JEE Challenger
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ChemMedAdv 2025 • Paper 2

Determine Solubility of Barium Iodate in Mixed Solution

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL200\text{ mL} of 0.010 M0.010\text{ M} barium nitrate with 100 mL100\text{ mL} of 0.10 M0.10\text{ M} sodium iodate is X×10−6 mol dm−3X \times 10^{-6}\text{ mol dm}^{-3}. The value of XX is ______.

Use: Solubility product constant (KspK_{\text{sp}}) of barium iodate =1.58×10−9= 1.58 \times 10^{-9}

Official Numerical Answer3.85 to 4.15

Topics & Concepts

EquilibriumSolubility Product

Step-by-Step Solution

To find the solubility of barium iodate, Ba(IO3)2\text{Ba(IO}_3)_2, in the mixed aqueous solution, we first determine the concentrations of the ions present after the precipitation reaction goes to completion.

Step 1: Calculate the initial millimoles of the ions

  • Volume of Ba(NO3)2\text{Ba(NO}_3)_2 solution =200 mL= 200\text{ mL}

  • Molarity of Ba(NO3)2\text{Ba(NO}_3)_2 solution =0.010 M= 0.010\text{ M} Millimoles of Ba2+=200 mL×0.010 M=2.0 mmol\text{Millimoles of }\text{Ba}^{2+} = 200\text{ mL} \times 0.010\text{ M} = 2.0\text{ mmol}

  • Volume of NaIO3\text{NaIO}_3 solution =100 mL= 100\text{ mL}

  • Molarity of NaIO3\text{NaIO}_3 solution =0.10 M= 0.10\text{ M} Millimoles of IO3−=100 mL×0.10 M=10.0 mmol\text{Millimoles of }\text{IO}_3^- = 100\text{ mL} \times 0.10\text{ M} = 10.0\text{ mmol}

  • Total volume of the mixed solution =200 mL+100 mL=300 mL= 200\text{ mL} + 100\text{ mL} = 300\text{ mL}

Step 2: Precipitation reaction and limiting reagent analysis

The precipitation reaction for barium iodate is: Ba(aq)2++2IO3(aq)−⇌Ba(IO3)2(s)\text{Ba}^{2+}_{(\text{aq})} + 2\text{IO}_{3(\text{aq})}^- \rightleftharpoons \text{Ba(IO}_3)_{2(\text{s})}

Since 2.0 mmol2.0\text{ mmol} of Ba2+\text{Ba}^{2+} reacts with 2×2.0=4.0 mmol2 \times 2.0 = 4.0\text{ mmol} of IO3−\text{IO}_3^-, Ba2+\text{Ba}^{2+} is the limiting reagent and precipitates out almost completely.

  • Millimoles of excess IO3−\text{IO}_3^- remaining =10.0−4.0=6.0 mmol= 10.0 - 4.0 = 6.0\text{ mmol}

The concentration of excess IO3−\text{IO}_3^- in the final 300 mL300\text{ mL} solution is: [IO3−]excess=6.0 mmol300 mL=0.020 M[\text{IO}_3^-]_{\text{excess}} = \frac{6.0\text{ mmol}}{300\text{ mL}} = 0.020\text{ M}

Step 3: Solubility equilibrium calculation

Let SS be the solubility of Ba(IO3)2\text{Ba(IO}_3)_2 in mol dm−3\text{mol dm}^{-3} (or mol L−1\text{mol L}^{-1}) in the presence of excess IO3−\text{IO}_3^- ions.

At equilibrium: [Ba2+]=S[\text{Ba}^{2+}] = S [IO3−]=0.020+2S≈0.020 M(since 2S≪0.020)[\text{IO}_3^-] = 0.020 + 2S \approx 0.020\text{ M} \quad (\text{since } 2S \ll 0.020)

The solubility product expression for Ba(IO3)2\text{Ba(IO}_3)_2 is: Ksp=[Ba2+][IO3−]2K_{\text{sp}} = [\text{Ba}^{2+}][\text{IO}_3^-]^2

Substituting the equilibrium concentrations into the KspK_{\text{sp}} expression: 1.58×10−9=S×(0.020)21.58 \times 10^{-9} = S \times (0.020)^2 1.58×10−9=S×(2.0×10−2)21.58 \times 10^{-9} = S \times (2.0 \times 10^{-2})^2 1.58×10−9=S×4.0×10−41.58 \times 10^{-9} = S \times 4.0 \times 10^{-4}

Solving for SS: S=1.58×10−94.0×10−4=0.395×10−5 mol dm−3=3.95×10−6 mol dm−3S = \frac{1.58 \times 10^{-9}}{4.0 \times 10^{-4}} = 0.395 \times 10^{-5}\text{ mol dm}^{-3} = 3.95 \times 10^{-6}\text{ mol dm}^{-3}

Conclusion

Comparing S=3.95×10−6 mol dm−3S = 3.95 \times 10^{-6}\text{ mol dm}^{-3} with X×10−6 mol dm−3X \times 10^{-6}\text{ mol dm}^{-3}, we get: X=3.95X = 3.95

Determine Solubility of Barium Iodate in Mixed Solution | Chemistry PYQ Solution - JEE Challenger