Question

Use the following solubility data to calculate a value of \(K_{\mathrm{sp}}\) for each compound: (a) \(\mathrm{CdCO}_{3} ; 1.0 \times 10^{-6} \mathrm{M}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2} ; 1.06 \times 10^{-2} \mathrm{M}\) (c) \(\mathrm{PbBr}_{2} ; 4.34 \mathrm{~g} / \mathrm{L}\) (d) \(\mathrm{BaCrO}_{4} ; 2.8 \times 10^{-3} \mathrm{~g} / \mathrm{L}\)

Short answer

\(K_{sp}\) values: (a) \(1.0 \times 10^{-12}\), (b) \(4.76 \times 10^{-6}\), (c) \(6.57 \times 10^{-6}\), (d) \(1.23 \times 10^{-10}\).

Step-by-step solution

Understand the problem

The exercise involves calculating the solubility product constant, denoted as \(K_{sp}\), for various compounds given their solubility data in water. Solubility is either provided in molarity (\(\text{M}\)) or grams per liter (\(\text{g/L}\)).

Calculate \(K_{sp}\) for \(\mathrm{CdCO}_{3}\)

Given solubility \(s = 1.0 \times 10^{-6} \text{ M}\), the dissociation of \(\mathrm{CdCO}_{3}\) in water is: \(\mathrm{CdCO}_{3}(s) \rightleftharpoons \mathrm{Cd}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\). The \(K_{sp}\) expression is \(K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{CO}_{3}^{2-}]\), which equals \(s \cdot s = (1.0 \times 10^{-6})^2 = 1.0 \times 10^{-12}\).

Calculate \(K_{sp}\) for \(\mathrm{Ca(OH)}_{2}\)

Given solubility \(s = 1.06 \times 10^{-2} \text{ M}\), the dissociation is \(\mathrm{Ca(OH)}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\). The \(K_{sp}\) expression is \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^{-}]^2 = s(2s)^2\). Thus, \(K_{sp} = 1.06 \times 10^{-2} \times (2.12 \times 10^{-2})^2 = 4.76 \times 10^{-6}\).

Convert solubility for \(\mathrm{PbBr}_{2}\) from grams to moles

The molar mass of \(\mathrm{PbBr}_{2}\) is \(207.2 + 2(79.90) = 367.0 \text{ g/mol}\). \(4.34 \text{ g/L}\) equals \(\frac{4.34}{367.0}\text{ moles/L} = 1.18 \times 10^{-2} \text{ M}\).

Calculate \(K_{sp}\) for \(\mathrm{PbBr}_{2}\)

From the previous calculation, \(s = 1.18 \times 10^{-2} \text{ M}\); the dissociation is \(\mathrm{PbBr}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{Br}^{-}(aq)\). \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Br}^{-}]^2 = s(2s)^2 = 1.18 \times 10^{-2} \times (2.36 \times 10^{-2})^2 = 6.57 \times 10^{-6}\).

Convert solubility for \(\mathrm{BaCrO}_{4}\) from grams to moles

The molar mass of \(\mathrm{BaCrO}_{4}\) is \(137.33 + 51.996 + 4(16) = 253.32 \text{ g/mol}\). \(2.8 \times 10^{-3} \text{ g/L}\) equals \(\frac{2.8 \times 10^{-3}}{253.32}\text{ moles/L} = 1.11 \times 10^{-5} \text{ M}\).

Calculate \(K_{sp}\) for \(\mathrm{BaCrO}_{4}\)

With \(s = 1.11 \times 10^{-5} \text{ M}\), the dissociation is \(\mathrm{BaCrO}_{4}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_{4}^{2-}(aq)\). \(K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_{4}^{2-}] = s^2 = (1.11 \times 10^{-5})^2 = 1.23 \times 10^{-10}\).

Key concepts

Solubility

Solubility refers to the ability of a substance (known as solute) to dissolve in a solvent, such as water, to form a homogeneous solution. This property is crucial because it determines how much of a substance can be dissolved at a given temperature and pressure. Solubility can be expressed in different units, with molarity (moles per liter) being common in chemical calculations.
To properly discuss solubility, it's important to know the form and behavior of the solute and solvent. While some compounds dissolve easily in water, others only dissolve partially or not at all.
The factors affecting solubility include:

• Nature of the solute and solvent: Like dissolves like; polar solutes generally dissolve in polar solvents.
• Temperature: For most solids, solubility increases with temperature.
• Pressure: Mainly affects gases, those more soluble at higher pressures.

In the case of our exercises, solubility data is key in calculating the solubility product constant, which allows us to understand dissolution at equilibrium.

Molarity

Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute per liter of solution, denoted as M (mol/L). This unit allows chemists to relate the amounts of substances in a chemical solution.
When given a solubility in terms of molarity, you can apply it directly in calculations, as seen in the original problem statement where \[ s = 1.0 \times 10^{-6} \, \text{M} \] for \( \mathrm{CdCO}_{3} \).
For solubility expressed in grams per liter, convert it to molarity by dividing the mass concentration by the molar mass of the compound. For example:

• For \( \mathrm{PbBr}_{2} \), first, find the molar mass (367.0 g/mol) and convert 4.34 g/L to molarity by \[ \frac{4.34 \mathrm{\, g}}{367.0 \, \mathrm{g/mol}} = 1.18 \times 10^{-2} \, \mathrm{M} \]

Molarity is pivotal in dissociation and solubility product calculations because it acts as a multiplier in solution equilibrium reactions.

Dissociation

Dissociation is the process by which an ionic compound separates into ions when it dissolves in a solvent like water. It's crucial for understanding how solubility leads to the formation of ions in solutions, influencing the properties of such solutions significantly.
For instance, consider \( \mathrm{Ca(OH)}_{2} \) dissociating in water according to the equation: \[ \mathrm{Ca(OH)}_{2} (s) \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2 \mathrm{OH}^{-} (aq) \]The stoichiometry indicates one mole of \( \mathrm{Ca(OH)}_{2} \) produces one mole of \( \mathrm{Ca}^{2+} \) and two moles of \( \mathrm{OH}^{-} \), showing how each formula unit breaks apart.
This breakdown into ions is fundamental in calculating the solubility product constant \( K_{sp} \) because it relates the concentration of ions at equilibrium to the compound's overall solubility. Understanding dissociation helps us predict how salts and minerals behave in aqueous solutions.

Solubility Product Constant (Ksp)

The solubility product constant, \( K_{sp} \), is an equilibrium constant that applies to the dissolution of a sparingly soluble ionic compound. It provides a quantitative measure of solubility, namely how much compound can dissolve to form its constituent ions in a competitive balance.
The formula for \( K_{sp} \) is derived from the equilibrium concentrations of the ions in solution. For \( \mathrm{CdCO}_{3} \), dissociating into \( \mathrm{Cd}^{2+} \) and \( \mathrm{CO}_{3}^{2-} \), the expression is:\[ K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{CO}_{3}^{2-}] = s^2 \]This indicates how these ions' concentrations relate to the solubility of the compound. Calculating \( K_{sp} \) involves squaring the molarity of the compound from its dissociation equation.
Each specific \( K_{sp} \) value reflects the degree of solubility in water, with larger values indicating higher solubility and smaller values suggesting low solubility. Thus, \( K_{sp} \) is crucial for understanding phase changes and predicting formation of precipitates in various chemical scenarios.