Posted by Jim_R on Thursday, July 16, 2009 at 7:35pm.
I'm not sure what you have done. For example, 0.75 what of the hydrate is added. And the 0.35 g trihydrate are mixed with what? In a separate beaker? With the acetic acid? Please clarify the problem.
Another thing! You haven't asked a question. What are you to calculate?
oh
This is an acid base experiment this is the last stage.
There are two parts:
part 1) 0.3470g sodium acetate trihydate is added to 15.00 ml acetic acid (concentration is 0.20)
then its mixed and a ph is taking--i got 4.30 PH
part II)
Added 0.7618 g of sodium acetate trihydrate to the above mixture. Ph was 4.85
Calculate the Ka for the weak acid.
Not really sure how to do it
HC2H3O2 ==> H^+ + C2H3O2^-
Ka = (H^+)(C2H3O2^-)/(HC2H3O2)
Calculate (H^+) from pH = -log(H^+).
Calculate (C2H3O2^-) from g of the trihydrate. That is moles = grams/molar mass and mole/L (L is 0.15) = M
Calculate (HC2H3O2) from L x M = ?
Plug in and calculate Ka.
For part 2,
determine moles trihydrate added and add that to the amount of trihydrate already there, then total moles/L = M of the (C2H3O2^-). Obtain (H^+) from pH, and (HC2H3O2) is the same as the first part. Calculate Ka.
The correct Ka is about 1.8 x 10^-5.
thanks
i got ka=2.5 x 10^-5 and 2.5 x 10-6
thanks again Jim
here's what i finally ended up with
I don't know why u used 59 grams
nac2h302 x 3h20---> na + c2h302 + 3h20
sodium acetate trihydrate=acetic ions
.3470 g nac2h302 x 3h20 /136.10= 2.550 x 10-3 mol acetic ions formed
2.550 x 10-3 M/.0150 L=.17
ka= (5.0 x 10-5)(.17)/(.20)=4.3 x10-5
part b
1.1088g/136.1=.54
ka =3.8 x10-5