In a saturated solution of the sparingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets is ${AgIO}_{3} (s) ⇌ {Ag}^{+} (aq) + {IO}_{3}^{-} (aq)$ If the solubility product constant $K_{sp} of {AgIO}_{3}$ at a given temperature $1 .0 \times 10^{- 8}$ what is the mass of ${AgIO}_{3}$ contained in 100 ml of its saturated solution?

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$1 .0 \times 10^{- 7} g$

$1 .0 \times 10^{- 4} g$

$28 .3 \times 10^{- 2} g$

$2 .83 \times 10^{- 3} g$

answer is C.

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Detailed Solution

Let s = solubility
${AgIO}_{3} ⇌ {Ag}^{+} + {IO}_{3}^{-} K_{sp} = [{Ag}^{+}] [{IO}_{3}^{-}] = s \times s = s^{2} Given K_{sp} = 1 \times 10^{- 8} ∴ s = \sqrt{K_{sp}} = \sqrt{1 \times 10^{- 8}} = 1 .0 \times 10^{- 4} mol / lit = 1 .0 \times 10^{- 4} \times 283 g / lit (∵ Molecular mass of Ag {IO}_{3} = 283) = \frac{1 .0 \times 10^{- 4} \times 283 \times 100}{1000} gm / 100 ml = 2 .83 \times 10^{- 3} gm / 100 ml$

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