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Topic: HCOOH  (Read 2164 times)

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Offline Rutherford

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HCOOH
« on: May 15, 2012, 11:34:48 AM »
When a solution of HCOOH c(H+)=4.15*10-3 is dilluted in water to 10 times bigger volume c(H+)=1.25*10-3mol/dm3. Calucalte the Ka.
HCOOH  ::equil:: H++HCOO-
  c-x            x    x
In first case x is 4.15*10-3, in second case x is 1.25*10-3.
Ka=x2/(c-x)
c after dilluting is equal to 0.1c before dilluting.
1. Ka=(4.15*10-3)2/(c-4.15*10-3)
2. Ka=(1.25*10-3)2/(c-1.25*10-3)
I get a negative result. Can't find a mistake.
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Offline Borek

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Re: HCOOH
« Reply #1 on: May 15, 2012, 11:56:26 AM »
Quote from: Raderford on May 15, 2012, 11:34:48 AM
2. Ka=(1.25*10-3)2/(c-1.25*10-3)

c or 0.1c?
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Offline Rutherford

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Re: HCOOH
« Reply #2 on: May 15, 2012, 11:57:58 AM »
Oh, now it looks so obvious when someone else tells it. Thanks, I got 1.91*10-4.
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