At what initial concentration would a solution of acetic acid (ka=1.8x10^-5) be 2% ionized?
Thanks
Responses
Chemistry - DrBob222, Thursday, February 12, 2009 at 10:00pm
Let's call acetic acid HAc. Then the ionization is
HAc ==> H^+ + Ac^-
We will call the initial concn of HAc x. If it is 2% ionized, then (H^+) = 0.02*x
(Ac^-) = 0.02*x
and (HAc) = x-(0.02*x) = 0.98*x
Now write the Ka expression and plug in those concns, then solve for x. Post your work if you get stuck. The answer is approximately 0.05 M
Chemistry - Needs help, please!, Thursday, February 12, 2009 at 11:05pm
How did you get 0.98? Why were you allowed to subtract 0.02 to get that? I'm still confused as to how to plug this in and solve, just because of the 0.98. ThanksD
The denominator is x-0.02x. That is 1.00x - 0.02x - 0.98x
Think about it this way.
100%-2%=98% and change % to a decimal by dividing by 100 to obtain 0.98. A third way to look at it is that the unknown concn of the acid is x. 2% (in decimal form it is 0.02) is what ionizes to give (H^+) = 0.02 times the acid concn or 0.02x and x-0.02x is what is left of the acid after it ionizes and that is the 0.98x. Or 0.98x is the amount of the unionized acid. If you still have trouble with it, make an ICE chart.