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 of tartaric acid with NaOH is: \[2 \mathrm{H}^{+} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-} + 2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-}\] The molarity of the tartaric acid solution is \(6.1625 \times 10^{-5} \: \mathrm{M}\).

Step-by-step solution

Write the balanced net ionic equation

First, we need to write the dissociation of each molecule in water. Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), loses two acidic hydrogens to form two \(\mathrm{H}^{+}\) ions and the tartrate ion: \[\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} \rightleftharpoons 2 \mathrm{H}^{+} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-}\] Sodium hydroxide, \(\mathrm{NaOH}\), dissociates completely in water to form \(\mathrm{Na}^{+}\) and \(\mathrm{OH}^{-}\) ions: \[\mathrm{NaOH} \longrightarrow \mathrm{Na}^{+} + \mathrm{OH}^{-}\] Now, writing the net ionic equation for the neutralization reaction by eliminating spectator ions (like \(\mathrm{Na}^{+}\)): \[2 \mathrm{H}^{+} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-} + 2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}^{2-}\]

Calculate moles of titrating NaOH

To find the molarity of the tartaric acid solution, we need to know the moles of NaOH added. We are given the volume and concentration of the NaOH solution, so we can calculate moles using the formula: Moles = Volume × Concentration Moles of NaOH = \(24.65 \: \mathrm{mL}\) × \(0.2500 \: \mathrm{M}\) = \(6.1625 \times 10^{-3} \: \mathrm{moles}\)

Calculate moles of tartaric acid

Now, we will use stoichiometry and the balanced net ionic equation to find the moles of \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\). From the balanced equation, we can see that the ratio of \(\mathrm{H}^{+}\) to \(\mathrm{OH}^{-}\) is 2:2, which means the moles of \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) must be half the moles of \(\mathrm{NaOH}\). Moles of \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) = \(6.1625 \times 10^{-3} \: \mathrm{moles} \: \mathrm{NaOH} × (1 \: \mathrm{mol} \: \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6} / 2 \: \mathrm{mol} \: \mathrm{NaOH}) = 3.08125 \times 10^{-3} \: \mathrm{moles}\)

Calculate the molarity of tartaric acid solution

Finally, we can calculate the molarity of the tartaric acid solution using the formula: Molarity = Moles / Volume Molarity of \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) = \(3.08125 \times 10^{-3} \: \mathrm{moles} / 50.00 \: \mathrm{mL}\) = \(6.1625 \times 10^{-5} \: \mathrm{M}\) The molarity of the tartaric acid solution is \(6.1625 \times 10^{-5} \: \mathrm{M}\).

Key concepts

Net Ionic Equation

Understanding the net ionic equation is crucial in chemistry, particularly in reactions like titration. This equation simplifies the reaction down to only those species that actually change during the reaction. In the case of the neutralization of tartaric acid with NaOH, we disregard the sodium ions since they are spectator ions and do not participate in the reaction. By focusing on the hydrogen ions from the acid and the hydroxide ions from the NaOH, we can see how they combine to produce water. This perspective helps clarify the underlying chemistry of the neutralization process and the stoichiometry involved. Remember, only components that undergo a change are included in the net ionic equation, which is essential for understanding whatever reaction you are analyzing.

Molarity Calculation

The molarity calculation is a fundamental concept in chemistry, representing the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In our exercise, we calculate the molarity of the tartaric acid to find out its concentration. The process involves measuring the volume of NaOH used in the titration and knowing its concentration. This information allows us to calculate the moles of NaOH and, through stoichiometry, the moles of tartaric acid. Dividing these moles by the volume of the tartaric acid solution yields the molarity. Calculating molarity is an essential skill, as it’s used to predict the outcomes of reactions and to prepare solutions with precise concentrations for various scientific applications.

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It's based on the conservation of mass and the balanced chemical equation. For the titration of tartaric acid with NaOH, stoichiometry allows us to deduce the mole ratio between the acid and the base, which is crucial for determining the concentration of the unknown solution. We use the known molarity and volume of NaOH to find the moles of base, which tell us the moles of acid through the mole ratio. Remember, stoichiometry doesn’t only involve mole-to-mole conversions but also volume-to-mole conversions as demonstrated in our exercise.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. In our textbook exercise, tartaric acid reacts with sodium hydroxide in a neutralization reaction that is quintessential in understanding how titration works. This kind of reaction is not only crucial in the laboratory for determining unknown concentrations but is also at play in everyday life, such as in the digestive system where excess stomach acid is neutralized by antacids. The principles you learn through titration and the chemical reasoning behind it help lay the foundation for a solid understanding of acid-base chemistry.