Question 65 of 69intermediate🔧 ApplyNumerical2 marks

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp=7.4×10−8K_{sp} = 7.4 \times 10^{-8}).

Correct Answer

When equal volumes of 0.002 M sodium iodate and 0.002 M cupric chlorate are mixed, the concentration of each ion is halved. Thus, the concentration of [Cu2+][\text{Cu}^{2+}] becomes 0.0010.001 M and [IO3−][\text{IO}_3^-] becomes 0.0010.001 M.

The ionic product (QQ) for cupric iodate, Cu(IO3)2\text{Cu}(\text{IO}_3)_2, is calculated as Q=[Cu2+][IO3−]2Q = [\text{Cu}^{2+}][\text{IO}_3^-]^2. Substituting the values, Q=(0.001)×(0.001)2=1.0×10−9Q = (0.001) \times (0.001)^2 = 1.0 \times 10^{-9}.

Comparing QQ with the given solubility product Ksp=7.4×10−8K_{sp} = 7.4 \times 10^{-8}, we find that Q<KspQ < K_{sp}. Therefore, the solution is unsaturated and no precipitation of copper iodate will occur.

Exercise: EXERCISES | Q: 6.69 | (Page 46)
For More Understanding

Explanation

To determine if precipitation occurs, we compare the ionic product (Q) of the resulting solution with the solubility product (Ksp) of the sparingly soluble salt. Since equal volumes are mixed, dilution halves the initial molarity of the ions. The calculated ionic product is less than the Ksp, indicating the solution can hold more salt without precipitating.

Solution Steps

  1. Step 1: Determine the concentration of ions after mixing. Equal volumes halve the concentration: [Cu2+]=0.0022=0.001[\text{Cu}^{2+}] = \frac{0.002}{2} = 0.001 M and [IO3−]=0.0022=0.001[\text{IO}_3^-] = \frac{0.002}{2} = 0.001 M.

  2. Step 2: Write the expression for the ionic product (QQ) of cupric iodate. For Cu(IO3)2⇌Cu2++2IO3−\text{Cu}(\text{IO}_3)_2 \rightleftharpoons \text{Cu}^{2+} + 2\text{IO}_3^-, Q=[Cu2+][IO3−]2Q = [\text{Cu}^{2+}][\text{IO}_3^-]^2.

  3. Step 3: Calculate QQ. Q=(0.001)×(0.001)2=10−3×10−6=1.0×10−9Q = (0.001) \times (0.001)^2 = 10^{-3} \times 10^{-6} = 1.0 \times 10^{-9}.

  4. Step 4: Compare QQ with KspK_{sp}. Since Q=1.0×10−9Q = 1.0 \times 10^{-9} is less than Ksp=7.4×10−8K_{sp} = 7.4 \times 10^{-8}, no precipitation will occur.

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