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 precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with \(\mathrm{NaOH}\). It requires \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) 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 between tartaric acid and NaOH is \(H_2C_4H_4O_6 + 2OH^- → C_4H_4O_6^{2-} + 2H_2O\). After titration, the moles of NaOH used are 0.0061625 mol. Using stoichiometry, we find that the moles of tartaric acid are 0.00308125 mol. Hence, the molarity of the tartaric acid solution is 0.061625 M.

Step 1: Write the balanced net ionic equation for the neutralization reaction.

First, we need to write the balanced net ionic equation for the reaction between tartaric acid and NaOH. The complete reaction between tartaric acid and NaOH is: \(H_2C_4H_4O_6 + 2NaOH → Na_2C_4H_4O_6 + 2H_2O\) Adjusting the equation for the net ionic form, weakening the strong electrolytes, we get: \(H_2C_4H_4O_6 + 2OH^- → C_4H_4O_6^{2-} + 2H_2O\)

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

The volume of NaOH used is given as 24.65 mL, and its concentration is 0.2500 M. Convert the volume to liters, and calculate the moles of NaOH using the formula: moles = molarity × volume moles of NaOH = \(0.2500 \frac{mol}{L} \times 0.02465 L = 0.0061625 mol\)

Step 3: Calculate the moles of tartaric acid using the stoichiometry of the balanced net ionic equation.

From the balanced net ionic equation, we can see that 1 mol of tartaric acid reacts with 2 mol of OH^- ions. Therefore, we can calculate the moles of tartaric acid using the stoichiometry: moles of tartaric acid = \(\frac{1}{2}\) × moles of NaOH = \(\frac{1}{2} \times 0.0061625 mol = 0.00308125 mol\)

Step 4: Calculate the molarity of the tartaric acid solution.

The volume of the tartaric acid solution is given as 50.00 mL. We can now calculate the molarity of tartaric acid using the formula: molarity = \(\frac{moles}{volume}\) molarity of tartaric acid = \(\frac{0.00308125 mol}{0.05000 L} = 0.061625 M\) So the molarity of the tartaric acid solution is 0.061625 M.