Question

$K_{sp}$ for lead iodate $\left[Pb\right(I{O}_3{)}_2]$ is $3.2\times {10}^{-14}$ at a given temperature. The solubility in mol $L^{-1}$ will be:

• A. $2.0\times {10}^{-5}$
• B. $(3.2\times {10}^{-7}{)}^{1/2}$
• C. $(3.8\times {10}^{-7})$
• D. $4.0\times {10}^{-6}$

Answer 1

The correct option is A $2.0\times {10}^{-5}$

Given for $\left[Pb\right(I{O}_3{)}_2]$, $K_{sp}=3.2\times {10}^{-14}$

Let its solubility be $S$.

$\left[Pb\right(I{O}_3{)}_2]\to P{b}^{+2}+2I{O}_3^{-}$

$S$ $2S$

$K_{sp}=S\times (2S{)}^2=4S^3=32\times {10}^{-15}$

Therefore, $S=2\times {10}^{-5}$ is the solubility in solution.