Numerical/Formula

1. TITLE

How to solve Predicting Precipitation ($Q_{sp}$ vs $K_{sp}$) in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This is a classic Ionic Equilibrium numerical focused on the Solubility Product. We use the Ionic Product ($Q_{sp}$) to predict if a solid will precipitate. The most typical mistake students make is forgetting that mixing two solutions changes their total volume, which inherently halves their initial concentrations!

3. MCQ

Question:
Equal volumes of $0.002\text{ M}$ solutions of sodium iodate ($\text{NaIO}3$) and cupric chlorate ($\text{Cu}(\text{ClO}_3)_2$) are mixed together. Calculate the ionic product ($Q{sp}${sp}$) for copper iodate, $\text{Cu}(\text{IO}3)_2$, and predict if precipitation will occur. (Given: $K{sp}${sp}$ for cupric iodate $= 7.4 \times 10^{-8}$)

A) $Q_{sp} = 1.0 \times 10^{-9}$, No precipitation occurs
B) $Q_{sp} = 8.0 \times 10^{-9}$, No precipitation occurs (Calculation mistake: Forgot dilution)
C) $Q_{sp} = 1.0 \times 10^{-6}$, Precipitation occurs (Sign/Formula error: Forgot to square $[\text{IO}3^-]$)
D) $Q{sp} = 4.0 \times 10^{-6}${sp} = 4.0 \times 10^{-6}$, Precipitation occurs (Unit/Dilution + Formula error combined)

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Initial concentration of $\text{NaIO}3 = 0.002\text{ M}$
* Initial concentration of $\text{Cu}(\text{ClO}_3)_2 = 0.002\text{ M}$
* $K{sp}${sp}$ of $\text{Cu}(\text{IO}_3)_2 = 7.4 \times 10^{-8}$
* Volumes mixed are equal ($V_1 = V_2$)

To Find:
* The Ionic Product ($Q_{sp}$) of $\text{Cu}(\text{IO}_3)_2$
* Whether precipitation occurs or not

Formula:
* $Q_{sp} = [\text{Cu}^{2+}][\text{IO}3^-]^2$
* Explain when to use this formula: Use this formula to calculate the ionic product at any arbitrary instant after mixing. If $Q{sp} > K_{sp}${sp} > K_{sp}$, precipitation occurs; if $Q_{sp} < K_{sp}$, no precipitation occurs.

Step 1: Calculate new concentrations after mixing equal volumes
When equal volumes ($V$) of two solutions are mixed, the total volume becomes $2V$. Therefore, the concentration of each ion is halved.
* $[\text{Cu}^{2+}] = \frac{0.002\text{ M}}{2} = 0.001\text{ M} = 10^{-3}\text{ M}$
* $[\text{IO}_3^-] = \frac{0.002\text{ M}}{2} = 0.001\text{ M} = 10^{-3}\text{ M}$

Step 2: Write the dissociation equation and calculate $Q_{sp}$
The sparingly soluble salt formed is copper iodate: $\text{Cu}(\text{IO}3)_2 \rightleftharpoons \text{Cu}^{2+} + 2\text{IO}_3^-$
* $Q{sp} = [\text{Cu}^{2+}][\text{IO}3^-]^2${sp} = [\text{Cu}^{2+}][\text{IO}3^-]^2$
* $Q{sp} = (10^{-3}) \times (10^{-3})^2${sp} = (10^{-3}) \times (10^{-3})^2$
* $Q_{sp} = 10^{-3} \times 10^{-6} = 1.0 \times 10^{-9}$

Step 3: Compare $Q_{sp}$ with given $K_{sp}$
* $Q_{sp} = 1.0 \times 10^{-9}$
* $K_{sp} = 7.4 \times 10^{-8}$
* $1.0 \times 10^{-9} < 7.4 \times 10^{-8}$
Because $Q_{sp} < K_{sp}$, the solution is unsaturated.

Answer: $Q_{sp} = 1.0 \times 10^{-9}$, No precipitation occurs (Option A)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Forgetting to halve the concentrations. When two solutions are mixed, the volume increases, which directly dilutes the initial concentrations.
  • ❌ Mistake 2: Forgetting to square the iodate concentration in the $Q_{sp}$ formula. The stoichiometric coefficient of $2\text{IO}_3^-$ must become the power in the formula.
  • ❌ Mistake 3: Confusing the precipitation condition. Thinking $Q_{sp} < K_{sp}$ causes a precipitate, when in reality it means the solution can still dissolve more solid.

6. QUICK FORMULA REFERENCE

Formulas Used:
* $Q_{sp} = [\text{M}^{p+}]^x [\text{X}^{q-}]^y$ — Use when calculating the instantaneous ionic product after mixing two solutions.
* $Q_{sp} > K_{sp}$ — Use when checking the condition for Precipitation to occur.
* $Q_{sp} < K_{sp}$ — Use when checking the condition for No precipitation (solution remains clear).

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: mixing unequal volumes (e.g., $10\text{ mL}$ of $0.1\text{ M}$ solution with $40\text{ mL}$ of $0.2\text{ M}$ solution) and using $M_1V_1 = M_2V_2$ to find final concentrations before calculating $Q_{sp}$."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 6 (Equilibrium), Exercise Problem 6.69.

9. CTA

More solved numericals → Equilibrium Complete Guide
Practice problems → [Topic Test: 6-Equilibrium]