Problem 102 - Chemistry The Central Science

Question

Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) , has two acidic hydrogens. The acid is often present in wines and a salt derived from the acid precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with NaOH. It requires 24.65 \(\mathrm{mL}\) of 0.2500 \(\mathrm{M}\) NaOH solution to titrate both acidic protons in 50.00 \(\mathrm{mL}\) of the tartaric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the tartaric acid solution.

Step-by-Step Solution

Answer

The balanced net ionic equation for the neutralization reaction is: \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{OH}^{-} \rightarrow \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^2- + 2 \mathrm{H}_{2}\mathrm{O}\), and the molarity of the tartaric acid solution is 0.061625 M.

Step 1: Write the balanced equation for the neutralization reaction

For the neutralization reaction, tartaric acid \(\mathrm{(H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6})\) reacts with sodium hydroxide \(\mathrm{(NaOH)}\). The balanced equation is: \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{NaOH} \rightarrow \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^2- + 2 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Na}^{+}\) The net ionic equation (excluding spectator ions) is: \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{OH}^{-} \rightarrow \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^2- + 2 \mathrm{H}_{2}\mathrm{O}\)

Step 2: Calculate the moles of NaOH used in the titration

We are given the volume (24.65 mL) and molarity (0.2500 M) of NaOH. To find the moles, we can use the formula: moles of solute = molarity × volume (in liters) moles of NaOH = (0.2500 mol/L) × (24.65 mL × (1 L / 1000 mL)) moles of NaOH = 0.0061625 mol

Step 3: Use stoichiometry to find the moles of tartaric acid

From the balanced equation, we know that 1 mole of tartaric acid reacts with 2 moles of NaOH. Using this stoichiometric ratio, we can find the moles of tartaric acid: moles of tartaric acid = (moles of NaOH) × (1 mol tartaric acid / 2 mol NaOH) moles of tartaric acid = 0.0061625 mol × (1/2) = 0.00308125 mol

Step 4: Calculate the molarity of the tartaric acid solution

To find the molarity, divide the moles of tartaric acid by the volume of the solution: Molarity of tartaric acid = (moles of tartaric acid) / (volume of tartaric acid in liters) Molarity of tartaric acid = 0.00308125 mol / (50.00 mL × (1 L / 1000 mL)) Molarity of tartaric acid = 0.061625 M In conclusion, the balanced net ionic equation for the neutralization reaction is: \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{OH}^{-} \rightarrow \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^2- + 2 \mathrm{H}_{2}\mathrm{O}\) And the molarity of the tartaric acid solution is 0.061625 M.

Key Concepts

Neutralization Reactions

Neutralization reactions are specific chemical reactions where an acid and a base react to form a salt and water. In these reactions, the acid's protons (H+) combine with the hydroxide ions (OH-) from the base to produce water (H2O). During the titration of tartaric acid with NaOH, the hydroxide ions from the sodium hydroxide (the base) neutralize the acidic protons from tartaric acid.

Understanding neutralization is crucial as it allows chemists to determine the concentration of an unknown solution through titration. It's important to note that the equivalence point is reached when the number of moles of H+ equals the number of moles of OH-, making the solution theoretically neutral. If we look at the reaction between tartaric acid and NaOH, every mole of tartaric acid will react with two moles of NaOH to neutralize completely, indicating that tartaric acid has two acidic hydrogens.

Net Ionic Equations

Net ionic equations are simplified chemical equations that show only the species participating directly in the chemical reaction, leaving out the spectator ions. Spectator ions are present in the reaction mixture but do not take part in the actual chemical change. In our tartaric acid titration example, the sodium ions (Na+) are spectator ions because they don’t change their oxidation state or combine with another element during the reaction.

When writing net ionic equations, it's vital to identify the compounds that are ionizing in solution and which ions are reacting. For the titration of tartaric acid with NaOH, once the equation is balanced, we exclude the Na+ ions to arrive at the net ionic equation. This emphasizes the actual chemical change occurring during the neutralization, which in this case, is the transformation of tartaric acid into its conjugate base and water.

Stoichiometry

The concept of stoichiometry revolves around the quantitative relationships between the reactants and products in a chemical reaction. It's a powerful tool in chemistry for predicting the amounts of substances consumed and produced in a given reaction.

In the context of the neutralization reaction between tartaric acid and NaOH, stoichiometry tells us that one mole of tartaric acid requires two moles of NaOH to neutralize completely. This stoichiometric ratio is vital for performing calculations in the titration. It allows us to calculate the moles of tartaric acid present once we determine the moles of NaOH used in the titration. Through these stoichiometric calculations, students can predict and understand reaction yields and design experiments accordingly.

Molarity Calculations

Molarity is a measure of concentration that exemplifies the number of moles of solute present in one liter of solution. It's a fundamental concept in chemistry, particularly in solutions and titrations, as it allows calculation of the number of moles present in a specific volume of solution.

The formula for molarity is \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). In the titration exercise, to find the molarity of the tartaric acid solution, we calculate the moles of tartaric acid (using the stoichiometric relationship with NaOH) and divide this by the volume of the tartaric acid solution in liters. Accurate molarity calculations are essential because they provide the quantitative basis for analyzing the results of a titration and for preparing solutions with precise concentrations for further experiments or industrial processes.