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Chapter 17: Problem 103

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

Expert verified
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 by step solution

01

(Write the chemical equation)

The dissolution of barium permanganate in water can be represented by the following equation: \[ \mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + 2\mathrm{MnO}_{4}^{-}(aq) \]
02

(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#
03

(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: \[ 2.5 \times 10^{-10} = (2.0 \times 10^{-8})[\mathrm{MnO}_{4}^{-}]^{2} \]
04

(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#
05

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

The dissociation of \(\mathrm{KMnO}_{4}\) in water can be represented as: \[ \mathrm{KMnO}_{4}(s) \rightleftharpoons \mathrm{K}^{+}(aq) + \mathrm{MnO}_{4}^{-}(aq) \]
06

(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: \[ [\mathrm{KMnO}_{4}] = 7.94 \times 10^{-2} \ \mathrm{M} \]

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Most popular questions from this chapter

Consider the equilibrium $$ \mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to $\mathrm{p} K_{a}$ for the acid.

A 20.0-mL sample of \(0.150 \mathrm{MKOH}\) is titrated with \(0.125 \mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the pH after the following volumes of acid have been added: $(\mathbf{a}) 20.0 \mathrm{~mL},(\mathbf{b}) 23.0 \mathrm{~mL},\( (c) \)24.0 \mathrm{~mL}\( (d) \)25.0 \mathrm{~mL},\( (e) \)30.0 \mathrm{~mL}$.

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(\mathrm{K}_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and $K_{f}=1.1 \times 10^{5}\( for \)\mathrm{AgCl}_{2}^{-},$ calculate the equilibrium constant for the following reaction: $$ \mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) $$

A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?
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