JEE Advanced 2025 Paper 2, Chemistry Section 3 Q2: Solubility Equilibria Of Sparingly Soluble Salts

JEE Advanced2025Paper 2Chemistry Section 3

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 $X \times 1 0^{- 6}$ mol dm $^{- 3}$ . The value of $X$ is _______.

Use: Solubility product constant ( $K_{s p}$ ) of barium iodate = $1.58 \times 1 0^{- 9}$

Your answer

Solution

Compute the initial mmol of each reactant in the mixed solution:

Ba(NO $_{3}$ ) $_{2}$ : $200 \times 0.010 = 2$ mmol; NaIO $_{3}$ : $100 \times 0.10 = 10$ mmol.

The precipitation reaction is

Ba(NO $_{3}$ ) $_{2}$ + 2 NaIO $_{3}$ $\to$ Ba(IO $_{3}$ ) $_{2}$ (s) + 2 NaNO $_{3}$

Ba $^{2 +}$ is the limiting reagent; it consumes $2 \times 2 = 4$ mmol of NaIO $_{3}$ , leaving 6 mmol NaIO $_{3}$ in 300 mL.

$[IO_{3}^{-}]_{excess} = \frac{6}{300} = 2 \times 1 0^{- 2}$ M.

Let $s$ = solubility of Ba(IO $_{3}$ ) $_{2}$ in this solution. With strong common-ion suppression by the excess IO $_{3}^{-}$ :

$K_{s p} = [Ba^{2 +}] [IO_{3}^{-}]^{2} \approx s \times (2 \times 1 0^{- 2})^{2}$

$1.58 \times 1 0^{- 9} = s \times 4 \times 1 0^{- 4}$

$s = 3.95 \times 1 0^{- 6}$ M.

$∴ X = 3.95$ .

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JEE Advanced 2025 Paper 2, Chemistry Section 3 Q2: Solubility Equilibria Of Sparingly Soluble Salts

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