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Topic: pH and Ka question  (Read 15624 times)

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Offline roundeggy

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pH and Ka question
« on: May 25, 2009, 05:22:13 PM »
A particular sample of vinegar has a pH of 2.90 assuming that acetic acid is the only acid that vinegar contains (Ka equals 1.8X 10-4) calculate the concentration of acetic acid in the vinegar...
 so i did 10^(-2.90) which gave me (1.26X10^-3) then i squared that and divided it by (1.26X10^-3-X) = Ka (1.8X10^-4) ... i figured i should multiple the Ka by the lower half of the fraction but for some reason i keep getting the value of X wrong. Please write back what you did clearly thanks!

[1.26X10^-3]^2/[1.26X10^-3-X]= 1.8X10^-4 wat is X

Offline Train

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Re: pH and Ka question
« Reply #1 on: May 25, 2009, 06:17:51 PM »
Why did you write (1.26*10-3 - x)?

Offline roundeggy

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Re: pH and Ka question
« Reply #2 on: May 25, 2009, 06:22:47 PM »
i wrote that for the bottom of the fraction to figure out what the concentration of {H+}... what would u have done?

Offline Train

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Re: pH and Ka question
« Reply #3 on: May 25, 2009, 06:45:54 PM »
You already know [H+] - from the pH.  As you wrote: [H+] = 10^(-2.90) = 1.26X10^-3

Why are you subtracting x from that number in the denominator?

Offline roundeggy

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Re: pH and Ka question
« Reply #4 on: May 25, 2009, 07:32:39 PM »
So you would just multiple Ka time the concentration then square root that answer?

Offline Train

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Re: pH and Ka question
« Reply #5 on: May 26, 2009, 01:12:20 AM »
No. 

You should write the expression for Ka.  After you have an expression for Ka, start trying to relate any unknowns until you only have 1 unknown left. 

How to find the relation you're looking for: Since acetic acid is a weak acid, write your equations as if it is initially undissociated and then partially dissociates into the conjugate base and H+.  You don't know how much acetic acid was initially present but if you assume all of the H+ is from acetic acid then you do know how much of it dissociated.  Your unknown, x, is the amount of acetic acid before it dissociated.

Offline AWK

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Re: pH and Ka question
« Reply #6 on: May 26, 2009, 01:31:23 AM »
A particular sample of vinegar has a pH of 2.90 assuming that acetic acid is the only acid that vinegar contains (Ka equals 1.8X 10-4) calculate the concentration of acetic acid in the vinegar...
 so i did 10^(-2.90) which gave me (1.26X10^-3) then i squared that and divided it by (1.26X10^-3-X) = Ka (1.8X10^-4) ... i figured i should multiple the Ka by the lower half of the fraction but for some reason i keep getting the value of X wrong. Please write back what you did clearly thanks!

[1.26X10^-3]^2/[1.26X10^-3-X]= 1.8X10^-4 wat is X

Ka equals 1.8X 10-5
AWK

Offline Borek

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Re: pH and Ka question
« Reply #7 on: May 26, 2009, 04:03:09 AM »
i wrote that for the bottom of the fraction to figure out what the concentration of {H+}... what would u have done?

You are trying to use ICE table not knowing what I, C and E stand for.

http://en.wikipedia.org/wiki/ICE_table
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline jpg28

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Re: pH and Ka question
« Reply #8 on: June 02, 2009, 10:47:14 AM »
I agree with AWK, the Ka for acetic acid is 1.8 x 10-5. Maybe that's the reason why the wrong value for x keeps on appearing. Try replacing the Ka with this one, and maybe the correct value for x might show up. This is just to reinforce AWK's post, by the way. Have a nice day and hope you'll be able to solve that. :)
« Last Edit: June 02, 2009, 12:08:57 PM by Borek »

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