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 didn't 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

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.

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

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

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

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]

.
.
.

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..

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

Yes, that's the answer.

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?

Ok, since the mole ratio is 1:1, you have the same number of moles of acetic acid reacting with the number of moles of NaOH.

The equation, if you want to know is:

NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O

So, you have the number of moles of acetic acid in 10 mL

Ohh so moles of acetic acid present in 10mL sampel is just same answer as moles of NaOH used to reach emd point , correct? Two answers are same , corret?

And do you know if only one hydrogen in acetic acid acts as an acid, write a balanced chemical reaction between acetic acid and sodium hydroxide.

CH3COOH+NaOH---->NaOOCCH3+H2O?

Yes, exactly!

Yes, only one does. The H in CH bond does not act as an acid, but the H in the COOH does. It's the case for all organic compounds, as far as I know.

The equation is already balanced :)

Thanks

I think its going to be last question

If initial volume of NaOH by mistake 1mL rather than 0mL.
What happens to the moles of NaOH? Higher , lower , no effect, why?
I think its lower

What effect does it have on the moles of acetic acid?
Lower

Is it correct?

Less amount of solution means less moles of NaOH used. In turn, you had less moles in 10 mL Since it's less, you have a more diluted acetic acid.

So, yes! :)

Thankssss

Thank you so much you are so nice person in the world!


Have a great night;)

You're welcome! It's 10.45 am here ;)