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 \(29.58 \mathrm{mL}\) of \(0.550 \mathrm{M} \mathrm{NaOH}\) is required to titrate the oxalic acid in the 4.554 -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 \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$

Short answer

The mass percent of oxalic acid in the mixture is 16.08%.

Step-by-step solution

Determine moles of NaOH

Calculate the moles of NaOH used in the titration using the formula: \( ext{moles of NaOH} = ext{volume (L)} \times ext{molarity (M)} \). Given: volume \(= 29.58\, \text{mL} = 0.02958\, \text{L} \) and molarity \(= 0.550\, \text{M} \). So, \( ext{moles of NaOH} = 0.02958 \times 0.550 = 0.016269\, \text{mol} \).

Calculate moles of oxalic acid

From the balanced chemical equation, 1 mole of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) reacts with 2 moles of NaOH. Therefore, the moles of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is \(\frac{0.016269}{2} = 0.0081345\, \text{mol} \).

Find the mass of oxalic acid

Using the molar mass of oxalic acid \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) which is \(90.03\, \text{g/mol}\), calculate the mass: \( ext{mass} = ext{moles} \times \text{molar mass} = 0.0081345 \times 90.03 = 0.732\, \text{g} \).

Determine mass percent of oxalic acid in the mixture

The mass percent of oxalic acid in the mixture is calculated as \( \left( \frac{\text{mass of oxalic acid}}{\text{total mass of mixture}} \right) \times 100\right)\). Thus, \( \left( \frac{0.732}{4.554} \times 100 \right) = 16.08\%\).

Key concepts

Understanding Oxalic Acid

Oxalic acid, known by its chemical formula \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is an organic compound. It falls under the category of dicarboxylic acids. This means it has two acidic carboxyl groups \((-COOH)\).

In its pure form, oxalic acid appears as a white crystalline solid. It is soluble in water, rendering it a popular choice for titration experiments.

Notably, oxalic acid serves as a primary standard in acid-base titrations due to its high purity and stability. This makes it ideal for accurately determining the concentration of solutions like sodium hydroxide.

Role of Sodium Hydroxide in Titration

Sodium hydroxide \(\mathrm{NaOH}\) plays a crucial role as a titrant in chemical reactions like titrations. This strong base, also known as caustic soda or lye, is highly effective in neutralizing acids.

In the context of titrations, \(\mathrm{NaOH}\) reacts with acids to form water and a salt, providing a way to determine the concentration of an acid in a solution.

The selection of sodium hydroxide for titrations is due to its strong base properties, which allow for a complete neutralization process. Also, it is readily available and is known for its reliability in chemical analyses.

Significance of the Chemical Equation in Titration

A balanced chemical equation is crucial to understanding how substances react. In our scenario, the chemical reaction between oxalic acid and sodium hydroxide is represented as:

\[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\]

This equation reveals that one mole of oxalic acid reacts with two moles of sodium hydroxide.

Understanding the molar ratios from the balanced chemical equation is essential for calculating the moles of reactants and products in a reaction. This knowledge forms the backbone of quantitative chemistry calculations.

Moles Calculation in Titration Procedures

Calculating moles is a fundamental step in titration analysis. Moles offer a way to measure the quantity of substance involved in a chemical reaction. In titration, determining the amount of moles helps gauge how much titrant is required to neutralize an analyte.

To find the number of moles of sodium hydroxide, the formula used is:\

• \( \text{moles of NaOH} = \text{volume (L)} \times \text{molarity (M)} \)

This calculation is based on the concentration and volume of the \(\mathrm{NaOH}\) used.

Next, using the balanced chemical equation, you can determine moles of oxalic acid. Given that one mole of oxalic acid reacts with two moles of \(\mathrm{NaOH}\), this ratio helps in calculating the moles of the acid. This forms the basis for further computations, such as finding the mass or mass percent of the analyte in a mixture.