Problem 39 - Chemistry The Molecular Science

Question

Vitamin \(\mathrm{C}\) is a monoprotic acid. To analyze a vitamin \(\mathrm{C}\) capsule weighing \(0.505 \mathrm{~g}\) by titration took \(24.4 \mathrm{~mL}\) of 0.110-M NaOH. Calculate the percentage of vitamin C \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6},\) in the capsule. Assume that vitamin \(\mathrm{C}\) is the only substance in the capsule that reacts with the titrant.

Step-by-Step Solution

Answer

The capsule contains approximately 93.56% vitamin C.

Step 1: Understand the Reaction

Vitamin C (ascorbic acid), denoted as \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6} \), reacts with \( \mathrm{NaOH} \) in a 1:1 molar ratio because it is a monoprotic acid. The balanced chemical equation is: \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6} + \mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{6}^- + \mathrm{Na}^+ + \mathrm{H}_{2} \mathrm{O}.\)

Step 2: Calculate Moles of NaOH Used

Given \( 24.4 \; \mathrm{mL} \) of \( 0.110 \; \mathrm{M} \) \( \mathrm{NaOH} \), convert the volume to liters and then use the molarity to find moles: - Volume in liters: \( 0.0244 \; \mathrm{L} \)- Moles of \( \mathrm{NaOH} \): \( 0.0244 \; \mathrm{L} \times 0.110 \; \mathrm{mol/L} = 0.002684 \; \mathrm{mol}.\)

Step 3: Determine Moles of Vitamin C

Since the reaction is 1:1, the moles of vitamin C equals the moles of \( \mathrm{NaOH} \):\[ \text{Moles of } \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6} = 0.002684 \; \mathrm{mol}. \]

Step 4: Calculate Mass of Vitamin C

Use the moles of vitamin C and its molar mass to find its mass. The molar mass of \( \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6} \) is approximately \( 176.12 \; \mathrm{g/mol} \):\[ \text{Mass } = 0.002684 \; \mathrm{mol} \times 176.12 \; \mathrm{g/mol} = 0.4727 \; \mathrm{g}. \]

Step 5: Calculate Percentage of Vitamin C in Capsule

Calculate the percentage of vitamin C in the capsule using the mass found and the initial mass of the capsule:\[ \text{Percentage} = \left( \frac{0.4727 \; \mathrm{g}}{0.505 \; \mathrm{g}} \right) \times 100\% \approx 93.56\%. \]

Key Concepts

Understanding Monoprotic Acids

Vitamin C, or ascorbic acid, is classified as a monoprotic acid. This means it can donate one hydrogen ion (proton) per molecule to a base during a chemical reaction. When it interacts with a base like sodium hydroxide (NaOH), the reaction occurs in a 1:1 ratio.
This characteristic is crucial in titration, a method used for determining the amount of an acid in a solution by adding a base of known concentration. Since Vitamin C has just one acidic proton, each mole of Vitamin C will react with exactly one mole of NaOH. This 1:1 stoichiometry simplifies calculations, as the moles of acid are equal to the moles of base.
Titrations involving monoprotic acids are foundational in chemistry because understanding them helps reveal the concentration of unknown solutions in various scientific analyses.

Molarity Calculation Steps

Molarity, the measure of concentration, is expressed as moles of solute per liter of solution. Calculating molarity involves a few straightforward steps, especially useful in titrations.

• First, convert the volume used in the titration from milliliters to liters. For example, if you have 24.4 mL, that's 0.0244 liters.
• Next, use the molarity of the titrant (in this case, 0.110 M NaOH) to determine the number of moles of NaOH used. Multiply the volume in liters by the molarity: \[0.0244 \; \text{L} \times 0.110 \; \text{mol/L} = 0.002684 \; \text{mol}.\]

The calculated moles of NaOH directly reveal the moles of Vitamin C reacted, thanks to the 1:1 relationship in monoprotic reactions. This calculated amount is pivotal in determining other metrics, like the mass and concentration of the acid.

Balancing Chemical Reactions

The essence of balancing chemical reactions is ensuring that the same number of each type of atom appears on both sides of the equation. This is vital for maintaining the law of conservation of mass.In the Vitamin C titration context, the equation is: \[\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6} + \mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{6}^- + \mathrm{Na}^+ + \mathrm{H}_{2} \mathrm{O}.\]Both sides have the same number of carbon, hydrogen, and oxygen atoms.

• The equation starts with one molecule of Vitamin C and one of NaOH.
• It ends with a deprotonated Vitamin C ion, a sodium ion, and water.

This balance helps in computing precise quantities in a chemical process, ensuring accurate calculations in lab settings. Understanding and balancing chemical reactions are essential skills in any chemist's toolkit, allowing for accurate experimental and analytical work.