Question

Suppose that \(25.0 \mathrm{~mL}\) of \(0.440 \mathrm{M} \mathrm{NaCl}\) is added to \(25.0 \mathrm{~mL}\) of \(0.320 \mathrm{MAgNO}_{3}\) (a) How many moles of \(\mathrm{AgCl}\) would precipitate? (b) What would be the concentrations of each of the ions in the reaction mixture after the reaction?

Short answer

0.0080 moles of AgCl would precipitate. The concentration of Na+ would be 0.220 M and NO3- would be 0.160 M after the reaction.

Step-by-step solution

Calculate Moles of Reactants

First, calculate the moles of NaCl and AgNO3 using their concentrations and volumes. The number of moles is found by multiplying concentration (in mol/L) by volume (in L). For NaCl: Moles of NaCl = 0.440 mol/L * 0.0250 L = 0.0110 mol. For AgNO3: Moles of AgNO3 = 0.320 mol/L * 0.0250 L = 0.0080 mol.

Determine Limiting Reactant and Calculate Moles of AgCl

Use the stoichiometry of the reaction to find the limiting reactant. The reaction is 1 NaCl + 1 AgNO3 -> 1 NaNO3 + 1 AgCl. Since both reactants react in a 1:1 ratio, the limiting reactant is AgNO3 with fewer moles (0.0080 mol). The amount of AgCl that would precipitate is equal to the moles of the limiting reactant, 0.0080 mol.

Calculate the Final Concentrations

After precipitation, the total volume of the solution is the sum of the initial volumes, which is 50.0 mL or 0.0500 L. Sodium (Na+) and nitrate (NO3-) ions remain in solution. Their initial moles haven't changed, but these are now dissolved in the final volume. Concentration of Na+ = Moles of NaCl / Total Volume = 0.0110 mol / 0.0500 L = 0.220 M. Concentration of NO3- = Moles of AgNO3 / Total Volume = 0.0080 mol / 0.0500 L = 0.160 M. As AgCl precipitates, there are no Ag+ or Cl- ions remaining in the solution.

Key concepts

Limiting Reactant

Understanding the concept of the limiting reactant is much like figuring out which ingredient will run out first when cooking. In a chemical reaction, the limiting reactant is the substance that is totally consumed when the chemical reaction is complete. The remaining reactants are called excess reactants and could react further if more of the limiting reactant were available.

To determine the limiting reactant, one must compare the mole ratios of the reactants used in the reaction with those in the balanced equation. The reactant that provides the fewest moles, relative to the stoichiometry, is the limiting reactant. In our exercise, NaCl and AgNO3 react in a 1:1 ratio according to the equation, and since AgNO3 has fewer moles, it limits the formation of AgCl and, therefore, dictates the amount of product formed.

Mole Concept

The mole concept serves as a bridge between the atomic world and the laboratory measurements. A mole represents 6.022 x 10²³ entities and helps chemists relate the mass of a substance to the number of atoms or molecules it contains. By using the molarity (moles of solute per liter of solution) and volume, one can calculate the number of moles of a substance in a solution. For example, in our exercise, we used the molarity of NaCl and AgNO3 solutions and their volumes in liters to calculate the moles of each reactant present before the reaction took place. This is a fundamental concept in stoichiometry and is crucial for theoretically predicting the amounts of products and reactants in chemical reactions.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in aqueous solution to form one or more insoluble products, known as precipitates. Precipitates form due to the low solubility product of the product compared to that of the reactants. In our exercise, when NaCl and AgNO3 solutions are mixed, a white solid of AgCl precipitates out of the solution because AgCl is not soluble in water. By using the stoichiometric relationships, one can predict the amount of precipitate formed, provided that the limiting reactant is known, as is the case here with AgNO3 being the limiting reactant.

Molarity

Molarity, defined as the number of moles of solute per liter of solution, is a way to express the concentration of a solution. It plays a crucial role in quantifying the reactants and products in a chemical reaction. After a reaction involving solutions, such as our precipitation reaction, the volume of the final solution can change, which in turn affects molarity. Calculating final molarities requires accounting for changes in volume and considering that precipitates are not part of the solution. In this particular exercise, the concentrations of Na+ and NO3- ions after the precipitation are calculated by dividing their respective moles by the total volume of the combined solutions.