Problem 112 - General Chemistry

Question

A \(71.2-g\) sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), was dissolved in \(1.00 \mathrm{~L}\) of solution. How would you prepare \(1.00 \mathrm{~L}\) of \(0.150 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) from this solution?

Step-by-Step Solution

Answer

Dilute 190 mL of the initial solution to 1.00 L.

Step 1: Determine Molar Mass of Oxalic Acid

First, calculate the molar mass of \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\). The atomic masses are approximately: Hydrogen (H) = 1.01 \(\mathrm{g/mol}\), Carbon (C) = 12.01 \(\mathrm{g/mol}\), and Oxygen (O) = 16.00 \(\mathrm{g/mol}\). The molar mass is \[(2 \times 1.01) + (2 \times 12.01) + (4 \times 16.00) = 2.02 + 24.02 + 64.00 = 90.04 \, \mathrm{g/mol}.\]

Step 2: Calculate the Initial Molarity

Next, determine the initial molarity of the solution using the given mass and volume. The number of moles of \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\) in the solution is \[\frac{71.2 \, \mathrm{g}}{90.04 \, \mathrm{g/mol}} = 0.790 \, \mathrm{mol}.\] Since these moles are dissolved in \(1.00 \, \mathrm{L}\) of solution, the molarity is \[0.790 \, \mathrm{M}.\]

Step 3: Use Dilution Formula

Now, use the dilution formula to find out how much of the initial solution is needed. The formula is \(C_1V_1 = C_2V_2\), where \(C_1 = 0.790 \, \mathrm{M}\), \(V_1\) is the volume needed, \(C_2 = 0.150 \, \mathrm{M}\), and \(V_2 = 1.00 \, \mathrm{L}\). Rearrange and solve for \(V_1\): \[V_1 = \frac{C_2V_2}{C_1} = \frac{0.150 \, \mathrm{M} \times 1.00 \, \mathrm{L}}{0.790 \, \mathrm{M}} = 0.190 \, \mathrm{L}.\]

Step 4: Prepare the Solution

Take \(0.190 \, \mathrm{L}\) (or \(190 \, \mathrm{mL}\)) of the initial solution and dilute it with water to make a total of \(1.00 \, \mathrm{L}\) solution. This will result in a \(0.150 \, \mathrm{M}\) solution of \(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\).

Key Concepts

Molarity Calculation

Molarity is a fundamental concept in chemistry that refers to the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. In the given exercise, we calculated the molarity of the oxalic acid solution by using the mass of oxalic acid that was dissolved:

• First, we determine the number of moles of oxalic acid using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
• The initial step involved calculating the molar mass of oxalic acid as 90.04 g/mol. With this, we find that the 71.2 g of oxalic acid is equivalent to 0.790 moles.
• Knowing that the solution's total volume is 1.00 L, the initial molarity can be calculated as 0.790 M using the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

Understanding molarity is crucial for preparing solutions of desired concentrations and for performing subsequent dilution calculations.

Dilution Formula

The dilution formula is used to determine how to prepare a solution of a desired concentration from a more concentrated stock solution. The formula for dilution is expressed as: \[ C_1V_1 = C_2V_2 \]

• \(C_1\) and \(V_1\) are the molarity and volume of the concentrated solution, respectively.
• \(C_2\) and \(V_2\) are the molarity and volume of the dilute solution, respectively.

In the exercise, we used this formula to find the volume \(V_1\) of the initial solution needed to prepare a 0.150 M oxalic acid solution:

• We know \(C_1 = 0.790 \, \mathrm{M}\) (initial molarity) and \(C_2 = 0.150 \, \mathrm{M}\) (desired molarity).
• Since the desired final volume \(V_2\) is 1.00 L, we rearranged the formula to solve for \(V_1\): \[ V_1 = \frac{C_2V_2}{C_1} = \frac{0.150 \, \mathrm{M} \times 1.00 \, \mathrm{L}}{0.790 \, \mathrm{M}} = 0.190 \, \mathrm{L} \]
• This means we must take 0.190 L (or 190 mL) of the concentrated stock solution and dilute it with water to achieve the desired concentration.

Utilizing the dilution formula allows chemists to efficiently and accurately mix solutions without repeated trial and error.

Oxalic Acid

Oxalic acid is an organic compound with the formula \( \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 \). It is a colorless crystalline solid that is soluble in water and forms a clear acidic solution. Oxalic acid is known for its uses in various industries and laboratory applications:

• In cleaning, it acts as a powerful cleaning agent for removing rust, bleaching, and other stains.
• It is used in the preparation of certain medicines and as a reagent in chemical laboratories.
• Oxalic acid occurs naturally in many plants and vegetables, such as spinach and rhubarb, but should be consumed in moderation due to its potential toxicity.

Safety is important when handling oxalic acid, as it can be harmful if ingested or inhaled in large amounts. It is essential to wear protective gear and work in a well-ventilated area during its handling. Understanding its properties and uses can help in safe and effective application in everyday and specialized tasks.