Question

The solubility of \(\mathrm{PbCl}_{2}\) in water is \(1.62 \times 10^{-2} \mathrm{M}\). Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{PbCl}_{2}\). Assume that there are no reactions other than the \(K_{\mathrm{sp}}\) reaction.

Short answer

K_sp = 1.7 × 10^{-5}

Step-by-step solution

Understand the Problem

The task is to find the solubility product constant, \(K_{\mathrm{sp}}\), for \(\mathrm{PbCl}_2\), using its solubility in water, which is given as \(1.62 \times 10^{-2}\ \mathrm{M}\). \(\mathrm{PbCl}_2\) dissolves in water as \(\mathrm{PbCl}_2(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq)\).

Key concepts

Solubility Calculations

Solubility calculations help us understand how much of a solute can dissolve in a solvent until the solution becomes saturated. In this case, we are interested in calculating the solubility product constant (K_{sp}) of lead(II) chloride (PbCl_2).
Given the solubility of PbCl_2 is 1.62 \times 10^{-2} mol/L, we start by writing the dissolution equation:\[ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq) \]From the stoichiometry of the equation, when PbCl_2 dissolves, it produces 1 mole of Pb^{2+} and 2 moles of Cl^{-} for each mole of PbCl_2 dissolved. Therefore, if the solubility of PbCl_2 is 1.62 \times 10^{-2} M, the concentration of Pb^{2+} ions is also 1.62 \times 10^{-2} M, and the concentration of Cl^{-} ions will be twice that, i.e., 3.24 \times 10^{-2} M.
The solubility product constant is calculated by the expression:\[ K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^{-}}]^2 \]Substituting the concentrations derived:\[ K_{sp} = (1.62 \times 10^{-2})(3.24 \times 10^{-2})^2 \]This calculation provides the K_{sp} value, a vital parameter in assessing the solubility equilibrium of ionic compounds.

Chemical Equilibrium

Chemical equilibrium is the state in which the concentrations of reactants and products remain constant over time, indicating that both the forward and reverse reactions occur at the same rate.
For PbCl_2 in water, the equilibrium is described as:\[ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq) \]Initially, no dissolved ions are present in the water. As PbCl_2 starts dissolving, the concentrations of Pb^{2+} and Cl^{-} ions increase. When the rate of dissolution equals the rate of precipitation, the system reaches equilibrium.
At equilibrium, the solubility product (K_{sp}) expression relates the concentrations of ions in a saturated solution. K_{sp} serves as an indicator of how much PbCl_2 can dissolve: a larger K_{sp} value implies greater solubility.
Understanding chemical equilibrium concepts is key to predicting how changes (such as temperature or common ions) affect the solubility of compounds like PbCl_2.

Ionic Compounds in Water

Ionic compounds, such as PbCl_2, are comprised of positively and negatively charged ions held together by ionic bonds. When these compounds are introduced into water, a polar solvent, the molecules of water interact with and disrupt the ionic bonds. This process is known as dissociation.
During dissociation of PbCl_2, the solid compound decomposes into its constituent ions:\[ \mathrm{PbCl_2}(s) \rightarrow \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq) \]In water, each positive lead ion (Pb^{2+}) is surrounded by the negative ends of the water molecules, and each chloride ion (Cl^{-}) is surrounded by the positive ends. This hydration process stabilizes the ions in solution, allowing the ionic compound to stay dissolved.
The capacity of water to dissolve ionic compounds depends on the strength of the ionic bonds compared to the interactions between ions and the surrounding water molecules. The solubility product, K_{sp}, helps quantify these interactions and provides insight into the conditions necessary for the compound to dissolve.