Question

A 50.00 -mL sample of solution containing \(\mathrm{Fe}^{2+}\) ions is titrated with a 0.0216 \(\mathrm{M} \mathrm{KMnO}_{4}\) solution. It required 20.62 \(\mathrm{mL}\) of \(\mathrm{KMnO}_{4}\) solution to oxidize all the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\) ions by the reaction $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\text { Acidic }}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q) \text{(Unbalanced)} $$ a. What was the concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution? b. What volume of 0.0150\(M \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution would it take to do the same titration? The reaction is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Cr}^{3+}(a q) +\mathrm{Fe}^{3+}(a q) \text {(Unbalanced)} $$

Short answer

a. The concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution is 0.04454 \(M\). b. The volume of 0.0150 \(M\) \(\mathrm{K}_{2}\) \(\mathrm{Cr}_{2}\) \(\mathrm{O}_{7}\) solution required for the same titration is 24.75\(\,mL\).

Step-by-step solution

Balance the given reactions

First, we need to balance both the given reactions before we proceed further. For the first reaction, \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)\) Balanced reaction: \(5\,\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_4^-(a q)+8\,\mathrm{H}^+(a q)\longrightarrow\mathrm{Mn}^{2+}(a q)+5\,\mathrm{Fe}^{3+}(a q)+4\,\mathrm{H}_2\mathrm{O}(l)\) For the second reaction, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)\) Balanced reaction: \(6\,\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_2\mathrm{O}_7^{2-}(a q)+14\,\mathrm{H}^+(a q)\longrightarrow 2\,\mathrm{Cr}^{3+}(a q)+6\,\mathrm{Fe}^{3+}(a q)+7\,\mathrm{H}_2\mathrm{O}(l)\) Now that both reactions are balanced, we can proceed with solving the exercise.

Find the concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution

We're given that a 50.00 -mL sample of solution containing \(\mathrm{Fe}^{2+}\) ions was titrated with a 0.0216 \(\mathrm{M} \mathrm{KMnO}_{4}\) solution, requiring 20.62 \(\mathrm{mL}\) of the \(\mathrm{KMnO}_{4}\) solution. First, we will find the moles of \(\mathrm{KMnO}_{4}\) used in the titration using its molarity and volume. Moles of \(\mathrm{KMnO}_{4}\) = Molarity \(\times\) Volume in Liters Moles of \(\mathrm{KMnO}_{4}\) = \(0.0216\,\mathrm{M}\times(20.62\,\mathrm{mL}\times(1\,\mathrm{L}/1000\,\mathrm{mL}))=4.45472\times10^{-4}\,\mathrm{mol}\) Now let's use stoichiometry to find the moles of \(\mathrm{Fe}^{2+}\). From the balanced equation, we have 1 mol of \(\mathrm{MnO}_4^-\) reacting with 5 mol of \(\mathrm{Fe}^{2+}\). Moles of \(\mathrm{Fe}^{2+}\) = Moles of \(\mathrm{KMnO}_{4}\) × \(\frac{5\,\mathrm{mol}\,\mathrm{Fe}^{2+}}{1\,\mathrm{mol}\,\mathrm{KMnO}_4}\) Moles of \(\mathrm{Fe}^{2+}\) = \(4.45472\times10^{-4}\,\mathrm{mol}\times5=2.22736\times10^{-3}\,\mathrm{mol}\) Now let's calculate the concentration of \(\mathrm{Fe}^{2+}\) ions in the solution by dividing the moles found by the volume of the sample in Liters. Concentration of \(\mathrm{Fe}^{2+}\) ions = \(\frac{\text{Moles of }\,\mathrm{Fe}^{2+}}{\text{Volume of the Sample in Liters}}\)= \(\frac{2.22736\times10^{-3}\,\mathrm{mol}}{(50\,\mathrm{mL}\times(1\,\mathrm{L}/1000\,\mathrm{mL})}\) Concentration of \(\mathrm{Fe}^{2+}\) ions = 0.04454 \(\mathrm{M}\) a. The concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution is 0.04454 \(\mathrm{M}\).

Find the volume of 0.0150\(M \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required for the same titration

We are given a solution with a molarity of 0.0150\(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and asked to find the volume required for the same titration. From the balanced reaction of the second equation, we have 1 mol of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) reacting with 6 mol of \(\mathrm{Fe}^{2+}\). Using stoichiometry, we can find the moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) required for the titration. Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) = Moles of \(\mathrm{Fe}^{2+}\) × \(\frac{1\,\mathrm{mol}\,\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7}{6\,\mathrm{mol}\,\mathrm{Fe}^{2+}}\) Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) = \(2.22736\times10^{-3}\,\mathrm{mol}\times\frac{1}{6}=3.71226\times10^{-4}\,\mathrm{mol}\) Now let's calculate the volume of 0.0150 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required for the titration. Volume of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution = \(\frac{\text{Moles of }\,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\text{Molarity of }\,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\,\text{solution}}\)= \(\frac{3.71226\times10^{-4}\,\mathrm{mol}}{0.015\,\text{M}}\) Volume of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution = 0.02475 \(\mathrm{L}\) (or 24.75\(\,\mathrm{mL}\)) b. The volume of 0.0150\(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required for the same titration is 24.75\(\,\mathrm{mL}\).

Key concepts

Equivalence Point

In redox titrations, just like in other types of titrations, the equivalence point is a crucial concept. It marks the stage at which the amount of oxidizing agent added equals the amount needed to completely react with the reducing agent in the solution.
This point is important because it helps us understand when the reaction is complete. In our exercise, the equivalence point is reached when all the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions.
When working with redox reactions, identifying the equivalence point helps in calculating the concentration of unknown substances.

Oxidation-Reduction Reaction

Redox reactions are chemical processes involving the transfer of electrons between two substances. This exercise demonstrates an oxidation-reduction reaction where the \(\mathrm{Fe}^{2+}\) ions lose electrons (undergo oxidation) to become \(\mathrm{Fe}^{3+}\) ions.
Meanwhile, the oxidizing agents, \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), gain electrons, undergoing reduction.

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.

These reactions are pivotal because they transform the chemical species involved, showing how interconnected chemical processes can be.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in a reaction. This concept is fundamental in redox titrations, including the example given, because it allows us to determine the amount of each substance required to reach the equivalence point.
In the exercise, stoichiometry helps us understand that 1 mole of \(\mathrm{MnO}_{4}^{-}\) reacts with 5 moles of \(\mathrm{Fe}^{2+}\). Subsequently, it aids in calculating the concentration of \(\mathrm{Fe}^{2+}\) ions in the solution.
Stoichiometry is like a recipe in chemistry, ensuring that the correct proportional ingredients are used to result in a desired outcome.

Balancing Chemical Equations

Balancing chemical equations is a key skill that ensures that the same number of atoms of each element are conserved before and after a reaction. It's a requirement to apply the law of conservation of mass.
While balancing, we count and adjust coefficients rather than altering formulas to make sure the equation is accurate.

• In the given exercise, the first equation involves \(5\) \(\mathrm{Fe}^{2+}\) ions reacting with \(1\) \(\mathrm{MnO}_{4}^{-}\).
• The second equation involves \(6\) \(\mathrm{Fe}^{2+}\) ions with \(1\) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\).

Understanding how to balance equations allows chemists to predict the amount of reactants needed and products formed in any given chemical reaction. Balancing stabilizes the reaction framework for accurate experimental predictions.