Problem 56 - Chemistry An Atoms First Approach

Question

A solution is prepared by mixing \(50.0 \mathrm{mL}\) of \(0.10\) \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) KCl. Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_{2}(s) \text { is } 1.6 \times 10^{-5} .\right]\)

Step-by-Step Solution

Answer

The equilibrium concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions are approximately \(0.049\,M\) and \(0.498\,M\), respectively.

Step 1: Determine the initial conditions of the reaction.

Before mixing, we have the following information: - Volume of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution = \(50.0\,\mathrm{mL}\) - Concentration of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution = \(0.10\,M\) - Volume of \(1.0\,\mathrm{M}\) \(\mathrm{KCl}\) solution = \(50.0\,\mathrm{mL}\) When the solutions are mixed, the volume will add up to \(100.0\,\mathrm{mL}\). To find initial concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) in the mixture, we can use the formula: Initial concentration = \(\frac{\text{moles of solute}}{\text{total volume in liters}}\)

Step 2: Calculate initial concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions.

For \(\mathrm{Pb}^{2+}\) ions: Moles of \(\mathrm{Pb}^{2+}\) = (concentration of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution) × (volume of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution) = \((0.10\,M) \times (50.0\,\mathrm{mL})\) Initial concentration of \(\mathrm{Pb}^{2+}\) = \(\frac{0.10\,M \times 50.0\,\mathrm{mL}}{100.0\,\mathrm{mL}}\) = \(0.050\,M\) For \(\mathrm{Cl}^{-}\) ions: Moles of \(\mathrm{Cl}^{-}\) = (concentration of \(\mathrm{KCl}\) solution) × (volume of \(\mathrm{KCl}\) solution) = \((1.0\,M) \times (50.0\,\mathrm{mL})\) Initial concentration of \(\mathrm{Cl}^{-}\) = \(\frac{1.0\,M \times 50.0\,\mathrm{mL}}{100.0\,\mathrm{mL}}\) = \(0.50\,M\)

Step 3: Set up an ICE table to find equilibrium concentrations.

We have the reaction: $$\mathrm{Pb}^{2+}(aq) + 2\mathrm{Cl}^{-}(aq) \rightleftharpoons \mathrm{PbCl}_{2}(s)$$ The ICE table for this reaction is: | | \(\mathrm{Pb}^{2+}\) | \(\mathrm{Cl}^{-}\) | \(\mathrm{PbCl}_{2}\) | |---------|-----------|-----------|------------| | Initial | \(0.050\,M\) | \(0.50\,M\) | \(0\) | | Change | \(-x\) | \(-2x\) | \(+x\) | | Equil. | \(0.050-x\) | \(0.50-2x\) | \(x\) | We can use the equilibrium expression: \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2 = (0.050-x)(0.50-2x)^2\) We are given the value of \(K_{sp} = 1.6 \times 10^{-5}\).

Step 4: Solve for x.

Now we can solve the equilibrium expression for \(x\): \(1.6 \times 10^{-5} = (0.050-x)(0.50-2x)^2\) As \(K_{sp}\) is very small, we can assume that \(x\) is very small compared to \(0.050\,M\) and \(0.50\,M\). So, we can approximate by dropping \(x\) from the expression: \(1.6 \times 10^{-5} = (0.050)(0.50-2x)^2\) Solve for \(x\): \(1.6 \times 10^{-5} = (0.050)(0.25-4x)^2\) \(x \approx 7.25 \times 10^{-4}\,M\)

Step 5: Calculate equilibrium concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\).

Using the approximated value of \(x\), we can calculate the equilibrium concentrations: Equilibrium concentration of \(\mathrm{Pb}^{2+}\) = \(0.050\,M - x \approx 0.050\,M - 7.25 \times 10^{-4}\,M \approx 0.049\,M\) Equilibrium concentration of \(\mathrm{Cl}^{-}\) = \(0.50\,M - 2x \approx 0.50\,M - 2(7.25 \times 10^{-4}\,M) \approx 0.498\,M\) At equilibrium, the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions are approximately \(0.049\,M\) and \(0.498\,M\), respectively.

Key Concepts

ICE Table

An ICE table is a simple tool to track the Initial concentrations, the Change that occurs, and the Equilibrium concentrations of reactants and products involved in a chemical reaction. When you mix solutions, changes happen at the molecular level until a state of equilibrium is reached.

Let's break it down:

• **Initial**: It records the starting concentrations before the reaction begins. For instance, in our example with ext{Pb}^{2+} ext{ and } ext{Cl}^{-}, the initial concentrations were 0.050 M and 0.50 M respectively.
• **Change**: As the system moves towards equilibrium, some amount (x) is reacted. The table denotes these changes as (-x) or (+x) based on whether a species is being consumed or formed.
• **Equilibrium**: Finally, after the reaction has proceeded to equilibrium, the concentrations are (the initial -/ + the change).

An ICE table helps you predict the equilibrium position when given initial conditions and equilibrium constants. It's all about simplifying complex equations to focus on changes in concentration.

Solubility Product Constant (Ksp)

The Solubility Product Constant, or K_{sp}, is a special equilibrium constant used to describe the saturation level of a sparingly soluble ionic compound in a solution.

When a solid dissolves in a liquid, it partially breaks down into its constituent ions until the solution becomes saturated and no more solid dissolves. At this point, the rate of dissolution equals the rate of precipitation, and equilibrium is reached. For PbCl_2(s), the equilibrium expression is: K_{sp} = [ ext{Pb}^{2+}][ ext{Cl}^{-}]^2.

Here is what you need to know about K_{sp}:

• It provides essential information about the solubility of a compound.
• A smaller K_{sp} value indicates lower solubility.
• It allows prediction of whether a precipitate will form when two solutions are mixed.
• The expression only includes dissolved ions, not the solid formed.

Understanding K_{sp} helps you make accurate predictions about a compound's behavior in a solution.

Equilibrium Concentrations

At equilibrium, the concentrations of reactants and products remain constant. This doesn't mean they are equal, but rather that their ratios remain stable within a closed system. Calculating these concentrations involves using the ICE table and the K_{sp}.

In our exercise, equilibrium concentrations were determined by:

• Starting with an assumption that x, the change, is small enough to simplify the calculations.
• Using the equilibrium expression K_{sp} = (0.050-x)(0.50-2x)^2 and substituting the given K_{sp} value.
• Solving for x to find how the initial concentrations decrease.
• Subtracting x from initial concentrations to get the equilibrium values for Pb^{2+} and Cl^{-}, which were approximately 0.049 M and 0.498 M, respectively.

Knowing the equilibrium concentrations lets you understand how much of each component remains or dissolves, valuable for predicting reaction outcomes.