Question

# What is the ${K}_{sp}$expression for the salt ${\mathrm{PbI}}_{2}$?

Open in App
Solution

## Step 1: Solubility productSolubility product which is denoted by ${K}_{sp}$ is a type of equilibrium constant at the saturated level of solution. It is dependent on the temperature of the system.Solubility is the amount of solute dissolved in a definite amount of solvent to form saturated solution at particular temperature.Step 2: Expression of ${\mathbit{K}}_{\mathbf{s}\mathbf{p}}$ for ${\mathbf{PbI}}_{\mathbf{2}}$A general equation of equilibrium constant for a reaction of dissociation of ${\mathrm{PbI}}_{2}$ as ${\mathrm{PbI}}_{2}⇌{\mathrm{Pb}}^{2+}+2{\mathrm{I}}^{-}$can be written as, $K=\frac{\left[P{b}^{2+}\right]{\left[{I}^{-}\right]}^{2}}{\left[Pb{I}_{2}\right]}$ Where, $K$is the equilibrium constant $\left[{\mathrm{Pb}}^{2+}\right]$,$\left[{\mathrm{I}}^{-}\right]$ and $\left[{\mathrm{PbI}}_{2}\right]$ are the concentration of ${\mathrm{Pb}}^{2+}$,${\mathrm{I}}^{-}$ and ${\mathrm{PbI}}_{2}$ respectively.But we know that ${K}_{sp}$ is the equilibrium constant at the saturated level. At saturation, no more ${\mathrm{PbI}}_{2}$will dissolve. Thus the concentration of ${\mathrm{PbI}}_{2}$ will be constant and can be taken as 1.Therefore the above equation can be written as, ${K}_{sp}=\left[P{b}^{2+}\right]{\left[{I}^{-}\right]}^{2}$Now consider the solubility of the salt ${\mathrm{PbI}}_{2}$ is $\mathrm{S}\mathrm{mol}/\mathrm{L}$. Then at saturation the solubility $\mathrm{S}$ for the reaction can be represented as below,${\mathrm{PbI}}_{2}⇌{\mathrm{Pb}}^{2+}+2{\mathrm{I}}^{-}\phantom{\rule{0ex}{0ex}}\mathrm{S}2\mathrm{S}$Therefore, ${K}_{sp}=S×{\left(2S\right)}^{2}\phantom{\rule{0ex}{0ex}}=4{S}^{3}$Thus the ${\mathbit{K}}_{\mathbf{s}\mathbf{p}}$ expression for the salt ${\mathbf{PbI}}_{\mathbf{2}}$ is ${\mathbit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\left[P{b}^{2+}\right]{\left[{I}^{-}\right]}^{\mathbf{2}}$or we can say ${\mathbit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{}\mathbf{=}\mathbf{4}{\mathbit{S}}^{\mathbf{3}}$.

Suggest Corrections
1
Related Videos
Solubility and Solubility Product
CHEMISTRY
Watch in App
Explore more