Question

25mL of acetic acid was placed in a 250mL flask and diluted using water to the mark (250mL) and mixed. 25mL of this solution is then placed into a seperate flask and titrated using 16.5mL of 0.106M NaOH. What is the molarity of acetic acid (M.W.= 60.05 g/mol)?

Answer 1

Number of moles of NaOH titrated= Molairty * Volume = 0.106*16.5 = 1.749 milli mol

So,

number of moles of acetic acid present also should be 1.749 milli mol

Let Molarity of acetic acid be M1. Initially we had 25 mL of this solution which was diluted to 250 mL but number of moles will be same before or after dilution.

Number of moles of acetic acid = M1 * 25

So,

M1*25 = 1.749

M1=0.07 M

Answer: 0.07 M

Approach 2:

Number of moles of NaOH titrated= Molairty * Volume = 0.106*16.5 = 1.749 milli mol

So,

number of moles of H+ present also should be 1.749 milli mol

Volume = 250 mL

[H+] = 6.996*10^-3 M

Ka of acetic acid = 1.77*10^-5

CH3COOH <---> H+ CH3COO-

Here [H+] = [CH3COO-] = 6.996*10^-3 M

Ka= [H+][CH3COO-]/[CH3COOH]

1.77*10^-5 = (6.996*10^-3) (6.996*10^-3)/ [CH3COOH]

[CH3COOH]= 2.77 M

Let original Molarity of acetic acid be M1. Initially we had 25 mL of this solution which was diluted to 250 mL but number of moles will be same before or after dilution.

Molarity after dilution = M1/10

we got,

M1/10 = 2.77

M1= 27.7 M

Answer : 27.7 M