Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: SaraW on September 30, 2009, 07:56:44 AM
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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:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.hotlinkfiles.com%2Ffiles%2F2111856_qa4gu%2FTitreringskurva.jpg&hash=cb7b6df9b72323248f01a9676972e0931a468db2)
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
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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?
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75 mL is wong - how many mL were added to reach end point pH you have mentioned?
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did i mention an end point pH? is that the equivalence point in this case?
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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
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Much better now.
For general information: http://www.titrations.info
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when it comes to the equilibrium constant
pKa(acid) = pH = 2,9
10^-pKa(Acid) = 10^-2,9 = 1.26*10^3
right?wrong?
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No idea what you are doing, why, nor what for.
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trying to get the Ka(equilibrium constant)
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ok so its wrong....any suggestions on how to make it better then?
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There is a characteristic place on the titration curve where pH=pKa.
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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
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No, its is not pH at the end point (nor equivalence point).
Do you know Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch)?
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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
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pKa = −log10Ka
So pKa is halfway to the equivalence point?
4.25= pKa?
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Yes :)
pH = pKa at 50% titration, when concentrations of HA and A- are identical.