91Ó°ÊÓ

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: \( 2 \mathrm{H}^{+} + 2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \)

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 \(\mathrm{NaOH}\) = \(0.2500 \mathrm{~M} \times 24.65 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.0061625 \mathrm{~mol}\)

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

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 \(0.061625 \mathrm{~M}\).