Problem 96 - Chemistry: The Central Science

Question

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4} .\) What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} M\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Step-by-Step Solution

Answer

To establish a concentration of \(2.0 \times 10^{-8} M\) for the \(Ba^{2+}\) ion in solution, a concentration of \(1.12 \times 10^{-1} M\) of \(KMnO_4\) is required.

Step 1: Write the balanced chemical equation and equilibrium constant expression for the dissolution of barium permanganate

First, we need to write the balanced chemical equation for the dissolution of barium permanganate in water: \[Ba(MnO_4)_2(s) \rightarrow Ba^{2+}(aq) + 2 MnO_4^-(aq)\] Next, we write the equilibrium constant expression for this reaction, which is the solubility-product constant (Kₛₚ): \[K_{sp} = [Ba^{2+}][MnO_4^-]^2\]

Step 2: Substitute the given values of Kₛₚ and [Ba²⁺] into the Kₛₚ expression

We are given that Kₛₚ = 2.5 x 10⁻¹⁰ and the concentration of Ba²⁺ in solution is 2.0 x 10⁻⁸ M. We can substitute these values into the Kₛₚ expression: \[2.5 \times 10^{-10} = (2.0 \times 10^{-8})[MnO_4^-]^2\]

Step 3: Solve for the concentration of MnO₄⁻ ions in solution

Now, we need to solve the equation for the concentration of MnO₄⁻ ions in the solution: \[\frac{2.5 \times 10^{-10}}{2.0 \times 10^{-8}} = [MnO_4^-]^2\] \[1.25 \times 10^{-2} = [MnO_4^-]^2\] To find the concentration of MnO₄⁻ ions, we take the square root of both sides of the equation: \[[MnO_4^-] = \sqrt{1.25 \times 10^{-2}}\] \[[MnO_4^-] = 1.12 \times 10^{-1} \, M\]

Step 4: Determine the concentration of KMnO₄ required

Since KMnO₄ and MnO₄⁻ have a 1:1 stoichiometric ratio, their concentrations will be equal in solution. Thus, the concentration of KMnO₄ required to establish a concentration of 2.0 x 10⁻⁸ M for the Ba²⁺ ion in solution is: \[[KMnO_4] = [MnO_4^-] = 1.12 \times 10^{-1} \, M\]

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentration of reactants and products remains constant over time, not that they are necessarily equal. In the case of the solubility of barium permanganate (\r\(Ba(MnO_4)_2\)), equilibrium is reached when the solid compound dissolves in water at the same rate at which the dissolved ions come together to form the solid compound.

\rIn an equilibrium state, the concentrations of all reactants and products are related by the equilibrium constant expression. For the dissolution of \r\(Ba(MnO_4)_2\), this constant is the solubility product constant (\r\(K_{sp}\)), which only includes the concentrations of the dissolved ions. The solid phase does not appear in the expression because its activity is constant and incorporated into the \r\(K_{sp}\) value. The \r\(K_{sp}\) reflects the extent to which a compound will dissolve, and thus, lower values indicate less soluble compounds.

Ionic Product

The ionic product refers to the mathematical product of the concentrations of the ions involved in the equilibrium, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. This product is compared to the solubility-product constant (\r\(K_{sp}\)) to predict whether a precipitate will form or dissolve. When the ionic product is equal to \r\(K_{sp}\), the system is at equilibrium, and no net change occurs in the concentrations of the ions in solution.

\rWhen solving for the ionic concentration of \r\(MnO_4^-\r\) in our barium permanganate problem, we calculated the ionic product to find that the concentration of \r\(KMnO_4\r\) required to maintain equilibrium at a certain \r\(Ba^{2+}\r\) concentration was \r\(1.12 \times 10^{-1} \r\) M. This demonstrates how understanding the ionic product can influence the addition of reagents in experimental design to maintain the desired conditions.

Concentration of Ions

The concentration of ions in a solution is crucial for calculating the ionic product and understanding chemical equilibria. It denotes the amount of an ion present in a unit volume of solution, usually expressed in moles per liter (M). The concentration can dictate the direction of the shift in equilibrium according to Le Chatelier's principle. For instance, increasing the concentration of \r\(Ba^{2+}\r\) ions would shift the equilibrium of our \r\(Ba(MnO_4)_2\r\) dissolution toward the formation of more \r\(MnO_4^-\r\) ions in accordance to the equilibrium constant expression.

\rIn the context of the problem, knowing that the concentration of \r\(Ba^{2+}\r\) ions is \r\(2.0 \times 10^{-8} M\r\), allowed us to work backwards to determine the concentration of \r\(KMnO_4\r\) needed. The control over ion concentrations is crucial for the precision of chemical processes in industrial and laboratory settings alike.