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JEE Advanced Chemistry Ionic Equilibrium 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Chemistry Question (2025) — Solution

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 X 10^ -6 ~mol dm ^ -3 . The value of X is . Use: Solubility product constant (K_ sp ) of barium iodate =1.58 10^ -9

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

[ NaIO _3 ]= 6 300 =2 10^ -2 M aligned Ba ( IO _3 )_2( ~s ) Ba ^ 2+ + & 2 IO _3^ - \\ s & (2 10^ -2 +2 ~s ) aligned aligned & K _ sp = [ Ba ^ +2 ] [ IO _3^ - ]^2 \\ & 1.58 10^ -9 = [ Ba ^ +2 ] (2 10^ -2 )^2 \\ & [ Ba ^ +2 ]= s =3.95 10^ -6 M \\ & X =3.95 aligned

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