Question

The solubility product of a sparingly soluble salt $Ca(I{O}_3{)}_2$ is $6.5\times {10}^{-6}.$ Its solubility value in $\left(mol{L}^{-1}\right)$ is:
$(1.63{)}^{\frac{1}{3}}=1.17$

• A. $1.2\times {10}^{-4}$
• B. $1.63\times {10}^{-6}$
• C. $1.2\times {10}^{-2}$
• D. None of the above

Answer 1

The correct option is C $1.2\times {10}^{-2}$

Given $K_{sp}=6.5\times {10}^{-6}$
The dissociation equation is $Ca\left(I{O}_3{)}_2\right(s)\rightleftharpoons C{a}^{2+}(aq)+2I{O}_3^{-}(aq)1001-ss2s$

$K_{sp}=\left[C{a}^{2+}\right][I{O}_3^{-}{]}^2$
$K_{sp}=s\times (2s{)}^2=4s^3$
$K_{sp}=4s^3=6.5\times {10}^{-6}$
$\therefore 4s^3=6.5\times {10}^{-6}$
$s^3=1.63\times {10}^{-6}$
$s=0.017M$