In a saturated solution of the sparingly soluble strong electrolyte ${AgIO}_{3}$ (Molecular mass = 283) the equilibrium which sets in is :
${AgIO}_{3} (s) ⟶ {Ag}^{+} (a q) + {IO}_{3}^{-} (a q)$
If the solubility product constant K_sp of ${AgIO}_{3}$ at a given temperature is 1.0 x 10^-8 , what is the mass of ${AgIO}_{3}$ contained in 100 mL of its saturated solution?

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1.0 X 10^-4 g

28.3 X 10^-2 g

1.0 X 10^-7g

2.83 X 10^-3 g

answer is D.

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Detailed Solution

Here, we have,

$S = \sqrt{K_{s p}} = 10^{- 4} M$

Weight,

$\begin{array}{l} W = 10^{- 4} \times 0.1 \times 283 \\ W = 2.83 \times 10^{- 3} g \end{array}$

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