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Chapter 17: Problem 96

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?

Short Answer

Expert verified
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 by step solution

01

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 (K鈧涒倸): \[K_{sp} = [Ba^{2+}][MnO_4^-]^2\]
02

Substitute the given values of K鈧涒倸 and [Ba虏鈦篯 into the K鈧涒倸 expression

We are given that K鈧涒倸 = 2.5 x 10鈦宦光伆 and the concentration of Ba虏鈦� in solution is 2.0 x 10鈦烩伕 M. We can substitute these values into the K鈧涒倸 expression: \[2.5 \times 10^{-10} = (2.0 \times 10^{-8})[MnO_4^-]^2\]
03

Solve for the concentration of MnO鈧勨伝 ions in solution

Now, we need to solve the equation for the concentration of MnO鈧勨伝 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 MnO鈧勨伝 ions, we take the square root of both sides of the equation: \[[MnO_4^-] = \sqrt{1.25 \times 10^{-2}}\] \[[MnO_4^-] = 1.12 \times 10^{-1} \, M\]
04

Determine the concentration of KMnO鈧� required

Since KMnO鈧� and MnO鈧勨伝 have a 1:1 stoichiometric ratio, their concentrations will be equal in solution. Thus, the concentration of KMnO鈧� required to establish a concentration of 2.0 x 10鈦烩伕 M for the Ba虏鈦� ion in solution is: \[[KMnO_4] = [MnO_4^-] = 1.12 \times 10^{-1} \, M\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5\) ? (b) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{-8}\).)

A solution containing an unknown number of metal ions is treated with dilute \(\mathrm{HCl} ;\) no precipitate forms. The \(\mathrm{pH}\) is adjusted to about 1, and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms.

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) because the solution will be quite basic.)

Benzenesulfonic acid is a monoprotic acid with \(\mathrm{p} K_{a}=2.25 .\) Calculate the \(\mathrm{pH}\) of a buffer composed of \(0.150 \mathrm{M}\) benzenesulfonic acid and \(0.125 \mathrm{M}\) sodium benzensulfonate.

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}\), \(0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\), and the \(\mathrm{pH}\) measured. \((\mathrm{a})\) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{pK}_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)
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