Question

In a saturated of the sparingly soluble strong electrolyte $Agl{O}_3$ (Molecular mass $=283$), the equilibrium which sets in is
$AgI{O}_3\left(s\right)\rightleftharpoons {Ag}^{+}\left(aq\right)+{IO}^{-}\left(aq\right)$
If the solubility product constant $K_{sp}$ of $Agl{O}_3$ at a given temperature is $1.0\times {10}^{-8}$, what is the mass of $Agl{O}_3$ contained in $100mL$ of its saturated solution?

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

Answer 1

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

Solution:- (D) $2.83\times {10}^{-3}$

$AgI{O}_3\rightleftharpoons {{Ag}^{+}}_{(aq.)}+{{IO}^{-}}_{(aq.)}$

Let the solubility of $AgI{O}_3$ be $s$.

Therefore,

$K_{sp}=\left[{Ag}^{+}\right]\left[{I{O}_3}^{-}\right]$

$\Rightarrow s\times s=1.0\times {10}^{-8}$

$\Rightarrow s=\sqrt{{10}^{-8}}={10}^{-4}mol/L$

Molecular mass of $AgI{O}_3=283g$

Therefore,

$s=\frac{283\times {10}^{-4}}{1000}\times 100=2.8\times {10}^{-3}gm/100mL$