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Chapter 19: Problem 67

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: $$\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+}$$ (a) Balance this equation. (b) If \(36.44 \mathrm{~mL}\) of a 0.01652 \(M \mathrm{KMnO}_{4}\) solution is required to completely oxidize \(25.00 \mathrm{~mL}\) of an \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate the molarity of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

Short Answer

Expert verified
The balanced equation is \(2\mathrm{MnO}_4^- + 5\mathrm{H}_2\mathrm{O}_2 + 6\mathrm{H}^+ \to 5\mathrm{O}_2 + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}\). The molarity of the \(\mathrm{H}_2\mathrm{O}_2\) solution is 0.0601 M.

Step by step solution

01

Identify Species in the Reaction

The given chemical reaction involves the reduction of permanganate ions \( \mathrm{MnO}_4^- \) (which are being reduced to \( \mathrm{Mn}^{2+} \)) and the oxidation of hydrogen peroxide \( \mathrm{H}_2\mathrm{O}_2 \) (to form \( \mathrm{O}_2 \)). Balancing the reaction requires that both mass and charge are balanced.
02

Balance Atoms in the Reaction

Start by balancing all the atoms except for O and H. For \( \mathrm{MnO}_4^- \) and \( \mathrm{Mn}^{2+} \), there is one manganese atom on both sides, so manganese is already balanced. We will then focus on balancing the oxygen and hydrogen atoms.
03

Balance Oxygen Atoms Using Water

The reaction takes place in an acidic medium, so we can add \( \mathrm{H}_2\mathrm{O} \) to balance oxygen. Each side needs 4 oxygen atoms from \( \mathrm{MnO}_4^- \) and 4 oxygen atoms from 2 molecules of \( \mathrm{H}_2\mathrm{O}_2 \), giving us a total of 8 oxygen atoms on each side.
04

Balance Hydrogen Atoms Using H+ Ions

In acidic solution, use \( \mathrm{H}^+ \) ions to balance hydrogen. On the right side, we have 4 oxygen from \( \mathrm{O}_2 \) and require 4 additional hydrogen atoms constituting 8 hydrogen ions, resulting in:\[2\mathrm{MnO}_4^- + 5\mathrm{H}_2\mathrm{O}_2 + 6\mathrm{H}^+ \rightarrow 5\mathrm{O}_2 + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}\]
05

Balance the Charges

Balance the charges across the equation. The left side has a total charge of -2 from the 2 permanganate ions, and 6 from \( 6\mathrm{H}^+ \) ions, giving a total charge of \( +4 \). The right side also needs to remain at a total charge of \( +4 \) with two \( \mathrm{Mn}^{2+} \) ions.
06

Calculate Moles of Permanganate

Given: Volume = 36.44 mL = 0.03644 L and Molarity = 0.01652 M. Find moles of \( \mathrm{KMnO}_4 \):\[n = Molarity \times Volume = 0.01652 \times 0.03644 = 0.000601 \text{ moles}\]
07

Use Stoichiometry to Find Moles of H2O2

The balanced equation shows that 2 moles of \( \mathrm{MnO}_4^- \) react with 5 moles of \( \mathrm{H}_2\mathrm{O}_2 \). Thus, \[\text{moles of } \mathrm{H}_2\mathrm{O}_2 = \frac{5}{2} \times \text{moles of } \mathrm{MnO}_4^-\]\[= \frac{5}{2} \times 0.000601 = 0.0015025 \text{ moles}\]
08

Calculate Molarity of H2O2 Solution

Using the moles of \(\mathrm{H}_2\mathrm{O}_2\) obtained and the volume of the \(\mathrm{H}_2\mathrm{O}_2\) solution, which is 25.00 mL or 0.025 L:\[\text{Molarity} = \frac{\text{moles of } \mathrm{H}_2\mathrm{O}_2}{\text{volume in liters}} = \frac{0.0015025}{0.025} = 0.0601 \text{M}\]

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Key Concepts

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

Balance Chemical Equations
When balancing chemical equations, the goal is to ensure that there are equal numbers of each type of atom on both sides of the equation. This is important for maintaining the conservation of mass principle.
The process typically involves a few key steps:
  • Identify the Species: Begin by identifying all the reactants and products in the equation. In the given unbalanced equation, we have permanganate ions \(\mathrm{MnO}_4^-\) and hydrogen peroxide \(\mathrm{H}_2\mathrm{O}_2\) as reactants, and oxygen \(\mathrm{O}_2\) and \(\mathrm{Mn}^{2+}\) ions as products.
  • Balance Atoms: Tackle one element at a time. Typically, elements other than hydrogen and oxygen are balanced first. The manganese is balanced in this reaction from the start.
  • Oxygen Balancing: Use water molecules \(\mathrm{H}_2\mathrm{O}\) to adjust the number of oxygen atoms, especially in reactions occurring in an acidic medium.
  • Hydrogen Balancing: Adjust the hydrogen using \(\mathrm{H}^+\) ions. Count how many are needed to balance the hydrogen atoms.
  • Charge Balance: Finally, make sure the charges are balanced on both sides by observing the number and types of ions involved.
By completing these steps, as shown in the original solutions, the balanced equation ensures that all atoms and charges are accounted for equally.
Potassium Permanganate
Potassium permanganate (\(\mathrm{KMnO}_4\)) is a strong oxidizing agent commonly used in titrations. It's recognizable by its deep purple color and is particularly useful in oxidation-reduction (redox) reactions.
Here's why it's important in titrations:
  • In acidic solutions, \(\mathrm{KMnO}_4\) can accept electrons from other substances, reducing itself to \(\mathrm{Mn}^{2+}\), which is colorless. This color change can help in visually determining the endpoint of a titration.
  • The reduction of permanganate involves gaining electrons: \(\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\).
  • Its role in this titration is to react with hydrogen peroxide, where the permanganate will facilitate the oxidation of \(\mathrm{H}_2\mathrm{O}_2\) to \(\mathrm{O}_2\).
This process is key because the amount of permanganate used can be related directly to the concentration of the substance being analyzed, like hydrogen peroxide in our exercise.
Molarity Calculation
Determining the molarity of a solution involves calculating the concentration of a solute in a given volume of solvent. Molarity (M) is measured in moles per liter (mol/L).
  • Formula for Molarity: \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\).
  • In the exercise, you are determining the molarity of the hydrogen peroxide solution. You first calculate the moles of potassium permanganate (\(\mathrm{MnO}_4^-\)) using the equation \(\text{moles} = \text{molarity} \times \text{volume in liters}\).
  • Since the titration involves a specific stoichiometric relationship (2:5) between \(\mathrm{KMnO}_4\) and \(\mathrm{H}_2\mathrm{O}_2\), you multiply the moles of \(\mathrm{MnO}_4^-\) by \(\frac{5}{2}\) to find the moles of hydrogen peroxide.
  • Finally, to find the molarity of \(\mathrm{H}_2\mathrm{O}_2\), divide its moles by the volume of the hydrogen peroxide solution.
This step-by-step calculation helps to accurately determine the concentration of a solution and is fundamental in titration analysis.
Oxidation-Reduction Reaction
Oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. One species gives up electrons (is oxidized), while another species gains electrons (is reduced).
In this exercise, the redox reaction features:
  • Reduction: Permanganate ions\((\mathrm{MnO}_4^-)\) accept electrons, reducing to \(\mathrm{Mn}^{2+}\).
  • Oxidation: Hydrogen peroxide \((\mathrm{H}_2\mathrm{O}_2)\) loses electrons, transforming into oxygen \((\mathrm{O}_2)\) gas.
  • Redox reactions can be identified by changes in oxidation states of the elements involved; in this reaction, manganese changes from +7 to +2 and oxygen is released as gas from hydrogen peroxide.
  • In the context of titration, redox reactions help in quantitative analysis by the measurement of chemical substances concentration in solution through electron exchange confirmation.
Understanding how electrons are exchanged in these reactions is crucial for mastering chemical equations and reactions, especially in analytical chemistry.

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