Problem 21 - Chemistry Matter and Change

Question

Challenge The \(K_{s p}\) of lead carbonate \(\left(P b C O_{3}\right)\) is \(7.40 \times 10^{-14}\) at 298 K What is the solubility of lead carbonate in \(g / L ?\)

Step-by-Step Solution

Answer

The solubility of lead carbonate in water is approximately \(2.30 \times 10^{-5}\) g/L.

Step 1: Write the Dissolution Equation

Lead carbonate dissociates in water according to the equation:\[ PbCO_3(s) \rightleftharpoons Pb^{2+}(aq) + CO_3^{2-}(aq) \]

Step 2: Express Ksp in Terms of Ion Concentrations

The solubility product constant (\(K_{sp}\)) expression for the dissociation of lead carbonate is formulated as:\[ K_{sp} = [Pb^{2+}][CO_3^{2-}] \]

Step 3: Set Up the Solubility Expression

Let's define the solubility of lead carbonate as \(s\) mol/L. Then, the concentration of \(Pb^{2+}\) and \(CO_3^{2-}\) ions at equilibrium will each be \(s\) mol/L.

Step 4: Substitute and Solve for Solubility

From the previous step, we substitute into the \(K_{sp}\) expression:\[ K_{sp} = s \times s = s^2 \]Given \(K_{sp} = 7.40 \times 10^{-14}\), the equation becomes:\[ s^2 = 7.40 \times 10^{-14} \]Solving for \(s\), we have:\[ s = \sqrt{7.40 \times 10^{-14}} \approx 8.60 \times 10^{-8} \text{ mol/L} \]

Step 5: Convert Molar Solubility to g/L

To find the solubility in grams per liter, we need to convert the molarity into grams. The molar mass of \(PbCO_3\) is calculated as:\[ \text{Molar Mass of } PbCO_3 = 207.2 + 12.01 + 3 \times 16.00 = 267.21 \text{ g/mol} \]Hence, the solubility in grams per liter is:\[ s \times \text{Molar Mass} = 8.60 \times 10^{-8} \times 267.21 \approx 2.30 \times 10^{-5} \text{ g/L} \]

Key Concepts

Dissolution Equation

In the study of solubility, understanding the dissolution equation is key. This equation describes how a compound breaks apart into its constituent ions when it dissolves in a solvent, usually water. For the compound lead carbonate (\(PbCO_3\)), its dissolution in water can be represented by the following equation:

\[ PbCO_3(s) \rightleftharpoons Pb^{2+}(aq) + CO_3^{2-}(aq) \]
This reaction shows that solid lead carbonate dissociates into lead ions \(Pb^{2+}\) and carbonate ions \(CO_3^{2-}\). As a dynamic equilibrium is established, the rate at which the solid dissolves equals the rate at which the ions recombine to form the solid.

Understanding the dissolution equation helps us determine how many ions are produced in the solution. This is important for calculating the solubility of the compound and predicting its behavior in different chemical environments.

Solubility Product Constant (Ksp)

The solubility product constant, better known as \(K_{sp}\), is vital for quantifying a compound's solubility in water. It provides a measure of how much of the compound can dissolve before the solution becomes saturated. For lead carbonate, the \(K_{sp}\) expression is determined by the concentrations of its ions at equilibrium:

\[ K_{sp} = [Pb^{2+}][CO_3^{2-}] \]
This formula uses the concentrations of the dissociated ions, each raised to the power of their coefficients in the balanced dissolution equation. \(K_{sp}\) is a fixed value at a specified temperature, here given as \(7.40 \times 10^{-14}\) at 298 K.

A low \(K_{sp}\) value indicates that the compound is not very soluble in water. Conversely, a higher \(K_{sp}\) suggests greater solubility. It plays a crucial role in predicting whether a precipitate will form when solutions containing the constituent ions are mixed.

Molar Solubility Calculation

Calculating molar solubility involves determining the maximum number of moles of solute that can dissolve per liter of solution until saturation is achieved. For lead carbonate, the molar solubility \(s\) can be calculated using its \(K_{sp}\) value.

Since each molecule of \(PbCO_3\) that dissolves produces one \(Pb^{2+}\) ion and one \(CO_3^{2-}\) ion, the concentrations of both ions at equilibrium will be equal to \(s\). Thus, the \(K_{sp}\) expression becomes:

\[ s^2 = K_{sp} \]
Given \(K_{sp} = 7.40 \times 10^{-14}\), solving gives:
\[ s = \sqrt{7.40 \times 10^{-14}} \approx 8.60 \times 10^{-8} \text{ mol/L} \]

To express this molar solubility in practical terms, it is often converted to grams per liter (g/L), considering the molar mass of the compound. Understanding molar solubility is crucial for various applications, including chemical reactions, formulation of products, and environmental science.