Question

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 $PbI_{2}$ ?

• 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 1

Answer: (B)

The equilibrium is $PbI_{2}\rightleftharpoons Pb^{2 +}+2I^{-}$

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 $\left[\right.Pb^{2 +}\left]\right.=1.3\times 10^{- 3}M$

$K_{s p}=\left[\right.Pb^{2 +}\left]\right.\left[\right.I^{-}\left]\right.^{2}=\left[\right.1.3\times 10^{- 3}M\left]\right.\left[\right.2.6\times 10^{- 3}M\left]\right.^{2}$

$K_{s p}=8.8\times 10^{- 9}.$