standardization of NaOH solution:
a)volume of 6.0M NaOH solution used: 20mL
b)approximate[NaOH] after dilution to 400mL: 0.5M
c)weight of empty flask: 86.525g
d)weight of flask plus KHP: 87.253g
e)weight of KHP: 0.728g
f)moles of KHP: 0.00356mol
g)moles of NaOH: same answer as moles of KHP
h)initial buret reading: 0mL
i)final buret reading: 28.49mL
j)volume of NaOH used up: 28.49mL
k)[NaOH]: 0.12496M
l)Average [NaOH]: 0.11421M (I just didnt write sample2 list on here)

Determination of percent Acetic Acid in an unknown vinegar:
Initial volume of NaOH in the buret: 0
Final volume of NaOH in the buret: 17.5
Volume of NaOH used to reach the end point: 17.5
Average [NaOH] from step 1:________________________
Moles of NaOH used to reach the end point: ______________
Moles of acetic acid present in 10mL sample:_______________

Please help on these three blank space for me and
step 1 is preparation of 0.3M NaOH solution-rinse your graduated cylinder with a few protions of distilled water.
just in case for step 2 is measure about 20mL of 6.0M NaOH solution.

Please help me

Unknown008

For the first blank, you have to insert the concentration of the NaOH you used for the titration in step 2.

Using volume and concentration, you can find the number of moles of NaOH used for the titration.

Acetic acid is CH3COOH. You have a monobasic acid, which means it reacts with one mole of OH^- to produce water. The mole ratio is therefore 1:1. You know the number of moles of NaOH used, it's the same number of moles, provided you titrated the NaOH against 10 mL of acid.

sarah1004

so third blank answer is just same as second blank answer?

sarah1004

second blank answer is
17.5mL X 1/1000mL X 0.11421M = .001999mole?

sarah1004

could you please help me the how to find moles of NaOH used to reach the end point
and
moles of acetic acid present in 10mL sample?

Thanks

Unknown008

Wait wait wait...

Could you post the questions as well? I'm getting quite confused about all this. Type in this format:

[Question]
[Answer you gave]

[Question]
[Answer you gave]

.
.
.

sarah1004

second blank answer is
17.5mL X 1/1000mL X 0.11421M = .001999mole?
(first blank average [NaOH] from step 1 is 0.11421)
I am sorry that cause its due tomorrow sorry..

Unknown008

I don't understand how you got your concentration as 0.11421 M.

Unknown008

Yes, that's the answer.

sarah1004

concentration is 0.11421M because before i took another test and then I used that number...
do you know how to find moles of acetic acid present in 10mL sample??