Solubility Product Ksp — When Does Precipitation Occur?

medium CBSE JEE-MAIN CBSE 2024 Board Exam 3 min read

Question

A solution contains 0.001 M Ba²⁺ ions. We slowly add Na₂SO₄ to this solution. At what concentration of SO₄²⁻ will BaSO₄ just start to precipitate?

Given: Ksp of BaSO₄ = 1.1 × 10⁻¹⁰ at 25°C.


Solution — Step by Step

BaSO₄ dissolves as:

BaSO4(s)Ba2+(aq)+SO42(aq)\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)

The Ksp expression is:

Ksp=[Ba2+][SO42]K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]

We use this expression both to predict solubility AND to predict precipitation — same equation, different question being asked.

Precipitation begins the moment the ionic product (IP) exceeds Ksp. Right at the threshold — when precipitation just starts — we have IP = Ksp. So we set:

[Ba2+][SO42]=Ksp[\text{Ba}^{2+}][\text{SO}_4^{2-}] = K_{sp}

This is the critical boundary. Below this, solution is unsaturated. Above this, BaSO₄ crashes out.

We know [Ba²⁺] = 0.001 M = 10⁻³ M, and Ksp = 1.1 × 10⁻¹⁰.

[SO42]=Ksp[Ba2+]=1.1×1010103[\text{SO}_4^{2-}] = \frac{K_{sp}}{[\text{Ba}^{2+}]} = \frac{1.1 \times 10^{-10}}{10^{-3}} [SO42]=1.1×107 M[\text{SO}_4^{2-}] = 1.1 \times 10^{-7} \text{ M}

Precipitation of BaSO₄ begins when [SO₄²⁻] just exceeds 1.1 × 10⁻⁷ M.

Any concentration of sulfate below this keeps the solution unsaturated — no solid forms. The moment we cross this threshold, the solution becomes supersaturated and BaSO₄ precipitates.


Why This Works

The ionic product (IP) is calculated exactly like Ksp — it’s just that IP uses the actual ion concentrations in solution at any given moment, while Ksp is the equilibrium value at saturation.

Think of Ksp as the maximum “carrying capacity” of the solution for those ions. As long as IP < Ksp, the solution can accommodate more dissolved ions without forming a solid. The moment IP > Ksp, the system is forced back to equilibrium by precipitating the excess as solid salt.

This is the same Le Chatelier principle you use everywhere else — the reaction shifts to consume excess products (ions) by forming the solid reactant (BaSO₄).

This principle has a direct real-world application: qualitative analysis in NCERT Class 12 uses selective precipitation based on Ksp differences. By controlling ion concentration, we can precipitate one ion while leaving another in solution. This concept carries high weightage in JEE Main and appeared in CBSE 2024 Board Exam Q8(b).


Alternative Method — Checking Any Given Mixture

Suppose we’re not finding the threshold but instead need to check whether a specific solution (say, 0.001 M Ba²⁺ mixed with 0.0002 M SO₄²⁻) will precipitate.

Calculate IP directly:

IP=[Ba2+][SO42]=(103)(2×104)=2×107IP = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (10^{-3})(2 \times 10^{-4}) = 2 \times 10^{-7}

Now compare with Ksp = 1.1 × 10⁻¹⁰:

Since IP = 2 × 10⁻⁷ >> Ksp, precipitation will occur.

This comparison approach — calculate IP, compare with Ksp — is the standard method for any “will a precipitate form?” question. The threshold calculation we did in the main solution is just this approach set equal to Ksp.


Common Mistake

The most frequent error: students confuse which ion concentration to use. If the question says “0.001 M BaCl₂”, some students write [Ba²⁺] = 0.0005 M because they’re thinking of the 2 in BaCl₂. Wrong — BaCl₂ is a strong electrolyte and fully dissociates: BaCl₂ → Ba²⁺ + 2Cl⁻, so [Ba²⁺] = [BaCl₂] = 0.001 M. The 2:1 ratio applies to Cl⁻, not Ba²⁺. Always write the full dissociation equation first before plugging into the Ksp expression.

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