Question

I did a titration of 0.08339 N of NAOH (buret). The acid used was 5% household vinegar (acetic acid) 25 mL was diluted with water in a 250 mL volumetric flask. Then 50 mL of that was put into a 250mL conical flask used in the experiment.
The end point occurred at 37.30 mL
I need to know the w/v % ??
Can you help please.
60.05 = g/mol acetic acid

Answer 1

25 ml 5% vinegar

25 * 5 = 125 / 100 = 1.25 gms of vinegar (A.A.)

which is diluted to 250 ml

i.e. w/v = 1.25 / 250 ml = 0.005%

out of which 50 ml is titrated against 0.08339 NaOH and volume required is 37.7 ml,

so,

we can calculate the w/v % of 50 ml as;

if 250 ml contains 0.005%

so;

50 ml = ?

by cross multiplying if

50 * 0.005 = 0.25 / 250 = 0.001 %

w/v = 0.001 %.

calculation as per titration

we can calculate the w/v of vinegar as;

moles = 0.08339 N * 37.7 = 3.14 moles of NaOH

required to neutralise the acid ,

so,

it may have 3.14 moles of Acetic acid

= moles = 3.14 * 60

= 188.62

so,

w/v = 188.62 gms of A.A./ 50 + 37.7

= 188.62 / 87.7

= 2.15 %