Question

Take approximately 20 mL of 0.5 M of HC2H3O2 (acetic acid) solution in a clean, small beaker. 2. Prepare 25 mL of three different acetic acid solutions with concentrations in the range between 0.05 and 0.2 M from the stock 0.5 M acetic acid solution using appropriate dilution techniques.

Answer 1

Given stock acetic acid solution concentration=0.5 M.

Now we have to make three diluted solutions of 25 ml.

1) 0.05 M, 25 ml

From dilution law, M1V1=M2V2

Here M1=0.5 M, V1=?, M2=0.05 M, V2=25 ml.

V1=M2V2/M1=(0.05x25)/0.5=2.5 ml.

So 2.5 ml of stock solution+22.5 ml water will give you 0.05 M solution.

2) 0.1 M solution.

similar calculation like above, V1=M2V2/M1

Here M2=0.1 M, V2=25ml and M1=0.5 M.

V1=(0.1x25)/0.5=5 ml.

So 5 ml of stock solution+20 ml of water will give you 0.1 M.

3) 0.2 M solution.

Here M2=0.2 M, V2=25 ml and M1=0.5 M,

V1=(0.2x25)/0.5=10 ml.

So 10 ml of stock solution+15 ml of water will give you 0.2 M solution.