Destroyer orgo 201

hunterpostbacst

10+ Year Member
I'm totally lost anyone who can reason this question?

loveoforganic

-Account Deactivated-
10+ Year Member
what's the question

Kneecoal

10+ Year Member
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

10+ Year Member
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 (because HF is not completely dissociated, a weak acid)

GOLF

7+ Year Member
yeh i did this problem the other day and i think their decimals are off by a spot

Kneecoal

10+ Year Member
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!