Question

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\) \(\left(\mathrm{Hg}_{2}^{2+}\right.\)is the cation in solution.)

Short answer

The solubility of the given compounds are: a. Ag3PO4: \(S \approx 7.53 \times 10^{-6} \ \mathrm{mol/L}\) b. CaCO3: \(S \approx 2.95 \times 10^{-5} \ \mathrm{mol/L}\) c. Hg2Cl2: \(S \approx 6.5 \times 10^{-7} \ \mathrm{mol/L}\)

Step-by-step solution

Write the dissociation reactions

For each compound, we'll write the dissociation reaction and the expression for Ksp. a. Ag3PO4: \(\mathrm{Ag}_{3} \mathrm{PO}_{4} \rightleftarrows 3\mathrm{Ag}^{+} + \mathrm{PO}_{4}^{3-}\) b. CaCO3: \(\mathrm{CaCO}_{3} \rightleftarrows \mathrm{Ca}^{2+} + \mathrm{CO}_{3}^{2-}\) c. H2Cl2: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2} \rightleftarrows \mathrm{Hg}_{2}^{2+} + 2\mathrm{Cl}^{-}\)

Find the relationship between solubility and Ksp

Now, we'll set up the Ksp expressions for each compound. We'll assume that the solubility is represented by "S" moles per liter. a. Ag3PO4: \(K_{sp} = [\mathrm{Ag}^{+}]^{3}[\mathrm{PO}_{4}^{3-}] = (3S)^3(S) = 27S^4\) b. CaCO3: \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}] = (S)(S) = S^2\) c. H2Cl2: \(K_{sp} = [\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^2 = (S)(2S)^2 = 4S^3\)

Solve for solubility (S) in moles per liter

a. For Ag3PO4: \(K_{sp} = 1.8 \times 10^{-18} = 27S^4\) \(S^4 = \frac{1.8 \times 10^{-18}}{27}\) \(S^4 = 6.67 \times 10^{-20}\) \(S = \sqrt[4]{6.67 \times 10^{-20}} \approx 7.53 \times 10^{-6} \ \mathrm{mol/L}\) b. For CaCO3: \(K_{sp} = 8.7 \times 10^{-9} = S^2\) \(S^2 = 8.7 \times 10^{-9}\) \(S = \sqrt{8.7 \times 10^{-9}} \approx 2.95 \times 10^{-5} \ \mathrm{mol/L}\) c. For Hg2Cl2: \(K_{sp} = 1.1 \times 10^{-18} = 4S^3\) \(S^3 = \frac{1.1 \times 10^{-18}}{4}\) \(S^3 = 2.75 \times 10^{-19}\) \(S = \sqrt[3]{2.75 \times 10^{-19}} \approx 6.5 \times 10^{-7} \ \mathrm{mol/L}\)

Key concepts

Chemical Equilibrium

In the context of solubility calculations, chemical equilibrium is the state reached when the rates at which a solute dissolves in a solvent and precipitates back out are equal, resulting in a constant concentration of solute in solution.

At this point, the system is said to be at a dynamic equilibrium; no visible change occurs, but on a molecular level, there's a continuous exchange between the solute and the solvent. Understanding this concept is crucial as it provides the basis for determining how much of a substance can be dissolved in a solvent at a given temperature.

When a solid substance is in contact with a liquid, it will try to dissolve into the liquid. However, not all substances dissolve completely, bringing into play the concept of solubility. Some compounds have low solubility, meaning only a small amount can dissolve in the liquid. These compounds eventually reach a state where any additional solid will simply remain undissolved, hence establishing an equilibrium with its ions in the solution.

Ksp (Solubility Product Constant)

Ksp, or the solubility product constant, is a special type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It is the product of the concentrations of the ions of a substance in a saturated solution, each raised to the power of its coefficient in the balanced equation for the dissociation reaction.

The Ksp value gives us insight into the solubility of the compound; a higher Ksp indicates greater solubility. In the example problems provided, Ksp values for various compounds are used to calculate their solubility. The Ksp equations established from the dissociation reactions provide a quantitative method to determine how much of the substance can dissolve to form a saturated solution.

Dissociation Reactions

Dissociation reactions are the processes through which ionic compounds separate into their individual ions when dissolved in water. This is significant for solubility calculations, as the dissociation reaction forms the basis for writing the solubility product (Ksp) expression.

Here's how it works: when an ionic compound dissolves, it splits into its constituent ions. For example, \( \text{CaCO}_3 \) dissociates into \( \text{Ca}^{2+} \) and \( \text{CO}_3^{2-} \) ions. The stoichiometry of the reaction, such as the 1:1 ratio for calcium carbonate, dictates how we write the Ksp expression. Different compounds dissociate into different numbers and types of ions, and therefore, each compound will have a unique Ksp expression based on its dissociation reaction.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated. It's essentially a measure of a substance’s solubility in a particular solvent. Calculating the molar solubility helps to understand how much of the substance can be expected to dissolve under certain conditions.

The calculated solubility, represented as 'S' in the problems, is extracted by manipulating the Ksp expression. The molar solubility is especially important when dealing with reactions that involve precipitation or when assessing the level of saturation of solutions in various chemical processes.

Improving Solubility Calculations

For better comprehension, keep in mind that molar solubility is dependent on the solubility product constant and the stoichiometry of the dissociation reaction of the compound. It's also worth considering factors like temperature, which can affect the solubility of a solute, although, in the given exercise, temperature effects are to be ignored.