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### Topic: Ka calculation  (Read 392 times)

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#### CT

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##### Ka calculation
« on: May 12, 2019, 12:12:56 PM »
One of my past exam papers asks the following:

A fraction of dissociated molecules of a weak monoprotic acid in its 0.09 mol/dm3 aqueous solution is only 10%. What is the Ka?

I used the formula Ka = [H+][A-]/[HA], plotted in and got the answer:
(0.09*0.1)(0.09*0.1)/(0.09) = 9x10-4

But the answer section says Ka = 10-3

Is my calculation wrong?

#### AWK

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##### Re: Ka calculation
« Reply #1 on: May 12, 2019, 12:21:04 PM »

I used the formula Ka = [H+][A-]/[HA], plotted in and got the answer:
(0.09*0.1)(0.09*0.1)/(0.09) = 9x10-4

But the answer section says Ka = 10-3

Is my calculation wrong?
Error. Your approximation work below 5 % dissociation.
AWK

#### CT

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##### Re: Ka calculation
« Reply #2 on: May 12, 2019, 01:01:23 PM »
I am not sure I understand. Can you elaborate?

#### Borek

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##### Re: Ka calculation
« Reply #3 on: May 12, 2019, 01:17:07 PM »
If 10% of the acid is dissociated, is the concentration of HA still 0.09 M?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### CT

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##### Re: Ka calculation
« Reply #4 on: May 12, 2019, 01:47:43 PM »
Ah, I got it now. [HA]=0.09M - [H+] = 0.081M
Thanks.