Question

In a saturated solution of the sparingly soluble strong electrolyte $AgI{O}_3$ (molecular mass = 283) the equilibrium which sets in is -
$AgI{O}_3\rightleftharpoons A{g}_{\left(aq\right)}^{+}+I{O}_{3\left(aq\right)}^{-}$.
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 $100ml$ 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

Answer: (B)

$2.83\times {10}^{-3}g$

The dissociation of $AgI{O}_3$ in the solution is represented by the following equilibrium:

$AgI{O}_3\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I{O}_3^{-}\left(aq\right)$

Let ${}^{\prime }{s}^{\prime }$ represent the solubility of $AgI{O}_3$.

The expression for the solubility product is $K_{sp}=\left[A{g}^{+}\right]\left[I{O}_3^{-}\right]$

Substitute values in the above expression:

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

Hence, $S={10}^{-4}mol/lit$

Converting the solubility in g/lit:

$S=\frac{{10}^{-4}\times 283}{1000}g/ml$

Converting solubility in g/100ml:

$S=\frac{{10}^{-4}\times 283}{1000}\times 100=2.83\times {10}^{-3}g/100ml$