Question

A volume of \(500 \mathrm{ml}\) of \(0.01 \mathrm{M}-\mathrm{AgNO}_{3}\) solution, \(250 \mathrm{ml}\) of \(0.02 \mathrm{M}-\mathrm{NaCl}\) solution and \(250 \mathrm{ml}\) of \(0.02 \mathrm{M}-\mathrm{NaBr}\) solution are mixed. The final concentration of bromide ion in the solution is \(\left(K_{\mathrm{sp}}\right.\) of \(\mathrm{AgCl}\) and \(\mathrm{AgBr}\) are \(10^{-10}\) and \(5 \times 10^{-13}\) respectively.) (a) \(0.01 \mathrm{M}\) (b) \(0.02 \mathrm{M}\) (c) \(0.005 \mathrm{M}\) (d) \(2.5 \times 10^{-5} \mathrm{M}\)

Short answer

The final concentration of bromide ion in the solution is \(0.005 \mathrm{M}\).

Step-by-step solution

Calculate Total Volume of the Solution

To find the final concentration of bromide ion, first calculate the total volume of the solution after mixing all the given solutions together. Total volume of the solution = Volume of AgNO3 solution + Volume of NaCl solution + Volume of NaBr solution = 500 ml + 250 ml + 250 ml = 1000 ml or 1 L.

Determine Moles of Bromide Before Mixing

Calculate the moles of bromide ion in the NaBr solution before mixing using the concentration and volume of NaBr solution. Moles of Br- = Molarity of NaBr x Volume of NaBr in liters = 0.02 M x 0.25 L = 0.005 moles.

Calculate the Final Concentration of Bromide Ion

Find the final concentration of bromide ion by dividing the moles of bromide ion by the total volume of the solution in liters. Final concentration of Br- = Moles of Br- / Total Volume in L = 0.005 moles / 1 L = 0.005 M.

Key concepts

Solution Concentration

When we perform experiments in chemistry, understanding the concentration of solutions is crucial as it tells us how much solute is present per unit of volume. The concentration can be expressed in several ways, but one common method is molarity, which is the number of moles of a solute per liter of solution. It allows us to compare the relative amounts of solutes in different solutions even if their volumes are not the same.

When combining solutions, as in our exercise, the final concentration of a particular ion depends on the total number of moles of that ion present in all solutions mixed and the total volume of the resulting mixture. For example, when mixing multiple solutions containing the same ion, the final concentration will be the sum of the individual contributions of that ion from each solution divided by the total volume after mixing.

Molarity Calculation

Calculating Moles and Volume

Molarity is such an essential concept in chemistry due to its direct relationship with the number of particulate entities involved in reactions. The molarity (M) can be calculated using the formula: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \].

To find the moles of a solute, you can rearrange this formula: \[ \text{moles} = \text{Molarity} \times \text{Volume in liters} \]. In the given exercise, we used this conversion to find the moles of bromide from the volume and concentration of the NaBr solution.

It’s essential to always convert the volume into the correct units (liters) when calculating molarity. The total volume impacts the final concentration, as seen in the subsequent dilution resulting from the mixture of different solutions.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, known as precipitates. This process is guided by the solubility product constant (\(K_{\text{sp}}\)), which is a measure of how much a compound can dissolve in solution before it starts precipitating. It gives us the maximal product of the ionic concentrations of a salt that can exist in a solution.

In the posed problem, if either silver chloride (\(AgCl\)) or silver bromide (\(AgBr\)) were to exceed their solubility products upon mixing the solutions, a precipitate would form. This concept is vital when analyzing the final concentration because the presence of a precipitate would mean that not all of the ions remain in solution, thus affecting the calculation of the final concentration.

Common Ion Effect

The common ion effect is a phenomenon where the solubility of an electrolyte is reduced when another solute that shares a common ion is added to the solution. This is because the added common ion shifts the equilibrium position of the solubility product reaction to the left, according to Le Chatelier's Principle, leading to less dissociation of the electrolyte and lower solubility.

In our exercise scenario, when \(AgNO_3\), \(NaCl\), and \(NaBr\) are mixed, the common ion effect does not directly influence the calculation of the final concentration of bromide ions, since we are not looking at the solubility of a bromide salt. However, in an actual laboratory setting, if a solution of silver nitrate was mixed with sodium chloride, the solubility of \(AgCl\) would decrease because of the introduction of additional \(Cl^-\) ions from the sodium chloride solution. Understanding this effect is critical when predicting whether a precipitate will form in a mixture of ionic solutions.