I'll show you how to do a and I expect you can do the rest but I can help you through them if necessary. Call benzoic acid HBz.
..........HBz ==> H^+ + Bz^-
I.........1.0M....0......0
C..........-x.....x......x
E........1.0-x....x......x
Then 6.3E-5 = (x)(x)/(1.0-x)
You can do the math but I get x = approx 8E-3M = (H^+) and convert that to pH.
Then %dissoc = [(H^+)/(HBz)]*100 = approx [(8E-3)/1]*100 = about 0.8%
You can do b and c. (H^+) becomes less as (HBz) becomes smaller ( as you would expect) BUT % dissoc increases. Why? Because HBz + H2O ==> H3O^+ + Bz^-
As H2O is added (making the solution more dilute) Le Chatelier's Principle says the reaction will shift to the right and of course that means it is dissociated more in more dilute solutions.
i need help solving this
Benzoic acid is a weak, monoprotic acid
(Ka = 6.3 × 10−5). Calculate the pH and the percent dissociation of each of the following solutions of benzoic acid. Then use Le Châtelier’s principle to explain the trend in percent dissociation of the acid as the solution becomes more dilute.
(a)1.0mol/L (b)0.10mol/L (c)0.01mol/L
1 answer
1 answer