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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 -


    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 -


    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


    ka =3.8 x10-5

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