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 $\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}$. The value of $\boldsymbol{X}$ is ____. Use: Solubility product constant $(K_{\mathrm{sp}})$ of barium iodate $=1.58 \times 10^{-9}$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Equilibrium

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_{sp})

of barium iodate

= 1.58 \times 1 0^{- 9}

Answer: 3.95

First, determine the limiting reagent:

Ba (N O_{3})_{2} + 2 NaI O_{3} \to Ba (I O_{3})_{2} + 2 NaN O_{3}

Initial moles:

Ba²⁺ is the limiting reagent. After precipitation:

For the solubility equilibrium:

Ba (I O_{3})_{2} (s) ⇌ B a^{2 +} + 2 I O_{3}^{-}

Let solubility = s

K_{s p} = [B a^{2 +}] [I O_{3}^{-}]^{2}

1.58 \times 1 0^{- 9} = s \times (2 \times 1 0^{- 2})^{2}

s = \frac{1.58 \times 1 0 ^{- 9}}{4 \times 1 0 ^{- 4}} = 3.95 \times 1 0^{- 6} M

Therefore, X = 3.95

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer