[h_loyali]

Hello everyone,

I had Acid-Base lab that need to determine Ka and concentraion but I'm confused on how to calculate concentration of CH3COO.

Equation is : H₂O + HC₂H₃O₂    H₃O + C₂H₃O₂
Concentration of Acetic Acid on bottle was 0.20 M

I tried to do some calculation but I couldn't figure it out so I came to this forum to get some help . I bolded those areas that I have no idea how to calculate.
It's divided in 3 parts :

Part A & B

HC₂H₃O₂ + NaOH  H₂O + Na + C₂H₃O₂

 A )  I did pipette 25 mL of 0.20 M acetic acid into beaker and added 11.52 ml NaOH and measured pH of solution which was 4.70 .

 B)  I did pipette 25 mL of 0.20 M acetic acid into beaker and added 18.40 ml NaOH and  pH of solution  was 5.10 .


Looks like my calculations are wrong and don't know the other ones so placed Q mark.

NaOH                         pH                          [H3O+]                         [CH3COO]/[Ch3COOH]                                        Ka           A) 11.52 mL                  4.70              10^-4.85= 1.9 *10-5                    1 ?                   [CH3COO]*[H3O+]/[Ch3COOH] = (1.9 *10-5)2 / 0.20M =1.90*10-9 ? B) 18.40 mL                  5.10              10^-5.10= 7.9 *10-6                     3 ?                    [CH3COO]*[H3O+]/[Ch3COOH] = (7.9 *10-6)2 / 0.20M =3.1*10-10 ?

Part C                                               Initial[CH3COOH]                       pH                       [H3O]                           [CH3COO]                  Step 1 (Acetic Acid only)                         0.20M                                 2.55                          1.86*10-3?                                ?                            Step 2 (+.63g of sodium acetate)            0.20M                                  4.20                        6.3*10-5?                                    ?                                        %Dissociation  :  ?

I would appreciate if you help me to understand and find each bolded one correctly.

Thanks.

[Borek]

[CH₃COO]*[H₃O+]/[Ch₃COOH] = (1.9*10^-5)² / 0.20M =1.90*10^-9

Apparently you are assuming here that [CH₃COO^-] = [H⁺]. That's (approximately) true when the only source of CH₃COO^- is the dissociation, that's not the case here. (You have even signaled it is not the case in the same line.)

Edit: not to mention other (related) mistake. Concentration of undissociated acid is not 0.2 M.

[h_loyali]

yes but how to find concentration of   [CH3COO] ?

[Borek]

In the table you suggested ratio of acetic acid to acetate is equal to 1 - what was the logic behind?

[h_loyali]

That is what our teacher said.

[Borek]

That hardly counts as a logic.

How much acetic acid was neutralized?