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Posted by on Thursday, July 23, 2009 at 11:12pm.

hi im having problems figuring this out.

consider the titration of 100 ml of 0.200 acetic acid by 0.100 koh. Calculate the ph if 250 ml of koh is added ?

I tried to find the limiting reagent and go from there but it was a no go

then i used the kb for find the poh and then the ph..i got 8.27

answer is 12.15

what is the correct method ?

  • chemistry - , Thursday, July 23, 2009 at 11:30pm

    You started correctly. I don't know where you went wrong.
    mols acetic acid initially = 0.1 x 0.2 = 0.02.
    mols KOH at 250 = 0.250 x 0.1 - 0.025
    So all of the acetic acid is used; an excess of 0.005 mols KOH remains. The volume (which probably is what you failed to take into account) is 100 + 250 = 350 mL.
    So (OH^-) = 0.005/0.350 = ??
    pOH and pH from there. I get 12.15, too.

  • chemistry - , Thursday, July 23, 2009 at 11:45pm

    thanks

    i found the lr

    then plugged did some extra stuff that i should have

    thanks again Jim

  • chemistry - , Thursday, July 23, 2009 at 11:56pm

    how bout if 200ml of koh is added ?

  • chemistry - , Friday, July 24, 2009 at 12:03am

    how bout if 200ml of koh is added ?


    the answer is 8.78

    im getting 12.82

  • chemistry - , Friday, July 24, 2009 at 12:21am

    At 200 mL, you are at the equivalence point.
    acetic acid KOH ==> Kacetate + H2O

    acetic acid initially = 0.02 moles
    KOH added 0.2 L x 0.1 M = 0.02 moles.
    So what do we have in solution. It is Kacaetate (potassium acetate) in 300 mL water. What determines the pH of salts in water. Its the hydrolysis, of course.
    So, acetate ion, which is C2H3O2^- hydrolyzes as follows:
    C2H3O2^- + HOH --> HC2H3O2 + OH^-
    Then Kb = Kw/Ka = (HAc)(OH^-)/(C2H3O2^-)
    Set up an ICE chart and calculate OH^- and from there pH.
    I get 8.79 using 1.75 x 10^-5 for Ka.

  • chemistry - , Friday, July 24, 2009 at 5:06pm

    thanks.

    Damn your smart dr.bob

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