Make 50.0 mL of a 0.500 M solution of acetic acid (CH3COOH) from the 1.00 M CH3COOH using a 50.0 mL graduated cylinder.
1. Show your calculations and explain how you made the acetic acid solution.
mL1 x M1 = mL2 x M2
50 x 0.5 = mL2 x 1.00
mL2 = 25 mL.
Add 25 mL of the 1.00 CH3COOH to the empty 50 mL graduated cylinder and add distilled water to the 50.0 mL mark. The result is 50.0 mL of 0.50M CH3COOH.