How to determine the volume of titrant required to reach the equivalance point????

Question

A 20.00 mL sample of a 0.270 M acetic acid solution is titrated with 0.400 M sodium hydroxide solution. The densities of the acetic acid solution and sodium hydroxide solutions are 1.025 g/mL and 1.104 g/mL, respectively.

Answer 1

20.00 mL sample of a 0.270 M acetic acid :

0.02000 Litres @ 0.270 mol/Litre = 5.4e-3 moles acetic acid

1 mol acetic acid reacts with 1 mole NaOH

5.4e-3 moles acetic acid = 5.4e-3 moles NaOH

5.4e-3 mol NaOH @ 0.400mol/litre = 0.0135-0 litres

your answer : 13.50 ml of NaOH

Answer 2

Density does not play any part in these problems

Write a balanced equation:

CH3COOH + NaOH → CH3COONa + H2O

1mol acid reacts with 1 mol base

C1V1 = C2V2

0.270*20 = 0.400*V2

V2 =  0.27*20/0.400

V2 = 13.5 ml

13.5ml base required