# The solubility product of

Question:

The solubility product of $\mathrm{PbI}_{2}$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $\mathrm{x} \times 10^{-6} \mathrm{~mol} / \mathrm{L}$. The value of $\mathrm{x}$ is _______________.(Rounded off to the nearest integer) [ Given $\sqrt{2}=1.41$ ]

Solution:

(141)

$\mathrm{K}_{\mathrm{SP}}\left(\mathrm{PbI}_{2}\right)=8 \times 10^{-9}$

$\mathrm{PbI}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{+2}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$

$\mathrm{S}+0.1 \quad 2 \mathrm{~S}$

$\mathbf{K}_{\mathrm{SP}}=\left[\mathrm{Pb}^{+2}\right]\left[\mathbf{I}^{-}\right]^{2}$

$8 \times 10^{-9}=(\mathrm{S}+0.1)(2 \mathrm{~S})^{2} \Rightarrow 8 \times 10^{-9} \simeq 0.1 \times 4 \mathrm{~S}^{2}$

$\Rightarrow \mathrm{S}^{2}=2 \times 10^{-8}$

$\mathrm{S}=1.414 \times 10^{-4} \mathrm{~mol} / \mathrm{Lit}$

$=\mathrm{x} \times 10^{-6} \mathrm{~mol} / \mathrm{Lit} \quad \therefore \mathrm{x}=141.4 \simeq 141$