Posted by Jim_R on Thursday, July 23, 2009 at 11:12pm.
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.
thanks
i found the lr
then plugged did some extra stuff that i should have
thanks again Jim
how bout if 200ml of koh is added ?
how bout if 200ml of koh is added ?
the answer is 8.78
im getting 12.82
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.
thanks.
Damn your smart dr.bob
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