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In a saturated solution of the sparingly soluble strong electrolyte $AgIO_3$ (molecular mass $= 283$) the equilibrium which sets in is
$AgIO_{3(s)} \rightleftharpoons Ag^+_{(aq)} +IO^-_{3(aq)}.$
If the solubility product constant $K_{sp}$ of $AgIO_3$ at a given temperature is $1. 0 \times 10^{-8},$ what is the mass of $AgIO_3$ contained in $100\, ml$ of its saturated saolution ?
$1.0 \times 10^{-4}\, g$
$28.3 \times 10^{-2}\, g$
$2.83 \times 10^{-3}\, g$
$1.0 \times 10^{-7}\, g$
Solution
The correct answer Is:
(c) $: \quad A g I O_{3} \rightleftharpoons A g^{+}+I O_{3}^{-}[S=$ Solubility $]$
$K_{s p}=S^{2}$
or, $S^{2}=1.0 \times 10^{-8}$ or, $S=1.0 \times 10^{-4} \mathrm{mol} / \mathrm{lit}$
$=1.0 \times 10^{-4} \times 283 \quad \mathrm{g} / \mathrm{lit}$
$=\frac{1.0 \times 10^{-4} \times 283}{1000} \mathrm{gm} / \mathrm{ml}$
$=\frac{1.0 \times 10^{-4} \times 283 \times 100}{1000} \mathrm{gm} / 100 \mathrm{ml}$
$=28.3 \times 10^{-4} \mathrm{gm} / 100 \mathrm{ml}$
$=2.83 \times 10^{-3} \mathrm{gm} / 100 \mathrm{ml}$
