# When solid lead iodide is added to water, the equilibrium concentration of $I^-$ becomes $2.6 \times 10^{-3}$M. What is the $K_{sp}$ for $PbI_2$

(a) $2.2 \times 10^{-9}$

(b) $8.8 \times 10^{-9}$

(c) $1.8 \times 10^{-8}$

(d) $3.5 \times 10^{-8}$

Answer: $8.8 \times 10^{-9}$

$PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$

On the basis of this equation, the concentration $Pb^{2+}$ ions will be half of the concentration of $I^-$ ions.
Thus,
$[I^-] = 2.6 \times 10^{-3}$ M and $[Pb^{2+}] = 1.3 \times 10^{-3}$ M

$\therefore$ solubility product $K_{sp} = [Pb^{2+}][I^-]^2 = (1.3 \times 10^{-3}) \times (2.6 \times 10^{-3})^2 = 8.8 \times 10^{-9}$

answered Nov 27, 2013