Question

What are \(\left[\mathrm{Pb}^{2+}\right]\) and \(\left[\mathrm{I}^{-}\right]\) in a saturated solution of \(\mathrm{PbI}_{2}\) ? The \(K_{\mathrm{sp}}\) for \(\mathrm{PbI}_{2}\) is \(9.8 \times 10^{-9}\).

Short answer

\(\left[\mathrm{Pb}^{2+}\right] = 1.34 \times 10^{-3}\, \text{mol/L}, \left[\mathrm{I}^{-}\right] = 2.68 \times 10^{-3}\, \text{mol/L}\).

Step-by-step solution

Understanding the Problem

We need to find the concentration of ions \(\mathrm{Pb}^{2+}\) and \(\mathrm{I}^{-}\) in a saturated solution of \(\mathrm{PbI}_2\). The solubility product constant (\(K_{sp}\)) is given as \(9.8 \times 10^{-9}\).

Setting the Dissociation Equation

The dissociation of \(\mathrm{PbI}_{2}\) in water can be represented as: \[\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^{-}(aq)\] This shows that for every mole of \(\mathrm{PbI}_2\) that dissolves, one mole of \(\mathrm{Pb}^{2+}\) and two moles of \(\mathrm{I}^{-}\) are produced.

Expressing Concentrations in Terms of Solubility

Let \(s\) be the solubility of \(\mathrm{PbI}_2\) in mol/L. This means \(\left[\mathrm{Pb}^{2+}\right] = s\) and \(\left[\mathrm{I}^{-}\right] = 2s\).

Writing the Solubility Product Expression

The solubility product expression for \(\mathrm{PbI}_{2}\) is: \(K_{sp} = \left[\mathrm{Pb}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\) Substituting the expressions we get: \(9.8 \times 10^{-9} = s \cdot (2s)^2\).

Solving the Solubility Product Equation

Replace \(\left[\mathrm{I}^{-}\right] = 2s\) into the expression: \[(2s)^2 = 4s^2\] Therefore, \[9.8 \times 10^{-9} = s \cdot 4s^2 = 4s^3\] Simplifying yields: \[(4s^3 = 9.8 \times 10^{-9}) \Rightarrow s^3 = \frac{9.8 \times 10^{-9}}{4} \Rightarrow s^3 = 2.45 \times 10^{-9}\] Taking the cube root gives: \(s = (2.45 \times 10^{-9})^{1/3} \approx 1.34 \times 10^{-3}\) mol/L.

Calculating Ion Concentrations

Now, we use \(s\) to find the concentrations of the ions: \[\left[\mathrm{Pb}^{2+}\right] = s = 1.34 \times 10^{-3} \, \text{mol/L}\] \[\left[\mathrm{I}^{-}\right] = 2s = 2 \times 1.34 \times 10^{-3} = 2.68 \times 10^{-3} \, \text{mol/L}\]

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, or \(K_{sp}\), is a crucial concept when discussing solubility equilibrium. It is a numerical value that indicates the extent to which a compound can dissolve in water. The \(K_{sp}\) is specific to each compound and varies with temperature. It provides insight into the maximum concentration of ions that a solution can hold while remaining at equilibrium.
To understand \(K_{sp}\), consider it as a threshold; when the product of the ionic concentrations exceeds this constant, precipitation occurs. In the case of \(\text{PbI}_2\), the \(K_{sp}\) given as \(9.8 \times 10^{-9}\) tells us about the relative solubility of lead(II) iodide in water. It indicates a very low solubility, meaning \(\text{PbI}_2\) will not dissolve much in water.
In general, if you know the \(K_{sp}\), you can calculate the solubility of the compound and the concentrations of its ions in a saturated solution, as seen in the exercise.

Ionic Concentrations

Ionic concentrations are the amounts of individual ions present in a solution, expressed in moles per liter (mol/L). When a solid salt dissolves in water, it dissociates into its constituent ions. The concentration of these ions is what we seek to find in solubility problems.
For \(\text{PbI}_2\), upon dissolving, it splits into one \(\text{Pb}^{2+}\) ion and two \(\text{I}^-\) ions for every formula unit. In a saturated solution, the concentration of \(\text{Pb}^{2+}\) ions is equal to the solubility \(s\) of \(\text{PbI}_2\), and the concentration of \(\text{I}^-\) ions is \(2s\) because there are two iodide ions produced for each unit of compound.
Calculating these concentrations allows us to understand the behavior of ions in solution and to predict how changes in conditions might affect solubility and precipitate formation.

Dissolution Equation

The dissolution equation represents how a solid compound breaks down into its ions in water. It is essential to understand these equations because they form the basis for calculating solubility and ionic concentrations.
For example, the dissolution of lead(II) iodide is represented by: \[\text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq)\]
This equation tells us that every mole of \(\text{PbI}_2\) that dissolves generates one mole of \(\text{Pb}^{2+}\) and two moles of \(\text{I}^-\) ions. Notice how the coefficients in the balanced equation help determine the relationship between the ions formed.
By setting up this equation, the connection between the solubility of the compound and the concentrations of its ions becomes more apparent, which is vital for calculations involving \(K_{sp}\) and predicting solubility.