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To investigate the concept of molarity and to determine the concentration of an acid with an unknown molarity.
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 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.
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Swirl the acetone in the beaker for a minute and then decant it into a container marked "acetone waste."
Record in the data table.
| Mass of 100 mL beaker (g) | Mass of beaker + CaCO3 before reaction (g) | Mass of CaCO3 before reaction (g) | Mass of beaker + CaCO3 after reaction (g) | Mass of CaCO3 after reaction (g) | Mass of CaCO3 used up in the reaction (g) |
|---|---|---|---|---|---|
| 54.67 | 68.45 | 13.78 | 59.24 | 4.57 | 9.21 |
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
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
| Parameter | Value |
|---|---|
| Moles of CaCO3 used up | 0.092 mol |
| Moles of HCl that reacted | 0.046 mol |
| Molarity of HCl solution | 0.92 M |
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.
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.
Determination of Hydrochloric Acid Concentration Using Molarity. (2024, Jan 05). Retrieved from https://studymoose.com/document/determination-of-hydrochloric-acid-concentration-using-molarity
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