Question

Challenge The K sp of lead carbonate (PbC O 3) is 7.40 × 1 0 -14 at 298 K. What is the solubility of lead carbonate in g/L?

Short answer

The solubility of lead carbonate (PbCO3) in water at 298 K is approximately 7.27 × 10^(-5) g/L.

Step-by-step solution

Write the Ksp expression for lead carbonate

The chemical formula for lead carbonate is PbCO3, and it dissociates into ions in the following manner: PbCO3 (s) ⟶ Pb²⁺ (aq) + CO3²⁻ (aq) The Ksp expression for this reaction is: Ksp = [Pb²⁺][CO3²⁻] Since the stoichiometry of the reaction is 1:1, we can express the concentrations of Pb²⁺ and CO3²⁻ ions as 'x', as both will have the same concentration.

Solve for the concentration of ions (x)

Set up the Ksp equation and solve for 'x'. Ksp = x² Given the value of Ksp = 7.40 x 10^(-14), we can solve for 'x': 7.40 x 10^(-14) = x² Now, solve for x: x = sqrt(7.40 x 10^(-14)) x ≈ 2.72 × 10^(-7) mol/L This is the concentration of Pb²⁺ and CO3²⁻ ions in the equilibrium.

Convert the concentration of ions to g/L

Now that we have the concentration of ions in mol/L, we can convert this into grams per liter using the molar mass of lead carbonate. The molar mass of PbCO3 is 267.2 g/mol. Solubility (g/L) = concentration (mol/L) × molar mass (g/mol) Solubility (g/L) = (2.72 × 10^(-7) mol/L) × (267.2 g/mol) Solubility ≈ 7.27 × 10^(-5) g/L Thus, the solubility of lead carbonate in water at 298 K is approximately 7.27 × 10^(-5) g/L.

Key concepts

Understanding Solubility

Solubility is an important concept in chemistry that denotes the maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature. When we talk about the solubility of a compound like lead carbonate (PbCO₃), we are referring to how much of this solid can dissolve in water to form a saturated solution.

A saturated solution is one in which the maximum amount of solute has dissolved, and any additional substance will remain undissolved in the solution. Solubility can be expressed in various units, including grams per liter (g/L) or moles per liter (mol/L). For calculating the solubility of PbCO₃, we begin by determining the concentration of the ions it produces upon dissolving.

Role of Chemical Equilibria in Solubility

Chemical equilibria play a crucial role in determining the solubility of ionic compounds like PbCO₃. When PbCO₃ dissolves in water, it sets up a dynamic equilibrium:

• The solid PbCO₃ dissolves into its constituent ions, lead (Pb²⁺) and carbonate (CO₃²⁻).
• At the same time, Pb²⁺ and CO₃²⁻ ions can recombine to form solid PbCO₃ again.

This balance between dissolution and precipitation continues until the system reaches equilibrium.

The solubility product constant (Ksp) describes the balance between the solid compound and its dissolved ions at equilibrium. For PbCO₃, the Ksp gives insight into the solubility-limit in a set volume of water. Therefore, Ksp is essential for quantitative determinations in solubility calculations.

The Concept of Molar Mass

Molar mass is a straightforward yet vital concept in chemistry that connects the mass of a substance to the amount of substance. For lead carbonate, the molar mass is calculated as follows:

• The atomic mass of lead (Pb) is approximately 207.2 g/mol.
• The atomic mass of carbon (C) is approximately 12.01 g/mol.
• The atomic mass of each oxygen (O) atom is about 16.00 g/mol, and there are three oxygen atoms in the carbonate ion (CO₃).

The total molar mass of PbCO₃ becomes the sum of these atomic masses, resulting in 267.2 g/mol.

Molar mass is critical as it allows us to convert between the number of moles of a substance and its mass, which is what we need when calculating solubility in g/L from mol/L.

PbCO3 Dissociation Explained

When considering the solubility of lead carbonate, it is essential to understand its dissociation in water. Lead carbonate is an ionic compound that dissolves according to this dissociation equation:

• PbCO₃(s) → Pb²⁺(aq) + CO₃²⁻(aq)

Each molecule of PbCO₃ that dissolves in water forms one lead ion (Pb²⁺) and one carbonate ion (CO₃²⁻).

This 1:1 stoichiometric relationship means that the concentration of the lead ions \([\text{Pb}^{2+}]\) and carbonate ions \([\text{CO}_3^{2-}]\) in a saturated solution is the same. This dictated by the stoichiometry helps to simplify calculations when determining solubility from the Ksp expression of PbCO₃.

Calculating Ions Concentration from Ksp

The concentration of ions in a saturated solution of an ionic compound such as PbCO₃ can be derived from its solubility product constant (Ksp). Given the Ksp value, calculations proceed as follows:

• First, set up the equilibrium expression for the dissociation of PbCO₃: \(K_{sp} = [\text{Pb}^{2+}][\text{CO}_3^{2-}]\).
• Since both ionic concentrations are equal at equilibrium, let them be represented by 'x': \(K_{sp} = x^2\).
• Solve for 'x' to find the concentration of ions: \(x = \sqrt{K_{sp}}\).

Applying the given Ksp of 7.40 × 10⁻¹⁴, we solve for x, resulting in \(x ≈ 2.72 \times 10^{-7}\) mol/L for both ion concentrations.

This concentration helps translate the dissolving potential of PbCO₃ in water and is a crucial step in understanding solubility in quantitative terms.