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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Advanced 2025 · Paper 2 · Chemistry · Ionic Equilibrium

The solubility of barium iodate in an aqueous solution prepared by...

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 moldm−3\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}. The value of X\boldsymbol{X} is \qquad . Use: Solubility product constant (Ksp)\left(K_{\mathrm{sp}}\right) of barium iodate =1.58×10−9=1.58 \times 10^{-9}The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL

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The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL
[NaIO3]=6300=2×10−2M\left[\mathrm{NaIO}_3\right]=\frac{6}{300}=2 \times 10^{-2} \mathrm{M}
Ba(IO3)2( s)⇌Ba2++2IO3−s(2×10−2+2 s)\begin{aligned} \mathrm{Ba}\left(\mathrm{IO}_3\right)_2(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}+ & 2 \mathrm{IO}_3^{-} \\ \mathrm{s} \qquad & \left(2 \times 10^{-2}+2 \mathrm{~s}\right)\end{aligned}
Ksp=[Ba+2][IO3−]21.58×10−9=[Ba+2]×(2×10−2)2[Ba+2]=s=3.95×10−6MX=3.95\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{IO}_3^{-}\right]^2 \\ & 1.58 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right] \times\left(2 \times 10^{-2}\right)^2 \\ & {\left[\mathrm{Ba}^{+2}\right]=\mathrm{s}=3.95 \times 10^{-6} \mathrm{M}} \\ & \mathrm{X}=3.95\end{aligned}
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