Problem 102 - Chemistry: The Central Science in SI Units, Global Edition

Question

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

Step-by-Step Solution

Answer

The balanced net ionic equation for the neutralization reaction between citric acid and potassium hydroxide (KOH) is: \[ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} + 3\mathrm{H^{+}} + 3\mathrm{OH^{-}} \to \mathrm{K}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \times \mathrm{H}_{2} \mathrm{O}\] After titrating the citric acid solution with \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) KOH, the molarity of the citric acid solution is found to be \(0.0387 \mathrm{M}\).

Step 1: Write the balanced chemical equation for the reaction

The chemical equation for the reaction between citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), and potassium hydroxide (KOH) can be written as: \[ 1 \times \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} + 3 \times \mathrm{KOH} \to \mathrm{K}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \times \mathrm{H}_{2} \mathrm{O}\]

Step 2: Write the balanced net ionic equation for the reaction

Rewrite the balanced chemical equation with the ionic species, and then remove the spectator ions to find the net ionic equation. Citric acid → \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} + 3\mathrm{H^{+}}\) 3 KOH → 3 \(\mathrm{K^{+}} + 3\mathrm{OH^{-}}\) The net ionic equation will be: \[ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} + 3\mathrm{H^{+}} + 3\mathrm{OH^{-}} \to \mathrm{K}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \times \mathrm{H}_{2} \mathrm{O}\]

Step 3: Use stoichiometry and the information given about the KOH solution to find the molarity of the citric acid solution

From the net ionic equation, we know that 3 moles of KOH neutralize 1 mole of citric acid. It's given that \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) KOH solution is required to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. First, calculate the moles of KOH used: moles of KOH = Molarity x Volume moles of KOH = \(0.500 \mathrm{M} \times 23.20 \times 10^{-3} L\) Moles of KOH = \(0.0116 \mathrm{~mol}\) From the stoichiometry, 3 mol of KOH neutralize 1 mol of citric acid. So, we can find the moles of citric acid using this ratio: moles of citric acid = \(\frac{1}{3}\) × moles of KOH = \(\frac{1}{3} \times 0.0116 \mathrm{~mol}\) moles of citric acid = \(0.00387 \mathrm{~mol}\) Now, calculate the molarity of the citric acid solution: Molarity of citric acid solution = \(\frac{\text{moles of citric acid}}{\text{volume of solution in L}}\) Molarity of citric acid solution = \(\frac{0.00387 \mathrm{~mol}}{100.00 \times 10^{-3} L}\) Molarity of citric acid solution = \(0.0387 \mathrm{M}\) The molarity of the citric acid solution is \(0.0387 \mathrm{M}\).

Key Concepts

Citric Acid

Citric acid, known chemically as \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \), is a triprotic acid. This means it has three acidic protons (H⁺ ions) that can be donated in reactions. It is commonly found in citrus fruits like oranges and lemons. Not only does citric acid add a sour taste to foods, but it also acts as a preservative and plays a role in metabolic pathways such as the citric acid cycle in biochemistry. In laboratory settings, citric acid is often used in titration experiments, where its concentration in a solution can be determined by neutralizing it with a base like potassium hydroxide (KOH). By understanding its chemical properties, we can apply citric acid in both food science and industry effectively.

Net Ionic Equation

To understand the reaction between citric acid and potassium hydroxide (KOH), we use the concept of a net ionic equation. This type of equation focuses solely on the chemical species that participate in the reaction, excluding spectator ions which remain unchanged.
In this case, we begin with the complete ionic equation where citric acid ( \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} \) ) reacts with KOH, breaking into ions in solution: 3 moles of \( \mathrm{H^{+}} \) combine with 3 moles of \( \mathrm{OH^{-}} \) from KOH, forming water. The net ionic equation simplifies to reflect the core reaction: \[ 3\mathrm{H^{+}} + 3\mathrm{OH^{-}} \to 3\mathrm{H_{2}O} \] This equation emphasizes the neutralization of acids by bases resulting in water (\( \mathrm{H_{2}O} \)). Removing spectator ions allows chemists to focus on the crucial transformation occurring in the reaction.

Stoichiometry

Stoichiometry is the calculation based process that allows chemists to predict the amounts of reactants or products involved in a chemical reaction. The stoichiometric coefficients in a balanced reaction tell us the ratio in which substances react.
In the case of the titration involving citric acid and KOH, the balanced chemical equation indicates that 1 mole of citric acid reacts with 3 moles of KOH. This ratio is critical for calculations because it directly relates the amount of acid to the amount of base used in the titration. Using the ratio, we can determine that when you use 3 moles of KOH, you are neutralizing 1 mole of citric acid. Understanding stoichiometry is thus essential for determining the precise amounts needed in reactions and ensuring complete neutralization in titrations.

Molarity Calculation

Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. Calculating molarity allows us to understand the strength of a solution, which is critical in many chemical reactions.
In the titration example, the volume of KOH used is 23.20 mL, and its concentration is 0.500 M. The moles of KOH are found using the formula: \( \text{moles} = \text{Molarity} \times \text{Volume} \). When converted into liters \((23.20 \times 10^{-3} L)\), the moles of KOH turn out to be 0.0116 mol. Since the stoichiometry establishes a 3:1 ratio of KOH to citric acid, the moles of citric acid can be calculated as \( \frac{1}{3} \times 0.0116 \). Finally, to calculate the molarity of citric acid, divide the moles of citric acid by the volume of the citric acid solution in liters \((0.100 L)\). This gives a solution with a molarity of 0.0387 M. Accurately calculating molarity is vital for the preparation and standardization of solutions in laboratories.