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Topic: Titration, weak unknown acid with strong base! Help please..  (Read 18999 times)

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Offline SaraW

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Titration, weak unknown acid with strong base! Help please..
« on: September 30, 2009, 07:56:44 AM »
I have a problem that needs solving...if anyone else has the time please take a look

Titration of a weak acid with a strong base. Im titrating an 50 ml of an unknown weak acid with an unknown concentration with NaOH (concentration NaOH 0,2 M) . I’ve been given a titration curve:



Im supposed to get the equivalence point out of this curve…Im guessing its between 8.5 – 9 (right?).
I need to calculate the unknown weak acids initial concentration and its Ka(equilibrium constant).

Please help

Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #1 on: September 30, 2009, 08:04:34 AM »
So.. Im thinking
C(acid)= (V NaOH)*(C NaOH)/(V(acid))

V NaOH: 75ml (from titration curve)
C NaOH: 0,2 M
V (acid): 50ml

0.2  * 0,075 =  0,3 M
0,05

Initial concentration of the weak acid is 0.3M
Does that sound right?
« Last Edit: September 30, 2009, 08:15:52 AM by SaraW »

Offline Borek

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #2 on: September 30, 2009, 08:30:33 AM »
75 mL is wong - how many mL were added to reach end point pH you have mentioned?
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Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #3 on: September 30, 2009, 08:36:30 AM »
did i mention an end point pH? is that the equivalence point in this case?

Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #4 on: September 30, 2009, 09:02:21 AM »
Ok let me try again...
C(acid)= (V NaOH)*(C NaOH)/(V(acid))

V NaOH: 50ml
C NaOH: 0,2 M
V (acid): 50ml

0.2  * 0,05 =  0,2 M
0,05

Initial concentration of the weak acid is then 0.2M

50 mL NaOH when pH 8,5-9

Offline Borek

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #5 on: September 30, 2009, 09:18:16 AM »
Much better now.

For general information: http://www.titrations.info
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Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #6 on: September 30, 2009, 09:49:57 AM »
when it comes to the equilibrium constant
pKa(acid)  =  pH  =  2,9

10^-pKa(Acid)  =  10^-2,9 = 1.26*10^3

right?wrong?

Offline Borek

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #7 on: September 30, 2009, 10:18:10 AM »
No idea what you are doing, why, nor what for.
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Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #8 on: September 30, 2009, 10:21:35 AM »
trying to get the Ka(equilibrium constant)

Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #9 on: September 30, 2009, 10:54:57 AM »
ok so its wrong....any suggestions on how to make it better then?

Offline Borek

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #10 on: September 30, 2009, 12:06:02 PM »
There is a characteristic place on the titration curve where pH=pKa.
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Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #11 on: September 30, 2009, 12:11:06 PM »
Ok i see 8.5-9.
i thought it was when NaOH startedbeing added(2.9).

does this then make sense..or is the way im calculating it wrong also..
pKa(acid)  =  pH  =  8.5

10^-pKa(Acid)  =  10^-8.5 = 3.16*10^9

Ka= 3.16*10^9

Offline Borek

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #12 on: September 30, 2009, 01:56:59 PM »
No, its is not pH at the end point (nor equivalence point).

Do you know Henderson-Hasselbalch equation?
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Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #13 on: September 30, 2009, 02:32:22 PM »
      pH = pKa + log10[conjugate base]
                             [conjugate acid]                       

how can i get pKa if I dont know which acid it is?

i know im dense...but i need a bit more help

Offline SaraW

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Re: Titration, weak unknown acid with strong base! Help please..
« Reply #14 on: September 30, 2009, 03:17:27 PM »
pKa = −log10Ka
So pKa is halfway to the equivalence point?
4.25= pKa?

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