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In a saturated solution of the sparingly soluble electrolyte ${AgIO}_{3}$ (relative molecular mass = 283) the equilibrium which sets in is
${AgIO}_{3} (s) ⇌ {Ag}^{+} (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{s p}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8} M$ what is the mass of ${AgIO}_{3}$ contained in 100 mL of its saturated solution?

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$28.3 \times 10^{- 2} g$

$2.83 \times 10^{- 3} g$

$1.0 \times 10^{- 7} g$

$1.0 \times 10^{- 4} g$

answer is B.

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Detailed Solution

For ${AgIO}_{3} (s) ⇌ {Ag}^{+} (aq) + {IO}_{3}^{-} (aq)$ we have $K_{sp} = [{Ag}^{+}] [{IO}_{3}^{-}] = s^{2}$ Hence $s = \sqrt{K_{sp}} = \sqrt{1.0 \times 10^{- 8} M^{2}} = 1.0 \times 10^{- 4} M$

Mass of ${AgIO}_{3}$ in 100 mL solution is
$m = (1.0 \times 10^{- 4} {molL}^{- 1}) (283 g {mol}^{- 1}) (0.1 L) = 2.83 \times 10^{3} g$

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