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Question

In a saturated solution of the sparingly soluble strong electrolyte $	ext{AgIO}_3$ (molecular mass = 283) the equilibrium which sets in is
$	ext{AgIO}_3(s) ightleftharpoons 	ext{Ag}^+(aq) + 	ext{IO}_3^-(aq)$
If the solubility product constant $K_{sp}$ of $	ext{AgIO}_3$ at a given temperature is $1.0 	imes 10^{-8}$, what is the mass of $	ext{AgIO}_3$ contained in 100 mL of its saturated solution? A). $1.0 	imes 10^{-4}	ext{ g}$ B). $28.3 	imes 10^{-2}	ext{ g}$ C). $2.83 	imes 10^{-3}	ext{ g}$ D). $1.0 	imes 10^{-7}	ext{ g}$

Edvaya · Accepted Answer

The correct answer is $2.83 	imes 10^{-3}	ext{ g}$. For the dissociation $	ext{AgIO}_3(s) ightleftharpoons 	ext{Ag}^+(aq) + 	ext{IO}_3^-(aq)$:
$K_{sp} = s^2$, where $s$ is the molar solubility.
$s = \sqrt{1.0 	imes 10^{-8}} = 1.0 	imes 10^{-4}	ext{ mol/L}$.
Moles of $	ext{AgIO}_3$ in 100 mL (0.1 L) of solution:
$n = s 	imes V = (1.0 	imes 10^{-4}	ext{ mol/L}) 	imes 0.1	ext{ L} = 1.0 	imes 10^{-5}	ext{ mol}$.
Mass = moles $	imes$ molar mass:
$	ext{Mass} = 1.0 	imes 10^{-5} 	imes 283 = 283 	imes 10^{-5}	ext{ g} = 2.83 	imes 10^{-3}	ext{ g}$.

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