Problem 141 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A volume of \(500 \mathrm{ml}\) of a \(0.1 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) added to \(500 \mathrm{ml}\) of \(0.1 \mathrm{M}\) solution of \(\mathrm{KCl}\). The concentration of nitrate ion in the resulting solution is (a) \(0.05 \mathrm{M}\) (b) \(0.1 \mathrm{M}\) (c) \(0.2 \mathrm{M}\) (d) Reduced to zero

Step-by-Step Solution

Answer

The concentration of nitrate ion in the resulting solution is 0.05 M (option a).

Step 1: Understand the Chemical Reaction

When solutions of \(\mathrm{AgNO}_{3}\) and \(\mathrm{KCl}\) are mixed, a chemical reaction will occur to form a precipitate of \(\mathrm{AgCl}\) and a residual solution containing \(\mathrm{KNO}_{3}\). The chemical equation for the reaction is: \[\mathrm{AgNO}_{3} + \mathrm{KCl} \rightarrow \mathrm{AgCl} \downarrow + \mathrm{KNO}_{3}\]. However, for the purpose of this problem, we only need to focus on the nitrate ions.

Step 2: Calculating the Moles of Nitrate Ions

Calculate the initial moles of nitrate ions present in the \(\mathrm{AgNO}_{3}\) solution using its molarity and volume. Moles of nitrate ion \(= 0.1 M \times 0.5 L = 0.05\) moles. Since the nitrate ion is a spectator ion and does not participate in the precipitate formation, its total moles remain unchanged.

Step 3: Determine the Final Volume of the Mixture

After the solutions are mixed, the final volume is the sum of the volumes of the individual solutions, which is \(500 \mathrm{ml} + 500 \mathrm{ml} = 1000 \mathrm{ml} = 1 \mathrm{L}\).

Step 4: Calculate the Final Concentration

The concentration of nitrate ion after the reaction can be found by dividing the moles of nitrate ion by the new volume of the mixture: \[\text{Final concentration} = \frac{0.05 \text{ moles}}{1 \text{ L}} = 0.05 \text{ M}\].

Key Concepts

Nitrate Ion Concentration

Understanding the concentration of nitrate ion in a chemical solution is crucial when studying chemical reactions, particularly in reactions involving soluble salts. The nitrate ion, NO₃^-, is commonly found in many compounds and can affect the overall chemistry of a solution. In our example, we examine a reaction where silver nitrate (AgNO₃) reacts with potassium chloride (KCl), to understand the nitrate ion concentration post-reaction.

To determine the concentration of nitrate ions after the reaction, we first identify that nitrate ions in this context are spectator ions—ions that do not participate directly in the reaction, as they do not form a precipitate or change oxidation states. This knowledge helps elucidate why the amount (moles) of nitrate ions remains unchanged through the reaction. By leveraging the concept of molarity, which signifies the moles of a solute per liter of solution, we're able to calculate the molarity of nitrate ions after the solutions are mixed.

Remember, no matter what happens in the reaction vessel, as long as there’s no addition or subtraction of nitrate ions specifically, the moles of nitrate ions you start with will be the moles you end up with. Therefore, even after the precipitation reaction, their concentration only changes due to the dilution effect caused by the increased volume of the resulting solution.

Precipitation Reactions

Precipitation reactions play a fundamental role in chemistry, especially in the field of analytical chemistry and are pivotal to understanding various biochemical and geological processes. During a precipitation reaction, ions in solution form insoluble compounds, which settle out of the solution as a solid precipitate.

In our case, when solutions of AgNO₃ and KCl mix, silver ions (Ag⁺) and chloride ions (Cl^-) combine to form silver chloride (AgCl), which is insoluble in water. This reaction is represented by the following ionic equation: \[AgNO_{3(aq)} + KCl_{(aq)} \rightarrow AgCl_{(s)} \downarrow + KNO_{3(aq)}\].

The down arrow symbol ('\downarrow') indicates that the product is precipitating out of the solution. Understanding the solubility rules helps to predict whether a precipitate will form when two ionic solutions are mixed. It's also crucial to recognize that while the visible effect of the reaction is the solid precipitate, the complete reaction includes the remaining ions in solution, which in our example includes the nitrate ions that remain in the aqueous phase.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of the solute divided by the volume of the solution in liters. Molarity calculations are vital in stoichiometry as they allow chemists to relate the quantities of reactants and products in a chemical reaction.

To perform molarity calculations, one must know the amount of the substance (in moles) and the volume of the solution (in liters). For example, in our exercise, we started with 500 ml of a 0.1 M solution of AgNO₃ and calculate the moles of nitrate ions. The formula used is \[ Molarity (M) = \frac{moles\text{ of solute}}{volume\text{ (L) of solution}} \].

After the reaction, the nitrate ions are still present but the volume has changed. We then recalculate the molarity given the new volume after the solutions are combined. This step is where understanding dilution is crucial—the concentration decreases as the volume increases, assuming the number of moles remains constant. Molarity is a core concept in preparing solutions and analyzing reactions in chemistry, essential for tasks ranging from laboratory experiments to industrial chemical production.