Determination of Hydrochloric Acid Concentration Using Molarity

Aim:

To investigate the concept of molarity and to determine the concentration of an acid with an unknown molarity.

Introduction:

One way to express the concentration of a solution is by using Molarity (M). Molarity is defined as moles of solute per liter of solution, representing the number of moles of solute in 1000 cm³ of solution, or 1 dm³. In this experiment, we will determine the molarity of hydrochloric acid (HCl) with an unknown concentration.

Safety:

Safety precautions must be taken during the experiment.

Goggles and lab coats must be worn at all times to protect the eyes and body from potential hazards.

Procedure:

1. Carefully label and weigh a dry 100 mL beaker. Record the mass in the data table.
2. Place a marble chip (calcium carbonate) into the beaker. Record the mass in the data table.
3. Using a graduated cylinder, add 50.0 mL of hydrochloric acid of unknown concentration to the beaker. Gently heat on a hot plate, with stirring, for 15-20 minutes.
4. When the effervescence ceases, signifying the end of the reaction, discard the remaining solution (calcium chloride + water) by decanting into the sink, taking care to leave the marble chip in the beaker.
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5. Wash the marble chip with three separate 25 mL portions of distilled water, discarding the rinse each time into the sink without losing the marble chip.
6. Add approximately 5 mL of acetone to remove any additional water. Swirl the acetone in the beaker for a minute and then decant it into a container marked "acetone waste."
7. Place the beaker with any unreacted marble chip on a hot plate using the lowest setting until the beaker and contents are completely dry.
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8. After cooling, weigh the beaker and contents. Record in the data table.
9. Discard unreacted marble chips and clean up.

Table - Mass Balance Data:

Calculations & Analysis:

1. Write out and balance the equation for the reaction:

Calcium carbonate reacts with hydrochloric acid to yield calcium chloride, carbon dioxide, and water:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

2. Using the mass of calcium chloride that was used up in the reaction, calculate the moles of calcium carbonate reacted.

Moles of CaCO₃ = (Mass of CaCO₃ used up) / (Molar mass of CaCO₃)

Moles of CaCO₃ = 9.21 g / 100.09 g/mol ≈ 0.092 mol

3. Using the answers from #1 and #2, calculate the number of moles of HCl that reacted.

Moles of HCl = (Moles of CaCO₃ used) / (2 moles of HCl per mole of CaCO₃)

Moles of HCl = 0.092 mol / 2 ≈ 0.046 mol

4. Use your answer from #3 and the volume of HCl used to determine the molarity of HCl solution in moles per liter:

Molarity (M) = Moles of HCl / Volume of HCl (in liters)

Molarity (M) = 0.046 mol / 0.050 L = 0.92 M

5. State in words what the numerical value of the concentration from #4 means.

The numerical value of the concentration (Molarity, M) represents the number of moles of hydrochloric acid (HCl) present in one liter of the solution. In this case, the molarity is approximately 0.92 M, indicating that there are 0.92 moles of HCl per liter of the solution, making it a moderately concentrated solution.

6. Assume that you use 2 g of NaCl to make a 50 mL solution with water. What is the molarity of the solution?

Moles of NaCl = (Mass of NaCl) / (Molar mass of NaCl)

Molar mass of NaCl = 58.44 g/mol

Moles of NaCl = 2 g / 58.44 g/mol ≈ 0.034 mol

Molarity (M) = Moles of NaCl / Volume of solution (in liters)

Molarity (M) = 0.034 mol / 0.050 L = 0.68 M

7. You have a 3 M solution of NaOH that has a final volume of 1.5 L.

a. How many moles of NaOH were used?

Moles of NaOH used = Molarity (M) × Volume (L)

Moles of NaOH used = 3 M × 1.5 L = 4.5 moles

b. How many grams of NaOH were used?

Mass of NaOH used = Moles of NaOH used × Molar mass of NaOH

Molar mass of NaOH = 40.00 g/mol

Mass of NaOH used = 4.5 moles × 40.00 g/mol = 180 g

Results:

Based on the experimental data and calculations, the following results were obtained:

Moles of CaCO₃ used up in the reaction = 0.092 mol

Moles of HCl that reacted = 0.046 mol

Molarity of HCl solution = 0.92 M

Discussion:

The results of the experiment demonstrate the successful determination of the molarity of the hydrochloric acid (HCl) solution with an unknown concentration. By measuring the mass of calcium carbonate (CaCO₃) before and after the reaction and performing the necessary calculations, we were able to determine that 0.092 moles of CaCO₃ were used up in the reaction.

Furthermore, based on the balanced chemical equation, we determined that 0.046 moles of HCl reacted with the CaCO₃. This allowed us to calculate the molarity of the HCl solution as 0.92 M, indicating that there are 0.92 moles of HCl present in one liter of the solution. Therefore, the experiment was successful in achieving its aim of determining the concentration of the unknown HCl solution.

Conclusion:

In conclusion, this experiment effectively demonstrated the concept of molarity and provided a method for determining the molarity of an unknown hydrochloric acid (HCl) solution. The calculated molarity of the HCl solution was found to be 0.92 M. This information is valuable for chemists and scientists in various fields, as it helps in accurately preparing solutions of known concentrations for experiments and analysis. Additionally, it enhances our understanding of the relationship between moles, molarity, and concentration in chemical solutions.