Question

Calculate the solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1 \times 10^{-54}\right)\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Short answer

The solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution is approximately \(1.7 \times 10^{-18}\: M\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the dissociation of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Pb}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\]

Write the expression for the \(K_{sp}\)

The \(K_{sp}\) expression for the dissolution of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is: \[K_{sp} = [\mathrm{Pb}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\]

Set up the ICE table

Initial concentrations: \([\mathrm{Pb}^{2+}] = 0.10\: M\), \([\mathrm{PO}_{4}^{3-}] = 0\: M\) Changes: Assume \(s\) is the concentration of \(\mathrm{PO}_4^{3-}\) at equilibrium. Then, \([\mathrm{Pb}^{2+}]\) will increase by \(3s\) and \([\mathrm{PO}_{4}^{3-}]\) will increase by \(2s\) since 3 moles of \(\mathrm{Pb}^{2+}\) forms for every 2 moles of \(\mathrm{PO}_{4}^{3-}\). Equilibrium concentrations: \([\mathrm{Pb}^{2+}] = 0.10 + 3s\: M\), \([\mathrm{PO}_{4}^{3-}] = 2s\: M\)

Substitute the equilibrium concentrations into the \(K_{sp}\) expression

Now, we can plug the equilibrium concentrations back into the \(K_{sp}\) expression: \(K_{sp} = (0.10 + 3s)^3(2s)^2\) Given that \(K_{sp} = 1 \times 10^{-54}\), we have: \(1 \times 10^{-54} = (0.10 + 3s)^3(2s)^2\)

Solve for the solubility

To solve for the solubility, we can approximate \(𝑠<<0.10\) because \(K_{sp}\) is so small that it is reasonable to assume that \(𝑠\) has a very small value. Therefore, \(1 \times 10^{-54} = (0.10)^3(2s)^2\) Now, solve for \(s\): \(s = \frac{\sqrt{1 \times 10^{-54}}}{0.10^{\frac{3}{2}} \times 2}\) \(s \approx 1.7 \times 10^{-18}\) The solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution is approximately \(1.7 \times 10^{-18}\: M\).

Key concepts

Chemical Equation

Understanding the chemical equation is fundamental when dealing with solubility problems. For this exercise, we focus on the dissolution of lead(II) phosphate, \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\). A chemical equation represents how substances react with each other and form products. This balanced equation shows the dissociation of the solid compound in water:\[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Pb}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\]This equation indicates that 1 mole of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) dissociates to produce 3 moles of \(\mathrm{Pb}^{2+}\) ions and 2 moles of \(\mathrm{PO}_{4}^{3-}\) ions in solution. The states of matter are also important here: \((s)\) denotes a solid, while \((aq)\) indicates the ions are in an aqueous, or water-based, solution. This dissociation is key to understanding how solubility and the solubility product, \(K_{sp}\), are calculated.

ICE Table

An ICE table is a powerful tool used to organize and visualize the changes occurring in a chemical reaction. ICE stands for Initial, Change, and Equilibrium, which represent the stages of concentration changes during the reaction.Let's examine the set up:

• **Initial concentrations:** Before dissociation, the concentration of \(\mathrm{Pb}^{2+}\) is \(0.10\: M\) due to the \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution, while \([\mathrm{PO}_{4}^{3-}]\) is \(0\: M\).
• **Change:** Assume that \(s\) is the solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\). As it dissolves, \([\mathrm{Pb}^{2+}]\) increases by \(3s\) and \([\mathrm{PO}_{4}^{3-}]\) by \(2s\).
• **Equilibrium concentrations:** At equilibrium, we have \([\mathrm{Pb}^{2+}] = 0.10 + 3s\) and \([\mathrm{PO}_{4}^{3-}] = 2s\).

The ICE table helps keep track of these values, making the calculations for the solubility product, \(K_{sp}\), more straightforward to handle.Following these steps simplifies the complex interaction and makes solving the problem easier.

Equilibrium Concentrations

Equilibrium concentrations play a crucial role in calculating the solubility of a compound. Once the initial amounts of reactants change, and the reaction reaches equilibrium, the concentration values found in the ICE table are set in the system. For the dissociation of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), we determine the equilibrium concentrations using the ICE table:

• **For Pb²⁺ ions:** \([\mathrm{Pb}^{2+}] = 0.10 + 3s\) at equilibrium.
• **For PO₄^3- ions:** \([\mathrm{PO}_{4}^{3-}] = 2s\) at equilibrium.

These concentrations directly feed into the solubility product expression, \(K_{sp}\). Understanding and accurately calculating these equilibrium values is fundamental for evaluating the solubility property of sparingly soluble salts.

Ksp Calculation

The solubility product constant, or \(K_{sp}\), is a measure of a sparingly soluble ionic compound's solubility. It represents the level at which a solute dissolves in solution to reach equilibrium. To calculate \(K_{sp}\), we insert the equilibrium concentrations from the ICE table into the \(K_{sp}\) expression:\[K_{sp} = (0.10 + 3s)^3 (2s)^2\]Given \(K_{sp} = 1 \times 10^{-54}\), we simplify the expression significantly, with an assumption that \(s\) is much smaller than \(0.10\). This simplification allows us to approximate:\[1 \times 10^{-54} = (0.10)^3 (2s)^2\]Solving for \(s\), the solubility, using this approach, gives us an approximate value of \(s \approx 1.7 \times 10^{-18}\: M\). This demonstrates that the presence of common ions in solution affects the overall solubility of the compound. Knowing \(K_{sp}\) and how to use it in these calculations is essential for anyone studying or working with chemical solutions.