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← JEE Advanced 2025 · Paper 2
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JEE Advanced 2025 · Paper 2, Chemistry question 10

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is X×10−6 mol dm−3\boldsymbol{X}\times10^{-6}\ \mathrm{mol\,dm^{-3}}. The value of X\boldsymbol{X} is _____.

Use: Solubility product constant (KspK_{sp}) of barium iodate =1.58×10−9= 1.58\times10^{-9}

Answer and worked solution

Official answer: 3.85

Precipitation. The mixture has 2 mmol BaX2+\ce{Ba^2+} and 10 mmol IOX3X−\ce{IO3-} in 300 mL. Ba(IOX3)X2\ce{Ba(IO3)2} precipitates, using 4 mmol IOX3X−\ce{IO3-}. That leaves 6 mmol IOX3X−\ce{IO3-} in 300 mL, i.e. [IOX3X−]=0.020[\ce{IO3-}]=0.020 M.

Solubility with the common ion.
s=Ksp[IOX3X−]2=1.58×10−94×10−4=3.95×10−6 mol dm−3.s=\frac{K_{sp}}{[\ce{IO3-}]^2}=\frac{1.58\times10^{-9}}{4\times10^{-4}}=3.95\times10^{-6}\ \text{mol dm}^{-3}.

Topics:Common ion effectSolubility productStoichiometry
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