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

Answer 1

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

$AgI{O}_3\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I{O}_3^{-}\left(aq\right)$
[s = Solubility]
$K_{sp}=x^x\times y^y\times s^{x+y}(\because x=No.ofA{g}^{+}ionsandy=No.ofI{O}_3^{-}ions)K_{sp}=s^2$
$\therefore s=1.0\times {10}^{-4}\text{mol/lit}=1.0\times {10}^{-4}\times 283\text{g/lit}=2.83\times {10}^{-3}\text{gm/100 ml}$