Question
ok im not sure how to do this
15.00ml of acetic acid is pipetted into a flask
0.35 grams of sodium acetate trihydrate is added and mixed
then 0.75 of the hydrate is added
1)i assumed that the volume(s) are additions. 15.00ml x .35g *m/1.45g and for the second part 15.00 + 0.35g *mol/1.45 g +0.75 g* mol/1.45g= x
2)not sure how to do the tryhydate. Do i minus it from the total mass of sodium hydrate ? or is it non reactive?
can u show me the proper method to do this ?
15.00ml of acetic acid is pipetted into a flask
0.35 grams of sodium acetate trihydrate is added and mixed
then 0.75 of the hydrate is added
1)i assumed that the volume(s) are additions. 15.00ml x .35g *m/1.45g and for the second part 15.00 + 0.35g *mol/1.45 g +0.75 g* mol/1.45g= x
2)not sure how to do the tryhydate. Do i minus it from the total mass of sodium hydrate ? or is it non reactive?
can u show me the proper method to do this ?
Answers
DrBob222
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.
DrBob222
Another thing! You haven't asked a question. What are you to calculate?
Jim_R
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
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
DrBob222
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.
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.
Jim_R
thanks
i got ka=2.5 x 10^-5 and 2.5 x 10-6
thanks again Jim
i got ka=2.5 x 10^-5 and 2.5 x 10-6
thanks again Jim
Jim_R
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
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