Problem 103 - Chemistry: The Central Science in SI Units, Global Edition

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?

Step-by-Step Solution

Answer

The required concentration of \(\mathrm{KMnO}_{4}\) to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in the solution is \(7.94 \times 10^{-2} \ \mathrm{M}\).

Step 1: (Write the chemical equation)

The dissolution of barium permanganate in water can be represented by the following equation: \[ \mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + 2\mathrm{MnO}_{4}^{-}(aq) \]

Step 2: (Write the equilibrium constant expression)

The solubility-product constant expression (Ksp) for this reaction is given by: \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{MnO}_{4}^{-}]^{2} \] #Step 2: Use the given \(\mathrm{Ba}^{2+}\) concentration to find the \(\mathrm{MnO}_{4}^{-}\) concentration#

Step 3: (Substitute the given concentration into the equation)

We are given the concentration of \(\mathrm{Ba}^{2+}\) as \(2.0 \times 10^{-8}\ \mathrm{M}\) and the Ksp as \(2.5 \times 10^{-10}\). Hence, we substitute these values into the equation to find the \(\mathrm{MnO}_{4}^{-}\) concentration: \[ 2.5 \times 10^{-10} = (2.0 \times 10^{-8})[\mathrm{MnO}_{4}^{-}]^{2} \]

Step 4: (Calculate the concentration of \(\mathrm{MnO}_{4}^{-}\))

To find the \(\mathrm{MnO}_{4}^{-}\) concentration, we solve for it in the equation: \[ [\mathrm{MnO}_{4}^{-}]^{2} = \frac{1.25 \times 10^{-10}}{2.0 \times 10^{-8}} \] \[ [\mathrm{MnO}_{4}^{-}] = \sqrt{6.25 \times 10^{-3}} \] \[ [\mathrm{MnO}_{4}^{-}] = 7.94 \times 10^{-2} \ \mathrm{M} \] #Step 3: Determine the concentration of \(\mathrm{KMnO}_{4}\) to produce the required \(\mathrm{MnO}_{4}^{-}\) concentration#

Step 5: (Write the reaction of \(\mathrm{KMnO}_{4}\) dissociation)

The dissociation of \(\mathrm{KMnO}_{4}\) in water can be represented as: \[ \mathrm{KMnO}_{4}(s) \rightleftharpoons \mathrm{K}^{+}(aq) + \mathrm{MnO}_{4}^{-}(aq) \]

Step 6: (Determine the required concentration of \(\mathrm{KMnO}_{4}\))

Since the dissociation of one mole of \(\mathrm{KMnO}_{4}\) produces one mole of \(\mathrm{MnO}_{4}^{-}\), in this case, the concentration of \(\mathrm{KMnO}_{4}\) will be equal to the concentration of \(\mathrm{MnO}_{4}^{-}\): \[ [\mathrm{KMnO}_{4}] = [\mathrm{MnO}_{4}^{-}] \] Therefore, the required concentration of \(\mathrm{KMnO}_{4}\) is: \[ [\mathrm{KMnO}_{4}] = 7.94 \times 10^{-2} \ \mathrm{M} \]

Key Concepts

Equilibrium

In chemistry, equilibrium refers to a balanced state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction.
For the dissolution of barium permanganate, represented as \( \mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + 2\mathrm{MnO}_{4}^{-}(aq) \), equilibrium is reached when the concentrations of \( \mathrm{Ba}^{2+} \) and \( \mathrm{MnO}_{4}^{-} \) ions in solution remain constant over time.
At this point, the system is said to be in dynamic equilibrium, where the dissolution and precipitation of barium permanganate occur at the same rate, maintaining a stable concentration of ions.The solubility product constant, \( K_{sp} \), is a specific type of equilibrium constant that applies to the dissolution of sparingly soluble salts.

• It provides a measure of how much solute can dissolve in a solution before it becomes saturated.
• In our example, the \( K_{sp} \) value of \( 2.5 \times 10^{-10} \) indicates a low solubility of barium permanganate in water.

Understanding equilibrium and the \( K_{sp} \) concept is crucial for calculating ion concentrations in such solutions.

Chemical Equations

Writing chemical equations is fundamental in understanding chemical equilibrium systems.
In the case of barium permanganate, the chemical equation \( \mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + 2\mathrm{MnO}_{4}^{-}(aq) \) captures the dissolution process.

This equation is essential as it represents the balanced chemical reaction, showing the stoichiometry or the proportion of substances involved.

• Reactants, found on the left, transition into products on the right.
• In this equation, one unit of barium permanganate dissociates to yield one barium ion and two permanganate ions.

Moreover, chemical equations allow us to construct expressions for equilibrium constants, like the solubility product, \( K_{sp} \).
This is done by using the concentrations of the products in solution. For barium permanganate, \( K_{sp} \) is expressed as \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{MnO}_{4}^{-}]^{2} \].

Here, it considers the stoichiometry where the concentration of \( \mathrm{MnO}_{4}^{-} \) is squared because two moles of \( \mathrm{MnO}_{4}^{-} \) are produced for each mole of \( \mathrm{Ba}^{2+} \) formed.

Dissociation

Dissociation refers to the process where molecules split into smaller particles, often ions, when dissolved in water.
For barium permanganate \( \mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2} \), dissociation occurs as it dissolves, producing \( \mathrm{Ba}^{2+} \) and \( \mathrm{MnO}_{4}^{-} \) ions.

Similarly, when \( \mathrm{KMnO}_{4} \) dissolves, it dissociates into \( \mathrm{K}^{+} \) and \( \mathrm{MnO}_{4}^{-} \) ions.

• Dissociation is vital for determining the concentration of various ions in solution, which in turn affects chemical equilibrium.
• The extent to which a compound dissociates directly influences the concentrations of ions available to participate in reactions.

The complete dissociation of \( \mathrm{KMnO}_{4} \) implies that the concentration of \( \mathrm{MnO}_{4}^{-} \) ions becomes directly equivalent to that of the \( \mathrm{KMnO}_{4} \) in solution.
This step is crucial in calculating how much \( \mathrm{KMnO}_{4} \) is needed to achieve the desired concentration of \( \mathrm{MnO}_{4}^{-} \).

Ion Concentration

Ion concentration is a pivotal concept when exploring solubility and equilibrium.
It refers to the amount of ions present in a solution.

For the reaction involving barium permanganate, we're interested in calculating the concentration of \( \mathrm{MnO}_{4}^{-} \) to maintain the equilibrium with \( \mathrm{Ba}^{2+} \) ions at a specified level.

• The given \( \mathrm{Ba}^{2+} \) concentration is used with the \( K_{sp} \) to find \( \mathrm{MnO}_{4}^{-} \) concentration.
• The expression \( K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{MnO}_{4}^{-}]^{2} \) helps us find this context-specific concentration.

Solving for \([\mathrm{MnO}_{4}^{-}]\), that is needed to balance the given \([\mathrm{Ba}^{2+}]\), we find it using:
\[ [\mathrm{MnO}_{4}^{-}] = \sqrt{ \frac{K_{sp}}{[\mathrm{Ba}^{2+}]} } \]
It's important to remember that the concentration of \( \mathrm{MnO}_{4}^{-} \) also directly determines the \( \mathrm{KMnO}_{4} \) concentration since one mole of \( \mathrm{KMnO}_{4} \) produces one mole of \( \mathrm{MnO}_{4}^{-} \).
Understanding ion concentration helps students predict how much of a compound is needed to achieve particular conditions in a solution.