Question

Consider a saturated solution of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s})\). If \(\left[\mathrm{Pb}^{2+}\right]\) is \(7.34 \times 10^{-14} \mathrm{M}\), find each of the following. a) \(\left[\mathrm{PO}_{4}^{3-}\right]\) b) the \(K_{\mathrm{sp}}\) of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s})\)

Short answer

\([PO_4^{3-}] = 4.894 \times 10^{-14}\, \text{M}\); \(K_{sp} = 9.474 \times 10^{-67}\).

Step-by-step solution

Understand the Dissolution Reaction

First, write the dissociation reaction for \(Pb_3(PO_4)_2\) in water. The balanced equation is: \[ Pb_3(PO_4)_2(s) \rightleftharpoons 3Pb^{2+}(aq) + 2PO_4^{3-}(aq) \] From this equation, note that the stoichiometry between \(Pb^{2+}\) and \(PO_4^{3-}\) is 3:2.

Relate Ion Concentrations Using Stoichiometry

Let the solubility of \(Pb_3(PO_4)_2\) be \(s\). From the stoichiometry of the dissolution reaction, \[ [Pb^{2+}] = 3s \] \[ [PO_4^{3-}] = 2s \] By given, \([Pb^{2+}] = 7.34 \times 10^{-14} \) M, thus \(3s = 7.34 \times 10^{-14} \), solve for \(s\).

Calculate Solubility, \(s\)

From \(3s = 7.34 \times 10^{-14}\): \\[ s = \frac{7.34 \times 10^{-14}}{3} = 2.447 \times 10^{-14} \text{ M} \]

Determine \([PO_4^{3-}]\) from Solubility

We have \([PO_4^{3-}] = 2s\): \\[ [PO_4^{3-}] = 2 \times 2.447 \times 10^{-14} = 4.894 \times 10^{-14}\, \text{M} \] So, \([PO_4^{3-}] \) is \(4.894 \times 10^{-14}\, \text{M} \).

Calculate \(K_{sp}\) Using Ion Concentrations

Since \(K_{sp}\) expression for \(Pb_3(PO_4)_2\) is \(K_{sp} = [Pb^{2+}]^3 [PO_4^{3-}]^2\), substitute the ion concentrations:\[K_{sp} = (7.34 \times 10^{-14})^3 \times (4.894 \times 10^{-14})^2\]Calculate the expression to find \(K_{sp}\).

Compute \(K_{sp}\)

Perform the calculations: \[ K_{sp} = (7.34 \times 10^{-14})^3 = 3.953 \times 10^{-40} \] \[ (4.894 \times 10^{-14})^2 = 2.396 \times 10^{-27} \] Combine these: \[ K_{sp} = (3.953 \times 10^{-40}) \times (2.396 \times 10^{-27}) = 9.474 \times 10^{-67} \] Thus, \(K_{sp} = 9.474 \times 10^{-67}\).

Key concepts

Dissolution Reaction

A dissolution reaction describes the process of a solid compound dissolving in a solvent, resulting in its constituent ions being dispersed in the solution. Understanding the dissolution reaction requires knowledge of the solubility equilibrium, illustrated with solid compounds breaking down into their ionic components.

In the case of lead phosphate, \( Pb_3(PO_4)_2 \), the dissolution reaction can be denoted as: \[ Pb_3(PO_4)_2(s) \rightleftharpoons 3Pb^{2+}(aq) + 2PO_4^{3-}(aq) \]

This equation represents the equilibrium state where the rates of dissolution and re-formation of the solid are balanced. It's essential to understand the stoichiometric ratios of the ions produced, which in this reaction are 3:2, meaning for every 3 lead ions produced, 2 phosphate ions are released.

These ratios help us determine the concentrations of the ions present in the solution once the solid reaches equilibrium.

Ion Concentration Calculations

Ion concentration calculations are crucial for understanding the extent to which a salt dissolves in a solvent, typically represented by molarity (M). We start with the stoichiometry derived from the dissolution reaction to express the relationship between the solubility \( s \) of the compound and its ionic species.

For \( Pb_3(PO_4)_2 \), if \( 3s = [Pb^{2+}] \), and knowing \([Pb^{2+}] = 7.34 \times 10^{-14} \text{ M}\), we can find \( s = \frac{7.34 \times 10^{-14}}{3} \), which gives \( s = 2.447 \times 10^{-14} \text{ M}\).

Next, using the relation \( [PO_4^{3-}] = 2s \), we determine \( [PO_4^{3-}] = 2 \times 2.447 \times 10^{-14} \), resulting in \( 4.894 \times 10^{-14} \text{ M} \).

These calculations are valuable for understanding the concentrations of ions in a saturated solution and are instrumental in further calculations like determining the solubility product constant.

Stoichiometry in Solutions

Stoichiometry in solutions refers to the quantitative relationship between reactants and products in a chemical reaction conducted in a solution. In this specific context, it relates to the ratios of ions formed when a solid salt dissolves.

Understanding this concept is crucial for calculating the individual ion concentrations in a solution based on a known amount of the dissolved compound. For \( Pb_3(PO_4)_2 \), stoichiometry tells us that the relationship between \( Pb^{2+} \) and \( PO_4^{3-} \) is a 3:2 ratio. This informs us that for every mole of \( Pb_3(PO_4)_2 \) that dissolves, three moles of \( Pb^{2+} \) ions and two moles of \( PO_4^{3-} \) ions are produced.

These stoichiometric coefficients are critical in ensuring that we accurately reflect the proportionate ion concentrations in our calculations. By understanding this concept, students can ascertain how changes in one part of a chemical equation can affect the entire system, especially in balanced chemical reactions.