For part of the experiment, you will need to prepare 100 mL of 0.025 M CH3CO2H (
ID: 945262 • Letter: F
Question
For part of the experiment, you will need to prepare 100 mL of 0.025 M CH3CO2H (acetic acid), starting from commercial ?glacial? acetic acid, which has a concentration of 17.4 M. Devise a method using a two step serial dilution to make 100 mL of 0.025M acetic acid solution. For the first step dilute 1.00 mL of glacial acetic acid to 50.0 mL to produce solution 1. You will then use solution 1 to make the 0.025 M concentration for step 2. Determine the volume of solution 1 needed to make 100 mL of 0.025 M acetic acid.
Explanation / Answer
1)
V1 = 1 mL
M1 = 17.4 M
V2 = 50 mL
M2 = ?
M1V1 = M2V2
M2 = M1V1/V2 = 17.4 M x 1 mL / 50 mL = 0.348 M
M2 = 0.348 M
2) M1 = 0.348 M
V1 = ?
M2 = 0.025 M
V2 = 100 mL
M1V1 = M2V2
V1 = M2V2/M1 = 0.025 M x 100 mL/ 0.348 M = 7.2 mL
Therefore,
7.2 mL of solution 1 needed to make 100 mL of 0.025 M acetic acid.
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