Question

A 1.00 -L. solution saturated at \(25^{\circ} \mathrm{C}\) with lead(lI) iodide contains 0.54 \(\mathrm{g}\) of \(\mathrm{Pbl}_{2}\) . Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\) .

Short answer

The solubility product constant (Ksp) for PbI2 at 25°C is approximately 8.04 × 10⁻⁹.

Step-by-step solution

Find the molar mass of PbI2

To determine the concentration, first calculate the molar mass of PbI2 using the periodic table. Molar mass of PbI2 = (207.2 g/mol for Pb) + (2 × 126.9 g/mol for I) Molar mass of PbI2 = 460.998 g/mol

Calculate the moles of PbI2 dissolved in the solution

Now we can find the moles of PbI2 dissolved in the solution using the mass of PbI2 and its molar mass. moles of PbI2 = mass / molar mass moles of PbI2 = 0.54 g / 460.998 g/mol = 0.00117 mol

Find the concentration of ions in the solution

Since the solution is 1.00 L, the concentration of PbI2 is equal to the moles of PbI2 dissolved in the solution. In a saturated solution of PbI2, the equilibrium reaction is: PbI2 (s) ⇌ Pb²⁺ (aq) + 2I⁻ (aq) From the stoichiometry of the reaction, each mole of PbI2 produces 1 mole of Pb²⁺ ions and 2 moles of I⁻ ions in the solution. Thus, the concentrations can be found as follows: [Pb²⁺] = moles of PbI2 / volume = 0.00117 mol / 1.00 L = 0.00117 M [I⁻] = 2 × moles of PbI2 / volume = 2 × 0.00117 mol / 1.00 L = 0.00234 M

Calculate Ksp using the concentrations of the ions

Now we can find the solubility product constant (Ksp) for PbI2 using the expression: Ksp = [Pb²⁺] × [I⁻]² Plug in the concentrations of the ions: Ksp = (0.00117 M) × (0.00234 M)² Ksp = 8.04 × 10⁻⁹ Thus, the solubility product constant for PbI2 at 25°C is approximately 8.04 × 10⁻⁹.

Key concepts

Molar Mass Calculation

Understanding molar mass is fundamental to solving a variety of chemistry problems. It is defined as the mass of one mole of a substance, expressed in grams per mole (g/mol). Each element in the periodic table has an atomic mass, and the molar mass of a compound can be calculated by adding up the atomic masses of its constituent elements, each multiplied by their respective number in the formula.

For example, lead(II) iodide (PbI₂) has one lead atom and two iodine atoms. Using the periodic table, the molar mass would be calculated as follows:
- Atomic mass of Lead (Pb) = 207.2 g/mol
- Atomic mass of Iodine (I) = 126.9 g/mol
The molar mass of PbI₂ is then:(1 × 207.2 g/mol) + (2 × 126.9 g/mol) = 460.998 g/mol

This calculated molar mass can then be used to figure out the number of moles in any given mass of the compound, which is crucial when dealing with concentrations and reactions in solution.

Equilibrium Constant

The equilibrium constant (K_eq), for a chemical reaction is a measure of the extent to which the reaction will proceed to form products under standard conditions. It is determined by the concentrations of the reactants and products at equilibrium. For a solubility equilibrium, this constant is specifically referred to as the solubility product constant (K_sp).

The K_sp is particularly useful for sparingly soluble salts, like lead(II) iodide. It represents the maximum product of ionic concentrations that can exist in a solution before precipitation occurs. The mathematical expression for the K_sp of PbI₂ is:K_sp = [Pb²⁺] × [I^-]²
The brackets indicate concentration in moles per liter (M). For lead(II) iodide in water at equilibrium, PbI₂ (s) ↔ Pb²⁺ (aq) + 2I^- (aq), the K_sp expression takes into account that the production of one Pb²⁺ ion is accompanied by the production of two I^- ions. This equilibrium expression is vital for predicting whether a salt will dissolve in a solution and at what concentration.

Stoichiometry

Stoichiometry lies at the heart of chemical reactions. It involves quantitatively analyzing the relationships between the amounts of reactants and products in a chemical reaction. Key to understanding stoichiometry is the balanced chemical equation, which provides the mole ratio of reactants and products.

Take lead(II) iodide's dissolution in water, represented by the equation:PbI₂ (s) ↔ Pb²⁺ (aq) + 2I^- (aq)
This indicates that each mole of PbI₂ produces one mole of Pb²⁺ ions and two moles of iodide ions. When we calculated the concentrations in our solution example, we used stoichiometry to understand that for every 0.00117 moles of dissolved PbI₂, we obtained 0.00117 moles of Pb²⁺ and 0.00234 moles of I^- in the 1.00 L solution. Such stoichiometric relationships enable chemists to predict the amount of reactants needed and products formed in any given chemical reaction.