Question

A solution is prepared by dissolving $0.5842\mathrm{g}$ oxalic acid $\left({\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right)$ in enough water to make $100.0\mathrm{mL}$ of solution. A $10.00-\mathrm{mL}$ aliquot (portion) of this solution is then diluted to a final volume of $250.0$ $\mathrm{mL}$. What is the final molarity of the oxalic acid solution?

Short answer

The final molarity of the oxalic acid solution is approximately $0.00260\mathrm{M}$.

Step-by-step solution

Calculate initial moles of oxalic acid

The molar mass of oxalic acid can be calculated as: ${\mathrm{M}}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}=2\times {\mathrm{M}}_{\mathrm{H}}+2\times {\mathrm{M}}_{\mathrm{C}}+4\times {\mathrm{M}}_{\mathrm{O}}$ ${\mathrm{M}}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}=2\times 1.01+2\times 12.01+4\times 16.00$ ${\mathrm{M}}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}\approx 90.04\mathrm{g}/\mathrm{mol}$ Now, we can calculate the initial moles of oxalic acid, using the given mass $0.5842\mathrm{g}$ as follows: ${\mathrm{n}}_{\mathrm{initial}}=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}$ ${\mathrm{n}}_{\mathrm{initial}}=\frac{0.5842\mathrm{g}}{90.04\mathrm{g}/\mathrm{mol}}$ ${\mathrm{n}}_{\mathrm{initial}}\approx 0.00649\mathrm{mol}$

Calculate the initial molarity

The initial molarity can be found using the initial moles and the volume of $100.0\mathrm{mL}$: ${\mathrm{M}}_{\mathrm{initial}}=\frac{{\mathrm{n}}_{\mathrm{initial}}}{{\mathrm{V}}_{\mathrm{initial}}}$ ${\mathrm{M}}_{\mathrm{initial}}=\frac{0.00649\mathrm{mol}}{0.100\mathrm{L}}$ ${\mathrm{M}}_{\mathrm{initial}}\approx 0.0649\mathrm{M}$

Find the new moles of oxalic acid after the aliquot

After taking $10.00\mathrm{mL}$ aliquot, we need to find the new moles of oxalic acid in that portion. ${\mathrm{n}}_{\mathrm{new}}={\mathrm{M}}_{\mathrm{initial}}\times {\mathrm{V}}_{\mathrm{aliquot}}$ ${\mathrm{n}}_{\mathrm{new}}=0.0649\mathrm{M}\times 0.0100\mathrm{L}$ ${\mathrm{n}}_{\mathrm{new}}\approx 0.000649\mathrm{mol}$

Calculate the final molarity

Finally, calculate the final molarity of the oxalic acid, using the aliquot moles and the final volume of $250.0\mathrm{mL}$: ${\mathrm{M}}_{\mathrm{final}}=\frac{{\mathrm{n}}_{\mathrm{new}}}{{\mathrm{V}}_{\mathrm{final}}}$ ${\mathrm{M}}_{\mathrm{final}}=\frac{0.000649\mathrm{mol}}{0.250\mathrm{L}}$ ${\mathrm{M}}_{\mathrm{final}}\approx 0.00260\mathrm{M}$ So, the final molarity of the oxalic acid solution is approximately $0.00260\mathrm{M}$.

Key concepts

Oxalic Acid

Oxalic acid is a colorless organic compound with the formula ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$. It is known for its use as a bleaching and cleaning agent. The structure of oxalic acid is simple, consisting of two carboxylic acid groups. This compound is typically handled with care due to its acidic nature.

To calculate properties like molarity, you often need to know its molar mass, which is determined by summing up the atomic masses of hydrogen, carbon, and oxygen in its chemical formula. The calculation involves the following molecular weights: hydrogen ($1.01\mathrm{g}/\mathrm{mol}$), carbon ($12.01\mathrm{g}/\mathrm{mol}$), and oxygen ($16.00\mathrm{g}/\mathrm{mol}$). By summing these, we find the molar mass of oxalic acid to be approximately $90.04\mathrm{g}/\mathrm{mol}$. This value is essential for subsequent calculations in solution preparation.

Solution Concentration

Solution concentration, in terms of chemistry, often refers to molarity. Molarity provides a way to express the concentration of a solute in a solution, measured in moles per liter ($\mathrm{mol}/\mathrm{L}$). In our exercise involving oxalic acid, solution concentration is calculated at several steps.

Initially, we begin with a specific amount of oxalic acid ($0.5842\mathrm{g}$), dissolved in enough water to make $100.0\mathrm{mL}$ solution. To find initial molarity, we first need to know the moles present, which depends on the molar mass of the solute.

• Weight of oxalic acid: $0.5842\mathrm{g}$
• Volume of solution: $100.0\mathrm{mL}$ or $0.100\mathrm{L}$
• Calculated initial moles: $0.00649\mathrm{mol}$
• Initial molarity: $0.0649\mathrm{M}$

Understanding this core concept helps determine how the concentration changes during the dilution process.

Moles Calculation

Calculating moles allows us to understand how much substance is present in a chemical reaction. In our exercise, moles are calculated using the formula: $\mathrm{n}=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}$. Here's how it's applied: *

We start with $0.5842\mathrm{g}$ of oxalic acid and use its molar mass to find the moles in the solution. With the formula in place, the steps to deduce the moles are straightforward:

• Molar mass from previous calculation: $90.04\mathrm{g}/\mathrm{mol}$
• Resulting moles: $0.00649\mathrm{mol}$

This value is essential for further concentration calculations, illustrating how much of the oxalic acid is involved in reactions or further dilutions.

Dilution Process

The dilution process involves reducing the concentration of a solution by adding more solvent. Our exercise demonstrates this by taking a $10.00\mathrm{mL}$ aliquot and diluting it to a total volume of $250.0\mathrm{mL}$.

To calculate the final molarity of the diluted solution, we begin with the aliquot's initial moles of oxalic acid. Following the formula for molarity: $\mathrm{M}=\frac{\mathrm{n}}{\mathrm{V}}$ we continue with a few straightforward steps:

• Initial moles from $10\mathrm{mL}$: $0.000649\mathrm{mol}$
• Volume after dilution: $0.250\mathrm{L}$

Finally, we determine the final molarity to be $0.00260\mathrm{M}$, showcasing how dilution decreases concentration while maintaining the same amount of solute present. This understanding of dilution is critical for performing accurate laboratory experiments.