Question

Acetic Acid Serial Dilution

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 0.025 M acetic acid solution. The glassware available is a 100 mL volumetric flask, 50.00 mL volumetric flask, 1.00 mL volumetric pipet, and a 10 mL graduated pipet. For the first step, dilute 1.00 mL of glacial acetic acid to 50.00 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.

0.025 M CH3CO2H

A. ___________

B.____________

Answer 1

Alternatively: Using the formula: M1V1 = M2V2         - equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1

            M2= molarity of solution 2, V2= volume of solution 2.

Solution A:

Let’s denote the stock solution as 1 and the solution to be prepared as solution 2

                17.4 M x 1.0 mL = M2 x 50 mL                     [Volume of solution 2 = 50 mL]

                Or, M2 = (17.4 M x 1.0 mL) / 50 mL = 0.348

Thus, the final molarity of solution = 0.348 M

Solution B:

Here, solution 1 is solution A. Solution 2 is the solution to be prepared, i.e. solution B.

Solution B, 100 mL (V2), 0.025 M solution.

Solution 1 (= solution A) molarity (M1) = 0.348

Now, 0.348 M x V1 = 0.025 M x 100 mL

Or, V1 = 7.183908045977011 = 7.2 mL (approx.)

Method of preparation: Transfer 7.2 mL of solution A using 10.0 mL graduated pipette into 100 mL standard volumetric flask. Make the volume of the flask upto the mark with distilled water. The resultant solution B is 0.025 M, 100 mL.