Question

In a saturated solution of the sparingly soluble strong electrolyte $AgI{O}_3$ (Molecular mass = 283) the equilibrium which sets in is
$AgI{O}_{3\left(s\right)}\rightleftharpoons A{g}_{\left(aq\right)}^{+}+I{O}_3^{-}\left(aq\right)$.
If the solubility product constant $K_{sp}$ of $AgI{O}_3$ at a given temperature is $1.0\times {10}^{-8},$ what is the mass of $AgI{O}_3$ 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 the solubility of $AgI{O}_3$ be 's'
The dissociation equation is
$AgI{O}_3\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I{O}_3^{-}\left(aq\right)t=0100ateqb1-sss$
$K_{sp}=\left[A{g}^{+}\right]\left[I{O}_3^{-}\right]K_{sp}=s^2$
$s^2=1.0\times {10}^{-8}$
$s={10}^{-4}mol{L}^{-1}$
Mass of $AgI{O}_3$ in 100 mL
$=\frac{concentration\times Molarmass}{1000}\times Volume=\frac{{10}^{-4}\times 283}{1000}\times 100=0.00283$
Mass of $AgI{O}_3=2.83\times {10}^{-3}g$