1
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. The value of X is . Use: Solubility product constant (Ksp) of barium iodate = 1.58 10-9
Answer
3.85 to 4.15
Solution
After mixing (300 mL): [Ba2+] = 200 0.010300 = 6.67 10-3 M; [IO3-] = 100 0.10300 = 3.33 10-2 M.
Ba(IO3)2 precipitates almost completely: Ba2+ (2 mmol) uses 4 mmol IO3-; left: 6 mmol in 300 mL → [IO3-] = 0.020 M.
Solubility in this solution: s = Ksp[IO3-]2 = 1.58 10-9(0.020)2 = 3.95 10-6 mol dm-3
Answer: 3.95
JEE Advanced 2025 Paper 2