- --- Qsp < Ksp unsaturated no precipitate; any solid present dissolves Qsp = Ksp saturated equilibrium; no net change Qsp > Ksp supersaturated precipitate forms until Qsp falls to Ksp

Mixing and dilution. When volume V₁ of one solution is mixed with V₂ of another, each ion is diluted into the total volume V₁ + V₂. The concentration after mixing is c × V₁/(V₁ + V₂). Mixing equal volumes halves each concentration. Forgetting this step is the commonest source of wrong predictions, especially for salts with squared terms, where a factor of two error becomes a factor of four.

Before and after. Qsp computed immediately after mixing, before any reaction, predicts the direction. If precipitation occurs, the ions are consumed in their stoichiometric ratio until Qsp equals Ksp. The ion in excess remains largely in solution, while the limiting ion is reduced to a very low concentration set by Ksp. This "after" state is the basis of gravimetric analysis and of selective precipitation.

Kinetic caveat. Qsp > Ksp shows that precipitation is thermodynamically favourable, but a supersaturated solution can persist if nucleation is slow. Scratching the glass, adding a seed crystal or waiting can trigger precipitation. In practice, for most common sparingly soluble salts, precipitation is rapid once Qsp is well above Ksp.

Precision of predictions. When Qsp and Ksp differ by less than a factor of about two, the prediction is uncertain, because activity effects and temperature variation can shift Ksp by that much. Predictions are safe when Qsp exceeds or falls short of Ksp by orders of magnitude.

Step-by-step reasoning

1. Write the Ksp expression for the possible precipitate. 2. Calculate each ion's concentration after mixing, using the total volume. 3. Substitute into Qsp, including powers. 4. Compare with Ksp and state the direction. 5. If precipitation occurs and the question asks, find the excess ion and use Ksp to calculate the remaining concentration of the other ion.

Visual explanation

Draw a number line of log Qsp with a vertical mark at log Ksp. Points to the left are labelled "dissolves"; points to the right "precipitates". An arrow shows that precipitation moves a point leftwards until it reaches the mark, while dissolution moves a point rightwards up to the mark.

Real-world analogy

Ksp is like the capacity of a lift. Qsp is the number of people currently trying to get in. If fewer than capacity, more can enter (dissolve); if more, some must step out (precipitate) until the load equals capacity.

Real-world example

Hard water contains Ca²⁺ and hydrogencarbonate ions. When water is heated in a kettle, hydrogencarbonate decomposes to carbonate, raising [CO₃²⁻] so that the ion product of CaCO₃ exceeds its Ksp. Calcium carbonate then precipitates as limescale on the heating element.

Why?

Why must dilution be included? Mixing does not change the number of moles of each ion, but it spreads them over a larger volume. Equilibrium depends on concentration, so the diluted concentrations are the ones that determine whether the solid can form.

Common misconception

"A precipitate forms whenever the two ions of an insoluble salt are mixed." If the concentrations are low enough that Qsp is below Ksp, no solid forms, however "insoluble" the salt is described as being.

Worked example

Question: 50.0 cm³ of 1.0 × 10⁻⁴ mol dm⁻³ AgNO₃ is mixed with 50.0 cm³ of 1.0 × 10⁻⁴ mol dm⁻³ NaCl. Will AgCl (Ksp = 1.8 × 10⁻¹⁰) precipitate? Repeat for 1.0 × 10⁻⁵ mol dm⁻³ BaCl₂ mixed with an equal volume of 1.0 × 10⁻⁵ mol dm⁻³ Na₂SO₄ (Ksp BaSO₄ = 1.1 × 10⁻¹⁰).

Reasoning: After mixing, [Ag⁺] = [Cl⁻] = 5.0 × 10⁻⁵ mol dm⁻³. Qsp = 2.5 × 10⁻⁹, which exceeds 1.8 × 10⁻¹⁰, so AgCl precipitates. For barium sulfate, [Ba²⁺] = [SO₄²⁻] = 5.0 × 10⁻⁶ mol dm⁻³, Qsp = 2.5 × 10⁻¹¹, which is below 1.1 × 10⁻¹⁰.

Answer: AgCl precipitates; BaSO₄ does not.

Quick check

1. A solution has Qsp for CaF₂ equal to 1.0 × 10⁻¹² while Ksp is 3.9 × 10⁻¹¹. What happens if solid CaF₂ is added? Answer: Some of the solid dissolves, because the solution is unsaturated with Qsp less than Ksp.

Exam focus

Show the diluted concentrations explicitly, then Qsp with powers, then a clear comparison sentence. Examiners often choose data where ignoring dilution reverses the conclusion, so this step is worth checking twice.

Advanced insight

After precipitation, finding the remaining concentrations is a limiting-reagent problem followed by an equilibrium problem. If the ions are mixed in exactly stoichiometric amounts, the remaining concentrations equal those of a saturated solution in pure water. Otherwise, the excess ion acts as a common ion, pushing the concentration of the limiting ion far below its pure-water solubility.

Summary

The ion product Qsp has the same form as Ksp but uses actual concentrations. If Qsp exceeds Ksp, a precipitate forms until they are equal; if it is smaller, the solution is unsaturated. Concentrations must be corrected for dilution on mixing before Qsp is calculated. Supersaturation is possible because nucleation can be slow.

Practice questions

1. State what happens when Qsp equals Ksp. Answer: The solution is saturated and at equilibrium, so there is no net precipitation or dissolution. 2. 10.0 cm³ of 0.0020 mol dm⁻³ Pb(NO₃)₂ is mixed with 10.0 cm³ of 0.0020 mol dm⁻³ KI. Will PbI₂ (Ksp = 9.8 × 10⁻⁹) precipitate? Answer: [Pb²⁺] = [I⁻] = 0.0010 mol dm⁻³; Qsp = 0.0010 × (0.0010)² = 1.0 × 10⁻⁹, below Ksp, so no precipitate. 3. In question 2, would the conclusion change if dilution were ignored? Answer: Yes: Qsp would be 0.0020 × (0.0020)² = 8.0 × 10⁻⁹, still below Ksp, but only just, showing how dilution matters. 4. Why can a solution with Qsp slightly above Ksp remain clear for a time? Answer: Precipitation requires nucleation of new crystals, which can be slow, so a supersaturated solution may persist.