[Fish200398]

Buffer of HC3H2O2 45 mL 0,2 M (ka = 1,3 10^-5) with NaC3H5O2 0,10 M 55 mL. Then added NaOH 40 mg. what is the ratio of HC3H5O2/C3H5O2- ?

NaOH will react with HC3H2O2, both quantity is 9 moles, so it will react completely become salt NaC3H5O2? and using [C3H5O2-]^2 = kw/ka . [NaC3H5O2] ? i am not sure about this, since my answer not in the options. how to find the ratio of  HC3H5O2/C3H5O2- ?

[Borek]

NaOH will react with HC3H2O2

yes

both quantity is 9 moles, so it will react completely

no

[AWK]

NaOH will react with HC3H2O2, both quantity is 9 moles

9 milimoles

Then added NaOH 40 mg

How many milimoles?
Write down a balanced reaction and do simple stoichiometry.

[Fish200398]

1 mmol NaOH react with 9 mmol HC3H5O2. form 1 mmol NaC3H5O2. left 8 mmol HC3H5O2. total of C3H5O2^{-} = 6,5 mmol?
HC3H5O2 / C3H5O2^{-} = 8/6,5 = 16/13 ? is it true?

[AWK]

CH₃COOH + NaOH = CH₃COONa + H₂O
18 mmol     9 mmol                     
after reaction
9 mmol       0 mmol     9 mmol    it does not matter
now addition
              +1 mmol NaOH 
After addition of NaOH
9 ±..           0              9 ±..

[Fish200398]

noooo HC3H5O2 is 9 mmol at first, and there is 1 mole of NaOH

[AWK]

You have 9 mmol of salt and propanoic acid at the beginning. I formed this salt from an additional 9 mmol of acid and 9 mmol of NaOH.

[Fish200398]

there is 1 mole of NaOH so, produce 1 mole salt

[AWK]

Then added NaOH 40 mg

this is 1 millimole
Read the text carefully.

[Fish200398]

so the aswer is 16/13?

[AWK]

No.
The acid is a little less and the amount of salt has increased the same but more.
Starting numbers are 9/9