Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Noname16 on December 06, 2008, 08:03:19 PM
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1.6gram sample with a weak acid HX (82g) is dissolved in 60ml of H2O, titrated with .250M NaOH. When 1/2 of the HX is neutralized the pH is 5.0; and at the equivalence point the pH is 9.0. Find the % of HX in the original sample and the Ka of HX.
Even with help i was only able to get the Ka=1.0E-5 (which is correct) but my % was 51.3% (which is wrong).
My steps:
HX --> H+ + X-
Ka=[H+] [X-] / [HX]
@ 1/2 neutralization
Ka= [H+][1/2X-] / [1/2HX]
Ka = [H+] and since pH=5 @ 1/2 neutralization
pH = -log [H+]
5 = -log [H+]
[H+] = 1E-5
Ka = 1E-5
then after this point my answers start going wrong. I found the Ka but don't know how to get the %HX.
Can anyone please help me on how to approach this. You don't even need to do the answer, just explain to me how to think going about the problem. Thanks
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How do you calculate pH at the equivalence point?
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Nevermind, this problem is impossible.
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I think it is possible. Slightly unorthodox, but possible. Answer my question please.
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To get the pH=9 @ the equivalence point
[OH-] = square root of [(0.1)(Kb)]
[OH-] = 1e-5
pOH = 5
pH = 14-5
pH=9
The (.1) is the concentration of [X-]
Getting the pH is pretty basic stuff, so if you had to ask how to calculate it, i don't think i will get much help on this answer. But thanks for the effort.
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The (.1) is the concentration of [X-]
Where did you got it from?
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Given : @ equivalent point pH=9
pKw=pH+ pOH
14 = 9 + pOH
pOH = 5
[OH-] = 1e-5
X- + H20 = HX + OH-
Kb = [HX] [OH-] / [X-]
1e-9 = (1e-5)^2 / [X-]
[X-] = 0.1
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Oops. I am terribly sorry. Somehow I have managed to misread the question :'( Probably shouldn't do three things at the same time :(
I feel like an idiot if it helps.
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THanks for the try. And thanks for Chembuddy. It's been added to my favorites. I see alot of use for it in my future.