Question

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7}\) \(\operatorname{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Short answer

The solubility product constant, \(K_{sp}\), for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) at \(25^{\circ}\,\mathrm{C}\) is approximately \(1.0 \times 10^{-26}\).

Step-by-step solution

Write the balanced chemical equation for the dissolution of the compound

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\,\mathrm{Ca^{2+}(aq)} + 2\,\mathrm{PO_4^{3-}(aq)}\)

Express the solubility product constant, \(K_{sp}\)

The \(K_{sp}\) expression for the dissolution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is given by: \[K_{sp} = [\mathrm{Ca^{2+}}]^3 [\mathrm{PO_4^{3-}}]^2\]

Calculate the concentrations of ions at equilibrium using the given solubility

We know that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7}\,\mathrm{mol/L}\). Therefore, at equilibrium: - For every 1 mole of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) that dissolves, 3 moles of \(\mathrm{Ca^{2+}}\) are formed. Thus, \([\mathrm{Ca^{2+}}]_eq = 3(1.6 \times 10^{-7})\,\mathrm{mol/L} = 4.8 \times 10^{-7}\,\mathrm{mol/L}\) - For every 1 mole of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) that dissolves, 2 moles of \(\mathrm{PO_4^{3-}}\) are formed. Thus, \([\mathrm{PO_4^{3-}}]_eq = 2(1.6 \times 10^{-7})\,\mathrm{mol/L} = 3.2 \times 10^{-7}\,\mathrm{mol/L}\)

Calculate the value of \(K_{sp}\) using the ion concentrations

Now that we have the equilibrium concentrations of the ions, we can plug the values into the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Ca^{2+}}]^3 [\mathrm{PO_4^{3-}}]^2 = (4.8 \times 10^{-7})^3 (3.2 \times 10^{-7})^2\] After evaluating the expression: \[K_{sp} \approx 1.0 \times 10^{-26}\] Thus, the solubility product constant, \(K_{sp}\), for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) at \(25^{\circ}\,\mathrm{C}\) is approximately \(1.0 \times 10^{-26}\).

Key concepts

Ksp Calculation

Understanding how to calculate the solubility product constant, or Ksp, is key to predicting the solubility of ionic compounds in water. Ksp is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. When you are given the solubility of a compound, like \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) \) in moles per liter, the Ksp calculation involves taking this solubility, finding the concentrations of the ions in solution at equilibrium, and then applying these to the dissolution reaction.

For the compound \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) \) the dissolution reaction is important because it establishes a ratio of ions produced from the solid. Each mole of this compound produces three moles of calcium ions and two moles of phosphate ions. The relationship is then used in conjunction with the solubility to calculate the concentrations of each ion, which are raised to the power of their stoichiometric coefficients and multiplied together to give the Ksp. This process is succinctly demonstrated by the step-by-step solution provided, resulting in a Ksp of approximately \( 1.0 \times 10^{-26} \).

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each species in a reaction mixture when the reaction has reached a state where the rate of the forward reaction equals the rate of the reverse reaction, a point known as chemical equilibrium. At this point, the concentrations of reactants and products remain constant.

It's crucial to recognize that the initial solubility value provided in the exercise is just the starting point. To find the equilibrium concentrations of \( \mathrm{Ca^{2+}} \) and \( \mathrm{PO_4^{3-}} \) ions, we must consider how they relate to the original solid's mole ratio. For example, if 1 mole of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) dissolves, it will result in 3 moles of \( \mathrm{Ca^{2+}} \) and 2 moles of \( \mathrm{PO_4^{3-}} \) ions. These newly calculated equilibrium concentrations are plugged into the Ksp expression to solve for the solubility product constant.

Dissolution of Compounds

The process of dissolution involves the breaking down of a compound into its component ions in a solvent, such as water. In the case of sparingly soluble ionic compounds, this process is reversible and occurs until a dynamic equilibrium is established.

The solubility of an ionic compound is determined by various factors, including temperature, the polarity of the solvent, and the presence of common ions in solution. The dissolution of \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) can be written as a reversible reaction where the solid, upon reaching its solubility limit in water, exists in equilibrium with its dissociated ions. This state of balance is where the solubility product constant plays a pivotal role and is tied directly to the equilibrium concentrations of the ions in solution.

Chemical Solubility

Chemical solubility describes the extent to which a compound can dissolve in a solvent, often water. Solubility can vary greatly depending on the compound and conditions such as temperature and pH. For many salts, including \( \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2} \) in this exercise, their low solubility in water is quantified by the solubility product constant.

Understanding the concept of solubility is crucial in various scientific fields, from pharmacology, where the effectiveness of a drug can depend on its solubility, to environmental science, where solubility influences the distribution of substances in ecosystems. When the dissolution process of a compound at a given temperature reaches a point where no more solid can dissolve, the resulting ion concentrations define the solubility of that compound under those specific conditions.