Problem 117 - Chemistry The Molecular Nature of Matter

Question

Suppose that \(50.0 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.048 \mathrm{M} \mathrm{NaCl}\) solution. (a) What mass of \(\mathrm{AgCl}\) will form? (b) Calculate the final concentrations of all of the ions in the solution that is in contact with the precipitate. (c) What percentage of the \(\mathrm{Ag}^{+}\) ions have precipitated?

Step-by-Step Solution

Answer

The mass of AgCl formed is the molar mass of AgCl multiplied by the moles of the limiting reactant. Final ion concentrations are calculated by subtracting used moles and dividing by total volume. The percentage of Ag⁺ ions precipitated is found by dividing the moles of Ag⁺ that precipitated by the initial moles and multiplying by 100%.

Step 1: Calculate moles of Ag⁺

Use the molarity and volume of the AgNO₃ solution to calculate the moles of Ag⁺ ions. Molarity (M) is moles of solute per liter of solution. So, moles = molarity × volume. Since the volume needs to be in liters, convert 50.0 mL to 0.0500 L. Moles of Ag⁺ = 0.12 M × 0.0500 L.

Step 2: Calculate moles of Cl⁻

Use the molarity and volume of the NaCl solution to calculate the moles of Cl⁻ ions. Again, convert 50.0 mL to 0.0500 L. Moles of Cl⁻ = 0.048 M × 0.0500 L.

Step 3: Determine limiting reactant

Compare the moles of Ag⁺ and Cl⁻ to find the limiting reactant. The limiting reactant will be completely consumed and will dictate how much AgCl can form.

Step 4: Calculate the mass of AgCl formed

Use the moles of the limiting reactant to calculate the mass of AgCl. The molar mass of AgCl is approximately 143.32 g/mol. Mass of AgCl = moles of limiting reactant × molar mass of AgCl.

Step 5: Calculate the final concentrations of ions

Subtract the moles of the limiting reactant used from the initial moles of both reactants. Divide by the total volume of the solution (sum of both volumes) to find the concentration of each ion.

Step 6: Calculate the percentage of Ag⁺ ions that precipitated

Use the moles of Ag⁺ that precipitated and the initial moles of Ag⁺ to calculate the percentage. Percentage = (moles precipitated / initial moles of Ag⁺) × 100%.

Key Concepts

Limiting Reactant

In stoichiometry, the limiting reactant is the reagent that is completely consumed in a chemical reaction, therefore determining the maximum amount of product that can be formed. Identifying the limiting reactant is crucial as it allows chemists to predict the amounts of products and remaining reactants after the reaction.

To find the limiting reactant, one must first calculate the number of moles of each reactant using the concept of molarity, which is the concentration of a solution expressed as moles of solute per liter of solution. For a given volume, the molarity can be multiplied by the volume in liters to find the total moles of each reactant (e.g., Moles of Ag⁺ = 0.12 M × 0.0500 L).

Once the moles are calculated, compare the ratio of the moles of reactants to the stoichiometry of the balanced chemical equation. The reactant that provides the lesser amount of product is the limiting reactant. In this case, you compare the moles of Ag⁺ and Cl⁻ to determine which one will be exhausted first, thereby limiting the formation of the product AgCl.

Molarity Calculations

The concept of molarity, often represented as 'M', is essential for many stoichiometry problems, including calculating reactant and product quantities and interpreting precipitation reactions. Molarity is defined as the number of moles of a solute divided by the volume of the solution in liters. It provides a way to easily work with the concentration of substances in solution.

To calculate molarity, you can rearrange the equation to solve for the desired variable: Molarity (M) = moles of solute / volume of solution (L). In practical exercises, such as the given problem, you must often convert milliliters to liters before performing the calculation, since molarity is based on liters. After identifying the molarities of two solutions mixed together, the next steps involve working out the new concentration once the reactions have taken place, factoring in the volume change due to the combination of the solutions.

Precipitation Reactions

A precipitation reaction occurs when two aqueous solutions are mixed together to form an insoluble solid called a precipitate. In the context of the provided exercise, when AgNO₃ and NaCl solutions are combined, AgCl, which is insoluble in water, precipitates out of the solution.

To predict and understand precipitation reactions, chemists use solubility rules, which tell us which compounds can dissolve in water and which cannot. After identifying the limiting reactant and calculating the moles of precipitate formed, we can also determine the final concentration of ions that remain in solution.

This is accomplished by subtracting the amount of each reactant that has been turned into the precipitate and then dividing by the total volume of the solution. To quantify the extent of the reaction, one can calculate the percentage of one of the reactants that has precipitated based on the initial amount present and the amount consumed to form the precipitate.