hi im having problems figuring this out.

Question

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 ?

Answer 1

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.

Answer 2

thanks

i found the lr

then plugged did some extra stuff that i should have

thanks again Jim

Answer 3

how bout if 200ml of koh is added ?

Answer 4

how bout if 200ml of koh is added ?

the answer is 8.78

im getting 12.82

Answer 5

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.

Answer 6

thanks.

Damn your smart dr.bob