Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$ \mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2} $$ (b) If a 1.00 -g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

Short Answer

Expert verified
The balanced chemical equation is \(2\mathrm{MnO}_{4}^{-} + 16H^{+} + 5\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow 2\mathrm{Mn}^{2+} + 8H_{2}O + 10\mathrm{CO}_{2}\) and the percent by mass of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) in the sample is 5.4%.

Step by step solution

01

Balancing the chemical equation

For balancing, it’s convenient to break the process into two half-reactions: one for oxidation and one for reduction. In our reaction \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is oxidized to \(\mathrm{CO}_{2}\). The individual half reactions appear as: \[\mathrm{MnO}_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4H_{2}O\] \[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\rightarrow 2\mathrm{CO}_{2}+2e^{-}\] To balance the electrons lost and gained, we multiply the second equation by 5 and the first by 2. Add these equations together to get the total ionic equation.
02

Adding the half-reactions

Adding two half-reactions, the electrons cancel out and we get the balanced redox equation: \[2\mathrm{MnO}_{4}^{-} + 16H^{+} + 5\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow 2\mathrm{Mn}^{2+} + 8H_{2}O + 10\mathrm{CO}_{2}\] which is the balanced equation in acid solution.
03

Determining mass of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)

First, let's calculate the moles of \(\mathrm{KMnO}_{4}\) used for titration using: moles = molarity × volume(L), where molarity = \(0.0100 M\) and volume = \(24.0 \mathrm{~mL}\) = \(0.024 L\). So, moles of \(\mathrm{KMnO}_{4}\) = \(0.0100 M × 0.024 L = 0.00024 mol\). From the balanced equation, it can be seen that 2 mol of \(\mathrm{MnO}_{4}^{-}\) reacts with 5 mol of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\). So, Moles of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) = \((5/2) × Mol \: \mathrm{of} \: \mathrm{MnO}_{4}^{-}\) = \((5/2) × 0.00024 mol = 0.0006 mol\). Using the molecular weight of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} (90.03 g/mol)\), mass can be calculated as: \(Mass = moles × molecular weight\) = \(0.0006 mol × 90.03 g/mol = 0.054 g\).
04

Calculation of the percent by mass

The percent by mass of \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) in the sample can be calculated using: \(Mass\% = (mass \: of\: \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}/total \: mass) × 100 = (0.054 g/1.00 g) × 100 = 5.4%\)

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a foundational skill in chemistry essential for understanding how different substances react. In the context of redox reactions, the process becomes slightly more complex, involving the balancing of both mass and charge.

Redox titrations, such as the one described in the example, require careful attention to the transfer of electrons between the reactants. The half-reaction method is often employed, splitting the overall reaction into two parts: oxidation (loss of electrons) and reduction (gain of electrons). By balancing the number of electrons transferred in each half-reaction, we can ensure that the overall equation is both mass and charge balanced. This method highlights the stoichiometric relationship between reactants and products, an essential concept in quantitative chemistry.
Acid-Base Titration
Acid-base titration is a quantitative analytical procedure used to determine the concentration of an unknown solution by reacting it with a standard solution of known concentration. In the provided exercise, a redox titration is performed using KMnO4 as the titrant to react with oxalic acid (H2C2O4).

The endpoint of the titration is reached when the correct stoichiometric ratio between the titrant and analyte is achieved, which corresponds to the completion of the reaction. The volume of titrant used at the equivalence point provides valuable information that can be used to calculate the amount of substance in the sample, which is a critical aspect of stoichiometry.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It involves calculations that are based on the laws of conservation of mass and charge.

In the given exercise, stoichiometry is applied to determine the quantity of oxalic acid in the sample, using the balanced chemical equation as a roadmap. By calculating the moles of KMnO4 (the titrant) used, and understanding the molar ratios from the balanced equation, we can derive the amount of oxalic acid that was present in the sample. This type of calculation is essential for various applications in both laboratory and industrial settings where precise chemical formulation is required.
Percent Composition
Percent composition is a measure of the relative amounts of each element within a compound, expressed as a percentage of the total mass. In a chemical analysis context, determining the percent composition of a sample can reveal its purity or the presence of specific ingredients.

In the exercise, after calculating the mass of oxalic acid in the sample, the percent by mass is obtained by dividing the mass of the identified substance (oxalic acid) by the total mass of the sample, and then multiplying by 100. This simple calculation gives us insight into the concentration of a particular component in a mixture and is a valuable tool in lab analyses and quality control processes.

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Most popular questions from this chapter

One of the half-reactions for the electrolysis of water is $$ 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g) $$ If \(0.845 \mathrm{~L}\) of \(\mathrm{H}_{2}\) is collected at \(25^{\circ} \mathrm{C}\) and \(782 \mathrm{mmHg}\), how many faradays of electricity had to pass through the solution?

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To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.

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