Q. When solid lead iodide is added to water, the equilibrium concentration of $I^{-}$ becomes $2.6\times 10^{- 3}$ M. What is the $K_{s p}$ for $PbI_{2}$ ?

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Solution:

The equilibrium is $PbI_{2}\rightleftharpoonsPb^{2 +}+2I^{-}$
On the basis of this equation, the concentration of $Pb^{2 +}$ ions will be half of the concentration of $I^{-}$ ions. Thus,
$\left[\right.I^{-}\left]\right.=2.6\times 10^{- 3}M$ and $\left[\right.Pb^{2 +}\left]\right.=1.3\times 10^{- 3}M$
$K_{s p}=\left[\right.Pb^{2 +}\left]\right.\left[\right.I^{-}\left]\right.^{2}=\left[\right.1.3\times 10^{- 3}M\left]\right.\left[\right.2.6\times 10^{- 3}M\left]\right.^{2}$
$K_{s p}=8.8\times 10^{- 9}.$