Kneecoal

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adding the heat removes a water. so take "H2O" from the hydroxy parts of the carboxylic acids. you're left with O-, which becomes the O in the middle.
 
OP
H

hunterpostbacst

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Apr 7, 2009
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Sorry guys. I was totally messed up.

My questions is not orgo, but general chem #201.

The questions is

How many ml of 0.12M KOH must be added to 60mL of 0.2M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3)

I thought pH = -log[H+] = 3.3, so [H+]=10^-3.3
HF <-> H+ + F-
0.2*0.06mol 0.2*0.06mol 0.2*0.06mol

So, OH- from KOH neutralizes H+ of HF.
[KOH]=[OH]= 0.12*x mol

So, [H+]-[OH-]=(0.2*0.06)-(0.12*x)=10^-3.3
so, x = 0.09582344
So, it's 95ml. This is my reasoning, but the solution is completely different from what I did above.
Anyone? Thanks:idea:


(because HF is not completely dissociated, a weak acid)
 

GOLF

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Jul 6, 2009
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yeh i did this problem the other day and i think their decimals are off by a spot
 

Kneecoal

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did you read the explanation?

for one thing, you should be able to tell that your answer is waaaaaay off because there's no way it should take that much of a strong base to still have that acidic a pH.

also, HF doesn't dissociate completely, so where you did HF --> H+ + F- isn't exactly true. while those ions are floating around, you still have a bit of regular ol' HF floating around with it too.

i'm sorry if this didn't help much, but the explanation pretty much er, explains it all!