JEE Advanced 2025 Paper 2 · Q10 · Buffers & Solubility Product
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 $X \times 10^{-6}$ mol/$dm^3$. The value of X is __________.
Use: Solubility product constant $K_{sp}$ of barium iodate = $1.58 \times 10^{-9}$.
Reveal answer + step-by-step solution
Correct answer:3.95
Solution
Initial moles after mixing: $Ba^{2+} = 0.20 \times 0.010 = 0.002$ mol; $IO_3^- = 0.10 \times 0.10 = 0.010$ mol. Total volume = 300 mL = 0.3 L. Stoichiometry of $Ba(IO_3)_2$ precipitation: 1 mol $Ba^{2+}$ + 2 mol $IO_3^- \to$ ppt. $Ba^{2+}$ is limiting (0.002 mol consumes 0.004 mol $IO_3^-$). After precipitation, leftover $IO_3^-$ = $0.010 - 0.004 = 0.006$ mol in 0.3 L -> $[IO_3^-] = 0.020$ M (excess). In presence of excess iodate, additional dissolution $s$ of $Ba(IO_3)_2$ satisfies $K_{sp} = [Ba^{2+}][IO_3^-]^2 \approx s \cdot (0.020)^2 = 1.58 \times 10^{-9}$. $s = 1.58 \times 10^{-9} / 4 \times 10^{-4} = 3.95 \times 10^{-6}$ M. So $X = 3.95$.
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