How do I prepare 250ml of 0.5M acetic acid by dilution of 3M CH3COOH?
mL1 x M1 = mL2 x M2
250 x 0.5 = mL2 x 3M
Solve for mL2.
To prepare 250ml of a 0.5M acetic acid solution by dilution of a 3M CH3COOH solution, you can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (3M CH3COOH)
V1 = initial volume of the solution (unknown)
C2 = final concentration of the solution (0.5M acetic acid)
V2 = final volume of the solution (250ml)
Rearranging the formula, we get:
V1 = (C2 * V2) / C1
Now, substitute the values into the formula and solve for V1:
V1 = (0.5M * 250ml) / 3M
V1 = (0.5 * 250) / 3
V1 = 125 / 3
V1 ≈ 41.67 ml
Therefore, you need to take approximately 41.67ml of the 3M CH3COOH solution and then add enough solvent (usually water) to bring the total volume up to 250ml in order to prepare 250ml of 0.5M acetic acid.