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.

Short answer

The balanced net ionic equation for the neutralization reaction between tartaric acid (\(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) is: \( 2 \mathrm{H}^{+} + 2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \) After titrating with \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\), the molarity of the tartaric acid solution is \(0.061625 \mathrm{~M}\).

Step-by-step solution

Write the balanced net ionic equation for neutralization reaction

The balanced chemical equation for the neutralization reaction between tartaric acid (\(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) is: \( \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{H}_{2} \mathrm{O} \) The total ionic equation of the reaction is: \( 2 \mathrm{H}^{+} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-} + 2 \mathrm{Na}^{+} + 2 \mathrm{OH}^{-} \rightarrow \mathrm{Na}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} + 2 \mathrm{H}_{2} \mathrm{O} \) As sodium and tartrate ions are spectator ions, we can omit them from the net ionic equation: \( 2 \mathrm{H}^{+} + 2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \)

Calculate the moles of sodium hydroxide used

To find the moles of sodium hydroxide used in the titration, we can use the molarity and volume of sodium hydroxide solution: Moles of \(\mathrm{NaOH}\) = Molarity of \(\mathrm{NaOH}\) × Volume of \(\mathrm{NaOH}\) Moles of \(\mathrm{NaOH}\) = \(0.2500 \mathrm{~M} \times 24.65 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.0061625 \mathrm{~mol}\)

Calculate the moles of tartaric acid

From the balanced net ionic equation, we can see that 2 moles of \(\mathrm{NaOH}\) reacts with 1 mole of tartaric acid. So, we can find the moles of tartaric acid in the solution: Moles of tartaric acid = \(\frac{1 \mathrm{~mol \, of \, H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}}{2 \mathrm{~mol \, of \, NaOH}} \times 0.0061625 \mathrm{~mol}\) of \(\mathrm{NaOH}\) = \(0.00308125 \mathrm{~mol}\) of tartaric acid

Calculate the molarity of tartaric acid

We can now find the molarity of tartaric acid in the solution by dividing the moles of the tartaric acid by the volume of the tartaric acid solution: Molarity of tartaric acid = \(\frac{0.00308125 \mathrm{~mol}}{50.00 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}} = 0.061625 \mathrm{~M}\) So, the molarity of the tartaric acid solution is \(0.061625 \mathrm{~M}\).