Problem 59 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

An amount of \(0.10\) moles of \(\mathrm{AgCl}(\mathrm{s})\) is added to one litre of water. Next, the crystals of NaBr are added until \(75 \%\) of the \(\mathrm{AgCl}\) is converted to \(\mathrm{AgBr}(\mathrm{s})\), the less soluble silver halide. What is \(\mathrm{Br}^{-}\) at this point? \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}=2 \times 10^{-10}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{AgBr}=4 \times 10^{-13}\) (a) \(0.075 \mathrm{M}\) (b) \(0.025 \mathrm{M}\) (c) \(1.5 \times 10^{-4} \mathrm{M}\) (d) \(0.027 \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of Br- at this point is approximately \(2 \times 10^{-6.5} \mathrm{M}\), which is closest to the provided choice (c) \(1.5 \times 10^{-4} \mathrm{M}\).

Step 1: Determine the moles of AgCl that reacted

Since 75% of the AgCl is converted to AgBr, we first need to find out how many moles of AgCl reacted. To do this, multiply the initial moles of AgCl by 75% or 0.75. So the moles of AgCl that reacted are: 0.10 moles * 0.75 = 0.075 moles.

Step 2: Calculate moles of AgBr formed

The moles of AgBr formed will be equal to the moles of AgCl that reacted, since the reaction is a 1:1 molar ratio. Hence, the moles of AgBr formed are also 0.075.

Step 3: Finding the concentration of Br- ions

When AgBr forms, it dissociates into Ag+ and Br- ions in the solution. The Ksp expression for AgBr is: \( K_{\mathrm{sp}} = [\mathrm{Ag}^+] [\mathrm{Br}^-] \). Since we need to find the concentration of Br- when 75% of AgCl has been turned into AgBr, we use the Ksp for AgBr which is \(4 \times 10^{-13}\). The concentration of Br- ions would be the same as the concentration of Ag+ ions, so we can set up the following equation: \( [\mathrm{Br}^-] = \frac{0.075}{1 \, \mathrm{L}} = 0.075 \, \mathrm{M} \), but considering the Ksp, we need to consider the solubility equilibrium.

Step 4: Write the solubility product expression for AgBr

Ksp for AgBr is given by \( K_{\mathrm{sp}} = [\mathrm{Ag}^+] [\mathrm{Br}^-] \). Assuming that the system has reached equilibrium, the Ag+ ion concentration is equal to the concentration of Br- ions.

Step 5: Solve for Br- concentration

Substitute the value of \( K_{\mathrm{sp}} \) for AgBr and let \([\mathrm{Br}^-] = [\mathrm{Ag}^+] = x\). Therefore, \(4 \times 10^{-13} = x \cdot x\). Solving for x, we get \(x^2 = 4 \times 10^{-13}\), and \(x = \sqrt{4 \times 10^{-13}} = 2 \times 10^{-6.5}\). The concentration of Br- ions is then \(2 \times 10^{-6.5} \mathrm{M}\), which should be calculated to compare with the given answers.

Step 6: Calculate the concentration of Br- in moles per liter

To convert the concentration of Br- from the previous step into moles per liter, we simply take the value of x: \(2 \times 10^{-6.5} \mathrm{M}\). This value must be compared with the provided answer choices to find the closest match.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the products and reactants remain constant over time. It's important to note that this doesn't mean the reactants and products are in equal concentrations, but rather that their ratios remain constant.

When chemicals react, they eventually form a mixture where no further change occurs in the concentrations of reactants and products. This point is what we refer to as equilibrium. In the case of solubility equilibria, such as with the conversion of AgCl to AgBr in the given exercise, the equilibrium concept is visualized as the point at which the solid silver halide's dissolution rate equals the rate at which the ions combine to form the solid compound again.

Understanding equilibrium is crucial when solving problems involving the solubility product constant, K_sp. The K_sp is an expression of the equilibrium concentrations of the ions in a saturated solution of a compound. It's a special case of the equilibrium constant that applies to solubility equilibria. In the textbook example, the problem relies on understanding that how much AgCl converts to AgBr is governed by the solubility of AgBr, as expressed by its K_sp.

Molar Concentration

Molar concentration, also known as molarity, represents the concentration of a solute in a solution. It is usually expressed in moles per liter (M). Molarity is a critical concept in chemistry as it allows scientists to quantify the amount of a substance in a given volume of liquid, which is particularly useful when mixing solutions or reacting chemicals in solution.

In practice, the molarity of a solution is calculated by taking the number of moles of solute and dividing it by the volume of the solution in liters. This is expressed mathematically as: \[ C = \frac{n}{V} \] where C is the molarity, n is the number of moles, and V is the volume in liters.

In our exercise, understanding molar concentration is necessary to solve for the concentration of Br⁻ ions. Initially, it might seem like the molarity of Br⁻ would simply be the moles of AgBr formed (0.075 moles) divided by the volume of the solution (1 liter), but this does not take into account the equilibrium nature of the system, which is where the K_sp of AgBr comes into play.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, known as a precipitates. These reactions are fundamental in analytical chemistry and also play significant roles in various fields ranging from biomedical to environmental science.

For a precipitation reaction to occur, the product of the reaction must have a very low solubility product (K_sp), leading to the formation of a solid precipitate from the ions in solution. The K_sp helps predict whether a precipitate will form under certain conditions. It is calculated as the product of the molar concentrations of the ions each raised to the power of their stoichiometric coefficients. For example, for a generic salt AB that dissociates into A⁺ and B⁻, the K_sp is expressed as: \[ K_{\mathrm{sp}} = [\mathrm{A}^+] [\mathrm{B}^-] \]

In the exercise provided, the silver halide conversion to AgBr is in essence a precipitation reaction. The amount of AgBr that can be formed is limited by its K_sp, which dictates the maximum concentration of Ag⁺ and Br⁻ ions that can exist in equilibrium in solution. The concept of precipitation reactions is essential to solve the problem correctly by understanding how the solubility limits the amount of ions in the solution.