1.Pippete 20ml of 0.1M acetic acid and 20ml of 0.1M NaOH into a 100ml beaker (Remember that the final volume of this solution is the sum of the volumes of the acetic acid and NaOH solutions that have been mixed)
2. Pippete 20ml of 0.1 acetic acid into another 100ml beaker and add 20ml of distilled water with a pipette. This produces 40ml of a solution in which the concentration of the acetic acid is half that of the stock solution
Prepare the solution by mixing the solutions prepared in steps 1 and 2. The total volume should be 80ml.
What are the molarities of CH3C00 in the buffer described above?
You shouldn't have any trouble. Final molarity = final moles/final liters of solution. Post your work or detailed questions if you get stuck.
To find the molarities of CH3COOH in the buffer described above, we first need to calculate the moles of acetic acid present in each solution.
In step 1:
Volume of acetic acid solution = 20 ml
Molarity of acetic acid solution = 0.1 M
Using the formula:
Moles = Molarity * Volume (in liters)
Moles of acetic acid in step 1 = 0.1 M * (20 ml / 1000 ml) = 0.002 moles
In step 2:
Volume of acetic acid solution = 20 ml
Molarity of acetic acid solution = 0.1 M
Moles of acetic acid in step 2 = 0.1 M * (20 ml / 1000 ml) = 0.002 moles
Next, we need to calculate the total number of moles of acetic acid in the final buffer solution:
Total moles of acetic acid = moles from step 1 + moles from step 2 = 0.002 moles + 0.002 moles = 0.004 moles
Finally, we can calculate the molarity of acetic acid in the buffer solution:
Molarity = Moles / Volume (in liters)
Volume of the final buffer solution = 80 ml = 0.08 L
Molarity of acetic acid in the buffer = 0.004 moles / 0.08 L = 0.05 M
Therefore, the molarity of CH3COOH in the buffer described above is 0.05 M.