Problem 108 - Chemistry and Chemical Reactivity

Question

You have a mixture of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) and another solid that does not react with sodium hydroxide. If 29.58 mL of \(0.550 \mathrm{M} \mathrm{NaOH}\) is required to titrate the oxalic acid in the \(4.554-\mathrm{g}\) sample to the second equivalence point, what is the mass percent of oxalic acid in the mixture? Oxalic acid and NaOH react according to the equation $$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$

Step-by-Step Solution

Answer

The mass percent of oxalic acid in the mixture is 16.07%.

Step 1: Calculate moles of NaOH used

First, calculate the number of moles of NaOH used in the titration. Use the formula \( \text{moles} = \text{volume} \times \text{concentration} \). The volume of NaOH is 29.58 mL, which is equivalent to 0.02958 L, and the concentration is 0.550 M.\[ \text{moles of NaOH} = 0.02958 \, \text{L} \times 0.550 \, \text{M} = 0.016269 \, \text{mol} \]

Step 2: Relate moles of NaOH to moles of oxalic acid

From the balanced chemical equation, one mole of oxalic acid reacts with two moles of NaOH. Therefore, the moles of oxalic acid is half the moles of NaOH.\[ \text{moles of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} = \frac{0.016269 \, \text{mol NaOH}}{2} = 0.0081345 \, \text{mol} \]

Step 3: Calculate the mass of oxalic acid

The molar mass of oxalic acid \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \) is approximately \(2(1.01) + 2(12.01) + 4(16.00) = 90.03 \, \text{g/mol}\). Use the moles calculated in Step 2 to find the mass.\[ \text{mass of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} = 0.0081345 \, \text{mol} \times 90.03 \, \text{g/mol} = 0.7321 \, \text{g} \]

Step 4: Calculate the mass percent of oxalic acid

To find the mass percent of oxalic acid in the mixture, use the formula: \( \text{mass percent} = \left(\frac{\text{mass of component}}{\text{total mass of mixture}}\right) \times 100\). The total mass of the mixture is 4.554 g.\[ \text{mass percent of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} = \left(\frac{0.7321 \, \text{g}}{4.554 \, \text{g}}\right) \times 100 = 16.07\% \]

Key Concepts

Mass Percent Calculation

Calculating mass percent is a straightforward way to determine the proportion of a component in a mixture. It's expressed as the mass of a specific component divided by the total mass of the mixture, multiplied by 100 to get a percentage.

In the case of the oxalic acid titration exercise, you first need to determine the mass of oxalic acid present in the mixture. This is done using stoichiometry and molarity concepts. Once you have the mass of oxalic acid, calculate the mass percent using:

• The formula: \( \text{mass percent} = \left( \frac{\text{mass of component}}{\text{total mass}} \right) \times 100 \)
• The total mass of the mixture in the problem is 4.554 grams.
• The mass of the oxalic acid calculated is 0.7321 grams.
• Substituting these values gives a mass percent of 16.07%.

Understanding how mass percent works helps assess the concentration of a component in any sample or mixture.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, as illustrated in the equation for the reaction of oxalic acid with sodium hydroxide.

The chemical equation given is \(\text{H}_2\text{C}_2\text{O}_4(\text{aq}) + 2 \text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{C}_2\text{O}_4(\text{aq}) + 2 \text{H}_2\text{O}(\ell)\). This balanced equation shows that one mole of oxalic acid reacts with two moles of sodium hydroxide to produce one mole of sodium oxalate and two moles of water.

• Each molecule participates in the reaction in a fixed ratio.
• This ratio is vital for calculations involving the amount of each substance reacting or being produced.
• Balanced equations ensure the conservation of mass.

Learning to interpret and write balanced chemical equations is essential in stoichiometry, helping you understand the proportions of substances involved in chemical reactions.

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions based on the balanced equation of the reaction. Using stoichiometry, we can determine the relationships between reactants and products in a chemical reaction.

In the exercise, after calculating the moles of sodium hydroxide used, stoichiometry helps relate these moles to moles of oxalic acid by using the balanced equation.

• For every mole of \(\text{H}_2\text{C}_2\text{O}_4\) reacting, two moles of \(\text{NaOH}\) are needed.
• Therefore, the moles of \(\text{NaOH}\) determine the moles of \(\text{H}_2\text{C}_2\text{O}_4\) through division by 2.
• This relationship simplifies the calculation of oxalic acid's amount in the mixture.

Understanding stoichiometry is crucial as it provides the tools needed to predict the amounts needed or produced in a chemical reaction, and to ensure reactions are conducted efficiently.

Molarity

Molarity is a measure of the concentration of a solution, specifically the number of moles of a solute present in one liter of solution. It is a fundamental concept in both chemistry and biology, providing a way to express the concentration of a solution.

In this exercise, the molarity of sodium hydroxide is given as 0.550 M, which indicates that there are 0.550 moles of sodium hydroxide in every liter of solution. To find the amount of sodium hydroxide used in the titration, this molarity is multiplied by the volume (in liters) of the solution used:

• Convert the volume from mL to liters by dividing by 1000, hence 29.58 mL becomes 0.02958 L.
• Moles of \(\text{NaOH}\) = \(0.550 \, \text{M} \times 0.02958 \, \text{L}\).
• This gives 0.016269 moles of \(\text{NaOH}\).

Using molarity allows chemists to prepare and use solutions accurately, ensuring precise outcomes in chemical analysis and reactions.