Best answer:
I have been taught that when the dissociation is less than 5% ( the 5% rule) it is OK to take [HA] = original concentration of the acid. X will always be smaller than [HA] , because dissociation is weak. In this case X must be less than 5% of 0.1M or 0.005M.
Using your method , and dividing by 0.1 , you must have...
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Best answer: I have been taught that when the dissociation is less than 5% ( the 5% rule) it is OK to take [HA] = original concentration of the acid. X will always be smaller than [HA] , because dissociation is weak. In this case X must be less than 5% of 0.1M or 0.005M.
Using your method , and dividing by 0.1 , you must have found
1*10^-2 = [x] ² / 0.1
[X]² = (1*10^-2) *0.1
[X]² = 1*10^-3
[X] = 0.032M which is 32% dissociation - which is far too high to ignore X in the denominator . You have to divide by (0.1-X) .
But if Ka = 1*10^-4
1*10^-4 = X² / 0.1
x² = 1*10^-5
x = 3.2*10^-3 M = 0.0032M
and % dissociation = 0.0032/0.1*100 = 3.2% and you can definitely use 0.1 in the denominator
As a rule of thumb , if Ka is smaller than 1*10^-4 (as shown above) then you can use this rule. In your case Ka = 1.0*10^-2 , (which is > 1*10^-4) so you clearly cannot use the rule , and you have to divide by (0.1 - X) and you end up with a quadratic equation. In my opinion , marking your answer incorrect was properly done. Sorry.
4 answers
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2 weeks ago