Problem 104 - Chemistry: The Central Science

Question

The concentration of hydrogen peroxide in a solution is determined by titrating a \(10.0\) -mL sample of the solution with permanganate ion. $$ \begin{array}{r} 2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{O}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ If it takes \(14.8 \mathrm{~mL}\) of \(0.134 \mathrm{M} \mathrm{MnO}_{4}^{-}\) solution to reach the equivalence point, what is the molarity of the hydrogen peroxide solution?

Step-by-Step Solution

Answer

The molarity of the hydrogen peroxide solution is \(0.4963 \, \mathrm{mol/L}\).

Step 1: Write down the balanced chemical equation

The balanced chemical equation is already given in the exercise: \(2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{O}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)\).

Step 2: Find the moles of permanganate ion used in the titration

First, we need to calculate the moles of MnO4- used in the titration. We are given the volume and concentration of the MnO4- solution used to reach the equivalence point. Moles of MnO4- = volume (in Liters) × concentration Moles of MnO4- = (14.8 mL) × (0.134 mol/L) Since the volume is in milliliters, we need to convert it to Liters: Moles of MnO4- = (0.0148 L) × (0.134 mol/L) = 0.0019852 mol.

Step 3: Find the moles of hydrogen peroxide

Now, we need to find the moles of hydrogen peroxide using the coefficients in the balanced chemical equation. For every 2 moles of MnO4-, there are 5 moles of H2O2: \(2 \, \mathrm{MnO}_{4}^{-} \thickspace\thickspace\thickspace : \thickspace5 \, \mathrm{H}_{2} \mathrm{O}_{2}\). Therefore, we can set up a proportion to determine the moles of H2O2: \(\frac{0.0019852 \, \mathrm{mol}\, \mathrm{MnO}_{4}^{-}}{2 \, \mathrm{mol}\, \mathrm{MnO}_{4}^{-}} = \frac{x \, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}_{2}}{5 \, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}_{2}}\). Solve for x moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\): \(x\, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}_{2} = 0.0019852 \, \mathrm{mol}\, \mathrm{MnO}_{4}^{-} \frac{5 \, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}_{2}}{2 \, \mathrm{mol}\, \mathrm{MnO}_{4}^{-}} = 0.004963 \, \mathrm{mol}\, \mathrm{H}_{2} \mathrm{O}_{2}\).

Step 4: Calculate the molarity of hydrogen peroxide

We are given the volume of peroxide in the solution,10.0 mL, so we need to convert it to Liters: Volume of hydrogen peroxide solution = 0.010 L. Now, we can calculate the molarity of hydrogen peroxide: Molarity of hydrogen peroxide = moles of hydrogen peroxide / volume of hydrogen peroxide solution Molarity of hydrogen peroxide = (0.004963 mol) / (0.010 L) = 0.4963 mol/L. The molarity of the hydrogen peroxide solution is 0.4963 mol/L.

Key Concepts

Hydrogen Peroxide

Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) is a chemical compound that plays a key role in this titration exercise. It is a pale blue liquid in its pure form and is known for its strong oxidizing properties. Hydrogen peroxide is often used as a disinfectant and a bleaching agent. In this context, we are interested in determining its concentration in a solution. During titration, hydrogen peroxide reacts with permanganate ions, as shown in the balanced chemical equation:\[2 \mathrm{MnO}_{4}^{-} + 5 \mathrm{H}_{2}\mathrm{O}_{2} + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+} + 5 \mathrm{O}_{2} + 8 \mathrm{H}_{2}\mathrm{O}.\]This equation tells us that it takes 5 moles of hydrogen peroxide to completely react with 2 moles of permanganate ions. Understanding this stoichiometry is crucial for calculating the concentration, or molarity, of the hydrogen peroxide solution, which we find by titrating it with a known concentration of permanganate ions.

Permanganate Ion

The permanganate ion (\(\text{MnO}_4^-\)) is another essential component of this titration process. It is a purple ion that acts as a strong oxidizing agent. Permanganate ions change to a nearly colorless state as they are reduced to manganese (II) ions, \(\text{Mn}^{2+}\), during the reaction with hydrogen peroxide. This color change is often used as an indicator in titration processes, making it easy to identify when the reaction has reached its completion (the equivalence point).

Titration allows us to determine how much permanganate ion is required to react with a given amount of hydrogen peroxide. In this exercise, it was given that 14.8 mL of 0.134 M permanganate ion solution was used. This data helps calculate the moles of permanganate ions in the solution, using the formula: \[\text{Moles of MnO}_4^- = \text{volume (in Liters)} \times \text{concentration}\]Understanding the behavior and properties of permanganate ions is crucial for accurately determining the molarity of hydrogen peroxide through titration.

Molarity Calculation

The concept of molarity is fundamental in chemistry, especially when quantifying the concentration of solutions. Molarity (\(M\)), is defined as the number of moles of solute per liter of solution. To find the molarity of hydrogen peroxide, we first calculate the moles of permanganate ion used in the titration and then use the stoichiometry from the balanced equation.

The exercise provided the volume and molarity of the permanganate ion solution. With this, we determined: - Moles of permanganate = \(0.0148 \text{ L} \times 0.134 \text{ mol/L} = 0.0019852 \text{ mol MnO}_4^-\)Using the balanced equation, we find that the moles of hydrogen peroxide is \[\text{Moles of } \text{H}_2 \text{O}_2 = 0.0019852 \text{ mol MnO}_4^- \times \frac{5}{2} = 0.004963 \text{ mol}\]Finally, to find the molarity of the hydrogen peroxide solution:\[\text{Molarity of } \text{H}_2 \text{O}_2 = \frac{0.004963 \text{ mol}}{0.010 \text{ L}} = 0.4963 \text{ mol/L}\]This process highlights the importance of understanding chemical equations and calculations in determining solution concentrations.