Question

In a saturated solution of the sparingly soluble strong electrolyte AgIO${}_3$ (molecular mass = 283) the equilibrium which sets in is $AgI{O}_{3\left(s\right)}\rightleftharpoons Ag{{}^{+}}_{aq}+IO{{}_3}^{-}{}_{\left(aq\right)}$.

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 solution ?

• A. $28.3\times {10}^{-2}$ g
• B. $2.83\times {10}^{-3}$ g
• C. $1.0\times {10}^{-7}$ g
• D. $1.0\times {10}^{-4}$ g

Answer 1

The correct option is B $2.83\times {10}^{-3}$ g

Let $S$ M be the solubility of the electrolyte.
$\left[A{g}^{+}\right]=\left[I{O}_3^{-}\right]=\left[AgI{O}_3\right]=S$

$K_{sp}=\left[A{g}^{+}\right]\left[I{O}_3^{-}\right]=S\times S=S^2$

$1.0\times {10}^{-8}=S^2$

$S=1.0\times {10}^{-4}M$

$1000$ mL of solution will contain $1.0\times {10}^{-4}$ moles of electrolyte.
$100$ mL of solution will contain $1.0\times {10}^{-5}$ moles of electrolyte.

Molar mass of electrolyte is $283$ g/mol.
$100$ mL of solution will contain $1.0\times {10}^{-5}\times 283=2.83\times {10}^{-3}$ g of electrolyte.