Solubility product Ksp — will precipitation occur when solutions are mixed

Question

50 mL of 0.002 M $\text{BaCl}_2$ is mixed with 50 mL of 0.004 M $\text{Na}_2\text{SO}_4$ . Will $\text{BaSO}_4$ precipitate? Given: $K_{sp}$ of $\text{BaSO}_4 = 1.1 \times 10^{-10}$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

When two solutions are mixed, the total volume doubles (50 + 50 = 100 mL). Each concentration is halved:

[\text{Ba}^{2+}] = \frac{0.002}{2} = 0.001 \text{ M} = 10^{-3} \text{ M}

[\text{SO}_4^{2-}] = \frac{0.004}{2} = 0.002 \text{ M} = 2 \times 10^{-3} \text{ M}

The ionic product (reaction quotient) for $\text{BaSO}_4$ :

Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = 10^{-3} \times 2 \times 10^{-3} = 2 \times 10^{-6}

$Q = 2 \times 10^{-6}$ vs $K_{sp} = 1.1 \times 10^{-10}$

Q \gg K_{sp}

Since $Q > K_{sp}$ , the solution is supersaturated with respect to $\text{BaSO}_4$ .

\boxed{\text{Yes, BaSO}_4 \text{ will precipitate.}}

Precipitation continues until $Q$ decreases to equal $K_{sp}$ .

Why This Works

The solubility product $K_{sp}$ represents the maximum ion product a solution can sustain at equilibrium. If the actual ion product $Q$ exceeds $K_{sp}$ , the solution cannot hold that many ions — the excess precipitates out as a solid until equilibrium is restored.

The three cases:

$Q < K_{sp}$ : unsaturated — no precipitation, more solid can dissolve
$Q = K_{sp}$ : saturated — equilibrium, no net change
$Q > K_{sp}$ : supersaturated — precipitation occurs

In our problem, $Q$ exceeds $K_{sp}$ by a factor of about $10^4$ , so a large amount of $\text{BaSO}_4$ precipitates.

Alternative Method

You can also compare the concentrations needed for precipitation. For $[\text{Ba}^{2+}] = 10^{-3}$ M, precipitation starts when:

[\text{SO}_4^{2-}] > \frac{K_{sp}}{[\text{Ba}^{2+}]} = \frac{1.1 \times 10^{-10}}{10^{-3}} = 1.1 \times 10^{-7} \text{ M}

Since $[\text{SO}_4^{2-}] = 2 \times 10^{-3} \gg 1.1 \times 10^{-7}$ , precipitation definitely occurs.

For JEE, the critical step is dilution. When solutions are mixed, always recalculate concentrations using the total volume before computing $Q$ . Forgetting the dilution is the single most common error in precipitation problems.

Common Mistake

Students often use the original concentrations (before mixing) to compute $Q$ . When you mix 50 mL of solution A with 50 mL of solution B, the volume becomes 100 mL, so all concentrations are halved. Using undiluted concentrations gives a $Q$ that is 4 times too high (for a 1:1 electrolyte). While it may not change the qualitative answer here, it matters in borderline cases.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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