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In the question it is given that the solubility product (\[{{K}_{sp}}\]) of lead iodide is \[1.4\times {{10}^{-8}}\].

The molecular formula of lead iodide is\[Pb{{I}_{2}}\].

\[Pb{{I}_{2}}\] ionizes in an aqueous solution as follows.

\[Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}}\]

Assume the solubility of \[Pb{{I}_{2}}\] is S. Then

\[\begin{align}

& Pb{{I}_{2}} \longrightarrow P{{b}^{2+}}+2{{I}^{-}} \\

& S \quad \quad \quad S \quad \quad \quad 2S \\

\end{align}\]

We have to calculate the solubility of lead iodide in 0.1 M KI (potassium iodide) solution.

We know that potassium iodide is a strong electrolyte and it will ionize completely in an aqueous solution and releases iodide ions (\[{{I}^{-}}\]=0.1 M).

Therefore the concentration of lead and iodide ions in an aqueous solution is as follows.

\[\begin{align}

& [P{{b}^{2+}}]=S\text{ and } \\

& [{{I}^{-}}]=(2S+0.1)M \\

\end{align}\](Here S = solubility of respective ions)

Now we have to substitute all these values in solubility product formula

\[\begin{align}

& {{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}} \\

& 1.4\times {{10}^{-8}}=\text{ }S\text{ }\times \text{ }{{(2S+0.1)}^{2}} \\

& S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}} \\

\end{align}\]

Therefore the molar solubility of lead iodide in 0.1 M KI solution is \[S=1.4\times {{10}^{-6}}mol.\text{ }{{L}^{-1}}\].

Iodide is going to release from lead iodide and potassium iodide into the aqueous solution. So, whenever we are going to take the concentration of iodide in the solution we have to include the concentration of iodide from both lead iodide and potassium iodide.

That’s why\[[{{I}^{-}}]=(2S+0.1)M\].

2S = from lead iodide

And 0.1 = form potassium iodide.