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 \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Short answer

The concentration of KMnO4 required to establish the concentration of \(Ba^{2+}\) ions in the solution as \(2.0 \times 10^{-8} M\) is approximately \(1.12 \times 10^{-2} M\).

Step-by-step solution

Dissolution Reaction of Ba(MnO4)2

Write the balanced dissolution equation for barium permanganate: \(Ba(MnO4)2 (s) \rightleftharpoons Ba^{2+} (aq) + 2 MnO4^− (aq)\)

Write the expression for the solubility product constant (K_sp)

According to the dissociation equation, the expression for the solubility-product constant will be: \(K_{sp} = [Ba^{2+}] [MnO4^-]^2 \)

Calculate concentration of MnO4^− ions

Given that the concentration of \([Ba^{2+}]\) ions is \(2.0 \times 10^{-8} M\), and there are two \(MnO4^-\) ions formed for each \(Ba^{2+}\), the concentration of \(MnO4^-\) ions will be: \([MnO4^-] = 2 \times [Ba^{2+}] = 2 \times (2.0 \times 10^{-8}) \)

Solve for the concentration of KMnO4

Now that we know the concentration of \(MnO4^-\) ions and \(Ba^{2+}\) ions, we can use the \(K_{sp}\) formula to find the concentration of KMnO4 required: \((2.5 \times 10^{-10}) = (2.0 \times 10^{-8})[MnO4^−]^2\) Divide both sides by \((2.0 \times 10^{-8})\): \(\frac{2.5 \times 10^{-10}}{2.0 \times 10^{-8}} = [MnO4^−]^2\) Calculate the square root to find the concentration of \(MnO4^-\) ions: \([MnO4^-] = \sqrt{\frac{2.5 \times 10^{-10}}{2.0 \times 10^{-8}}} = 1.12 \times 10^{-2} M\) In this problem, the concentration of KMnO4 is equal to the concentration of \(MnO4^-\) ions, thus: The concentration of KMnO4 required to establish the concentration of \(Ba^{2+}\) ions in the solution as \(2.0 \times 10^{-8} M\) is approximately \(1.12 \times 10^{-2} M\).

Key concepts

Barium Permanganate Dissolution

Barium permanganate, \(\text{Ba}(\text{MnO}_4)_2\), is a slightly soluble compound that goes into solution as it dissociates into its ions. When we speak of the dissolution of barium permanganate, we refer to this process where the solid compound separates into its individual components — barium ions, \(\text{Ba}^{2+}\), and permanganate ions, \(\text{MnO}_4^-\). The balanced reaction for this dissolution can be written as:

• \(\text{Ba}(\text{MnO}_4)_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{MnO}_4^- (aq)\)

Here, we observe that one barium permanganate molecule produces one barium ion and two permanganate ions. This stoichiometry is crucial because it helps us determine the ratio in which ions are present in the solution. Understanding the dissolution at the ionic level is key to solving problems related to solubility and chemical equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction. For the dissolution of barium permanganate, once equilibrium is achieved, the rate of barium permanganate dissolving equals the rate of ions recombining to form the solid.When this equilibrium is in place, we can talk about the solubility product constant, \(K_{sp}\), which is a measure of the degree of solubility. For barium permanganate, \(K_{sp} = [\text{Ba}^{2+}][\text{MnO}_4^-]^2\). Special note should be taken that \(K_{sp}\) does not include solids, only the dissolved ions. It tells us how much of each ion is present in a saturated solution just at the point of equilibrium. Equilibrium concepts not only apply to solubility but are fundamental in various chemical processes:

• Predicting the direction of reactions.
• Understanding how changes in conditions affect system balance.

Using this understanding, we can predict how much of dissolve ion would be in the solution when other parameters are known.

Ionic Concentration Calculation

Calculating ionic concentration involves understanding relations between the dissociated ions. Here, we are particularly interested in finding out how much of \(\text{MnO}_4^-\) ion concentration we get when a certain amount of \(\text{Ba}^{2+}\) is present.After establishing the dissociation equation \(\text{Ba}^{2+} (aq) + 2 \text{MnO}_4^- (aq)\), it's clear that the amount of \(\text{MnO}_4^-\) ions is twice that of \(\text{Ba}^{2+}\) ions. This is because two permanganate ions are produced from each barium permanganate that dissolves.Steps to calculate include:

• Start with the concentration of \(\text{Ba}^{2+}\) ions.
• Determine the ion ratio from the chemical formula.
• Multiply the \(\text{Ba}^{2+}\) concentration by the stoichiometric factor to get \(\text{MnO}_4^-\).

For instance, if the \([\text{Ba}^{2+}]\) is \(2.0 \times 10^{-8} \text{ M}\), then \([\text{MnO}_4^-]\) will be \(2 \times (2.0 \times 10^{-8})\), leading to further calculations based on this value to find the concentration of compounds needed, like \(\text{KMnO}_4\). These steps ensure precise understanding and accurate determination of solubility and related parameters.