Problem 89 - Chemistry: Atoms First

Question

What volume of a 0.3300-M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid? \(\mathrm{C}_{2} \mathrm{O}_{4} \mathrm{H}_{2}(a q)+2 \mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

The volume of 0.3300-M sodium hydroxide needed is 0.06818 L or 68.18 mL.

Step 1: Identify the reaction stoichiometry

According to the balanced chemical equation, 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide. The stoichiometry shows a 1:2 molar ratio between oxalic acid and sodium hydroxide.

Step 2: Calculate moles of oxalic acid

Use the concentration and volume of the oxalic acid solution to find the moles. Moles of oxalic acid = molarity (M) × volume (L) = 0.1500 M × 0.01500 L.

Step 3: Calculate moles of sodium hydroxide required

Based on the stoichiometry, it will take twice as many moles of sodium hydroxide to neutralize the oxalic acid. Moles of sodium hydroxide required = moles of oxalic acid × 2.

Step 4: Calculate volume of sodium hydroxide solution

Use the molarity of the sodium hydroxide solution and the moles needed to find the volume. Volume (L) = moles of sodium hydroxide / molarity of sodium hydroxide.

Key Concepts

Stoichiometry in Acid-Base Reactions

Stoichiometry is a section of chemistry that involves using relationships from balanced chemical equations to calculate the quantities of reactants and products involved in chemical reactions. In an acid-base titration, such as the reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄), stoichiometry is used to determine the amount of acid that can be neutralized by a given amount of base, or vice versa.

For the given exercise, the stoichiometry is derived from the balanced chemical equation indicating that two moles of sodium hydroxide are required to neutralize one mole of oxalic acid. This 1:2 molar ratio is integral to the procedure because it dictates the relative volumes and concentrations needed to reach the endpoint of the titration, which is typically indicated by a color change if an indicator is used or by reaching a particular pH value.

Molarity Calculation

Molarity, expressed in moles per liter (M), is a measure of concentration for solutions. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. In a titration problem, the molarity of a solution is used to find out how much of the solution is needed to react completely with a given quantity of another substance.

In this exercise, the molarity calculation is crucial to finding the moles of oxalic acid. By using the formula \( \text{Moles} = \text{Molarity} (M) \times \text{Volume} (L) \), you can calculate the amount of oxalic acid that will participate in the reaction. This forms the foundation for subsequently calculating the required volume of the sodium hydroxide solution to achieve neutralization.

Titration Calculations

Titration calculations are used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. The goal is to reach the equivalence point where stoichiometrically equivalent amounts of the reactants have been mixed.

In the context of the given problem, after calculating the moles of oxalic acid present, you then use the stoichiometry of the reaction (1:2 molar ratio) to calculate the moles of sodium hydroxide needed. Once the moles of sodium hydroxide are known, you can determine the volume of the sodium hydroxide solution required using its known molarity, by rearranging the molarity formula to \( \text{Volume} (L) = \frac{\text{Moles of sodium hydroxide}}{\text{Molarity of sodium hydroxide}} \). This step-by-step approach in titration calculations helps to clearly understand and perform the acid-base neutralization process.