# chemistry

posted by .

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 ?

• chemistry -

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.

• chemistry -

Another thing! You haven't asked a question. What are you to calculate?

• chemistry -

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

• chemistry -

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.

• chemistry -

thanks

i got ka=2.5 x 10^-5 and 2.5 x 10-6

thanks again Jim

• chemistry -

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