Question

A solution is prepared by dissolving 0.090 g of ${\mathbf{PbI}}_2$ in 1.00 L of hot water and cooling the solution to 25°C. Will solid precipitate result from this process, according to the solubility product expression? Explain.

Short answer

In the formation of the given solution,${\mathrm{Q}}_{\mathrm{sp}}<{\mathrm{K}}_{\mathrm{sp}}$ .

So, no precipitate will be formed during this process.

Step-by-step solution

The relation between Qspand Ksp

The value ${\mathbf{Q}}_{\mathrm{sp}}$represents the solubility quotient of a solution, and it represents the current state of the solution (not necessarily the saturated solution).

The value ${\mathbf{K}}_{\mathbf{sp}}$represents the solubility product for a saturated solution.

The relation between ${\mathbf{Q}}_{\mathrm{sp}}$and ${\mathbf{K}}_{\mathbf{sp}}$ helps in determining the formation of a precipitate.

The Qsp value of lead iodide

The amount of lead iodide that is dissolved is 0.090 g.

The molar mass of lead iodide is 461.01 g/mol.

Its solubility value is calculated as shown below:

$\begin{array}{c}\mathrm{s}=\frac{0.090\mathrm{g}}{461.01\mathrm{g}/\mathrm{mol}}\times \frac{1}{1.0\mathrm{L}}\\ =1.95\times {10}^{-4}\mathrm{mol}/\mathrm{L}\end{array}$

The dissociation of barium chromate solution is as shown below:

$\begin{array}{c}{\mathrm{PbI}}_2\left(\mathrm{s}\right)\to {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^{2+}\right]{\left[2{\mathrm{I}}^{-}\right]}^2\\ =4{(1.95\times {10}^{-4})}^3\\ =2.96\times {10}^{-11}\end{array}$

Explanation

The solubility product value of lead iodide is$1.4\times {10}^{-8}\mathrm{at}25°\mathrm{C}$ .

Hence, ${\mathrm{Q}}_{\mathrm{sp}}<{\mathrm{K}}_{\mathrm{sp}}$.

So, no precipitate will be formed during this process.