Question

If I have 25 mL of CH3COOH and I titrate 23.9 mL of NaOH into it, then add another 25 mL of the acetic acid. What is the number of moles of CH3COOH and the resulting salt? Additionally what is the concentration of each?

Concentration of all acids are 0.1 M.

Answer 1

Answer :

Initial concentration of acid = 0.1M

First 25ml of 0.1M CH3COOH was titrated with 23.9 ml NaOH

CH3COOH + NaOH ----> CH3COONa + H2O

Moles of CH3COOH = 0.1M *0.025L = 0.0025 moles

As moles ratio of CH3COOH :CH3COONa is 1:1

Moles of CH3COONa = 0.0025 moles

After titration 25 ml of 0.1M another acetic acid is added.

So in the solution there are 0.0025 moles of acetic acid and 0.0025 moles of sodium acetate salt in the solution .

Total volume of the solution = 25 ml +23.9ml +25 ml = 73.9 ml = 0.0739L

So, [CH3COOH] = 0.0025 moles /0.0739L =0.034 M

[CH3COONa] = 0.0025 moles /0.0739L =0.034 M