JEE Advanced 2025 Chemistry question
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 mol dm⁻³. The value of X is . Use: Solubility product constant () of barium iodate =
Answer: 3.95
Solution
Ba²⁺ = 2 mmol and IO₃⁻ = 10 mmol in 300 mL. Precipitation of Ba(IO₃)₂ uses essentially all 2 mmol Ba²⁺ and 4 mmol IO₃⁻, leaving 6 mmol IO₃⁻: IO_3^-\frac{6}{300}$ =
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