Question

The ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$for $\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$. Calculate the solubility of $\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$in 0.048 M NaI.

Short answer

Answer

The solubility of $\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}$in 0.048 M NaI solution is $\mathbf{6}\mathbf{.}\mathbf{07}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L}$.

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility

Calculations

The dissolution equilibrium of$\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}$is

$$\begin{gathered} \mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{P}{\mathit{b}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{l}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)} \\ {\mathit{K}}_{\mathbf{sp}}\mathbf{=}{\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}}^{\mathbf{3}}\mathbf{\left[}{\mathbf{l}}^{\mathbf{-}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{2}}^{\mathbf{-}}\mathbf{\right]}}^{\mathbf{2}} \end{gathered}$$

Let ‘s’ mol/L represent the molar solubility of $\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}$.

role="math" localid="1657867494489" $$\begin{gathered} \mathbf{\left[}\mathbf{P}{\mathbf{b}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{\text{'}}\mathit{s}\mathbf{\text{'}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left[}{\mathbf{l}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{\text{'}}\mathbf{2}\mathit{s}\mathbf{\text{'}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left[}{\mathbf{l}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{2}\mathit{s}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{048}\mathit{M} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{048}\mathbf{}\mathit{M} \end{gathered}$$

The solubility of $\mathit{P}\mathit{b}{\mathit{l}}_{\mathbf{2}}$is

$$\begin{gathered} {\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}{\left[P{b}^{2+}\right]}^{\mathbf{3}}{\left[l^{-}\right]}^{\mathbf{2}} \\ \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[s\right]{\left[0.048\right]}^{\mathbf{2}} \\ \mathit{s}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{07}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{M} \end{gathered}$$