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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: CT on May 12, 2019, 12:12:56 PM

Title: Ka calculation
Post by: CT 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?

Title: Re: Ka calculation
Post by: AWK on May 12, 2019, 12:21:04 PM

Quote from: CT on May 12, 2019, 12:12:56 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.

Title: Re: Ka calculation
Post by: CT on May 12, 2019, 01:01:23 PM

I am not sure I understand. Can you elaborate?

Title: Re: Ka calculation
Post by: Borek on May 12, 2019, 01:17:07 PM

If 10% of the acid is dissociated, is the concentration of HA still 0.09 M?

Title: Re: Ka calculation
Post by: CT on May 12, 2019, 01:47:43 PM

Ah, I got it now. [HA]=0.09M - [H+] = 0.081M
Thanks.