Question

When \(5.0 \mathrm{mL}\) of \(0.012 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) are mixed with \(5.0 \mathrm{mL}\) of \(0.030 \mathrm{M} \mathrm{KI},\) a yellow precipitate of \(\mathrm{PbI}_{2}(\mathrm{s})\) forms. a. How many moles of \(\mathrm{Pb}^{2+}\) are initially present?\(\quad\quad\) __________ moles b. How many moles of \(I^{-}\) are originally present?\(\quad\quad\) __________ moles c. In a colorimeter the cquilibrium solution is analyzed for \(\mathrm{I}^{-}\), and its concentration is found to be \(7 \times\) \(10^{-3}\) mole/liter. How many moles of \(\mathbf{I}^{-}\) are present in the solution ( \(10 \mathrm{ml}\) )? __________ moles d. How many moles of I - precipitated?\(\quad\quad\) __________ moles e. How many moles of \(\mathrm{Pb}^{2+}\) precipitated?\(\quad\quad\) __________ moles f. How many moles of \(\mathrm{Pb}^{2+}\) are left in solution?\(\quad\quad\) __________ moles g. What is the concentration of \(\mathrm{Pb}^{2+}\) in the equilibrium solution?\(\quad\quad\) __________ moles/liter h. Find a value for \(K_{\mathrm{sp}}\) of \(\mathrm{PbI}_{2}\) from these data.\(\quad\quad\) __________

Short answer

a. \(6.0 \times 10^{-5}\) moles b. 0.00015 moles c. \(7 \times 10^{-5}\) moles d. \(7 \times 10^{-5}\) moles e. \(3.5 \times 10^{-5}\) moles f. \(2.5 \times 10^{-5}\) moles g. \(2.5 \times 10^{-3}\) moles/liter h. \(1.225 \times 10^{-8}\)

Step-by-step solution

a. Moles of Pb²⁺ initially present

To find the moles of Pb²⁺ initially present, we can use the formula: Moles = Molarity × Volume Moles of Pb²⁺ = 0.012 mol/L × 0.005 L Moles of Pb²⁺ = \(6.0 \times 10^{-5}\) moles

b. Moles of I⁻ originally present

To find the moles of I⁻ originally present, we use the same formula as in part a: Moles of I⁻ = 0.030 mol/L × 0.005 L Moles of I⁻ = 0.00015 moles

c. Moles of I⁻ present in the solution

Given the concentration of I⁻ in the solution is 7 × 10⁻³ mol/L and the total volume of the solution is 10 mL (0.010 L). We use the formula: Moles = Concentration × Volume Moles of I⁻ = \(0.007\) mol/L × \(0.010\) L Moles of I⁻ = \(7 \times 10^{-5}\) moles

d. Moles of I⁻ precipitated

We can find the moles of I⁻ precipitated by subtracting the moles of I⁻ present in the solution from the moles of I⁻ initially present: Moles of I⁻ precipitated = Initial moles - moles in solution Moles of I⁻ precipitated = \(0.00015 - 7 \times 10^{-5}\) Moles of I⁻ precipitated = \(7 \times 10^{-5}\) moles

e. Moles of Pb²⁺ precipitated

The stoichiometry of the reaction is 1:2 (1 mole of Pb²⁺ reacts with 2 moles of I⁻). Therefore, the moles of Pb²⁺ precipitated are half the moles of I⁻ precipitated. Moles of Pb²⁺ precipitated = Moles of I⁻ precipitated ÷ 2 = \(7 \times 10^{-5} ÷ 2\) = \(3.5 \times 10^{-5}\) moles

f. Moles of Pb²⁺ left in the solution

To find the moles of Pb²⁺ left in the solution, subtract moles of Pb²⁺ precipitated from initial moles of Pb²⁺. Moles of Pb²⁺ left = Initial moles of Pb²⁺ - moles of Pb²⁺ precipitated = \(6.0 \times 10^{-5} - 3.5 \times 10^{-5}\) = \(2.5 \times 10^{-5}\) moles

g. Concentration of Pb²⁺ in the equilibrium solution

Using the moles of Pb²⁺ left and the total volume of the solution, we calculate the concentration of Pb²⁺: Concentration of Pb²⁺ = moles / volume = \(2.5 \times 10^{-5}\) moles ÷ 0.010 L = \(2.5 \times 10^{-3} \) mol/L

h. Value for Kₛₚ of PbI₂

The solubility product constant, Kₛₚ, for PbI₂ is given by: Kₛₚ = [Pb²⁺] × [I⁻]² Given that the [Pb²⁺] = 2.5 × 10⁻³ M and [I⁻] = 7 × 10⁻³ M: Kₛₚ = \(2.5 \times 10^{-3}\) × \((7 \times 10^{-3})^2\) = \(2.5 \times 10^{-3}\) × \((49 \times 10^{-6})\) = \(1.225 \times 10^{-8}\)

Key concepts

Stoichiometry Calculations

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. In stoichiometry calculations, we often use the mole concept, which provides a way of quantifying substances based on the number of particles they contain. One common stoichiometry problem involves determining the amount of reactants required to produce a certain amount of product, or vice versa.

As an example from the exercise, when calculating the moles of a given substance, the equation \( \text{Moles} = \text{Molarity} \times \text{Volume} \) is employed. Knowing the molarity (moles per liter) and volume in liters, the number of moles can be easily found. This is especially useful when dealing with reactions in solution, as our exercise demonstrates. Once the number of moles of each reactant is known, we can use the stoichiometry of the reaction to determine how much of each reactant will actually take part in the reaction and form the product. This reaction stoichiometry is based on the balanced chemical equation, which defines the ratio in which the reactants combine to form products.

Solubility Product Constant (Ksp)

The solubility product constant (\(K_{sp}\)), is vital for understanding chemical equilibria involving solid substances dissolving in solution. It is a special kind of equilibrium constant that applies to the dissolution of sparingly soluble salts. The \(K_{sp}\) value is an indication of the degree to which a compound will dissolve and thereby indicates its solubility.

The general form for the solubility product expression for a salt \(AB_{n}\) can be written as \(K_{sp} = [A^+]^m[B^-]^n\), where \(m\) and \(n\) are the stoichiometric coefficients of the ions in the chemical equation, and \([A^+]\) and \([B^-]\) represent the molar concentrations of these ions at equilibrium. The exponents are determined by the ratio in which ions appear in the balanced dissolution equation. The exercise provided shows an application of the \(K_{sp}\) concept, where the solubility product of \(PbI_2\) is calculated using the equilibrium concentrations of its ions, \(Pb^{2+}\) and \(I^-\).

Molarity and Concentration

Molarity is a measure of concentration expressed as moles of a substance per liter of solution. It quantifies the concentration of a solute in a solution and is a fundamental concept in chemistry when working with solutions in the laboratory. Molarity (\(M\)) is calculated by dividing the number of moles of solute by the volume of the solution in liters: \(M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\).

Molarity is especially important when trying to understand how substances will react in a solution since the rate and extent of the reactions often depend on the concentration of the reactants. In the context of the provided exercise, understanding molarity aids in calculating the initial amounts of reactants, the extent of their reaction, and the concentration of the ions remaining in solution upon reaching equilibrium. These calculations are crucial for quantifying the resultant precipitate and determining the remaining concentration of reactants.