[imcuteali2]

(1) Calculate the pH of a 0.10 M solution of HOCl, Ka = 3.5 x 10^-8.
A. 4.23
B. 8.46
C. 3.73
D. 1
E. 3.23

Answer A, 4.23 because Ka=3.5 x 10^-8= [H][OCL-]/ [HOCL]
Initial HOCL=.1M
use ice table procudure to obtain:
Ka=3.5x10^-8= (x)^2.(.10-x)
Since K value is so small we can ignore x term in the denominator.
x= 5.9x10^-5

check if error is under 5%
(5.9x10^-5/.10 )x 100%= .059%
therefore pH= -log(concentration)= 4.23

(2) The following question refers to a 0.70 M solution of hypochlorous acid, HClO.
If the molarity was decreased to 0.3 M, which of the following statements would be true?

A. The % dissociation would not change.
B. The % dissociation would increase.
C. The % dissociation would decrease.
D. The equilibrium constant would stay the same.
E. Two of these.

I think it is E because even though the concentration decreased, the overall percentage of dissociation would stay the same. Also, the question does not say that the temperature is changing therefore the answer is also D...therefore 2 of these are correct....Am I correct on this?

Thank you for you *delete me*

[Borek]

pH= -log(concentration)= 4.23

OK

I think it is E because even though the concentration decreased, the overall percentage of dissociation would stay the same. Also, the question does not say that the temperature is changing therefore the answer is also D...therefore 2 of these are correct....Am I correct on this?

It is E, but you are wrong. Dissociation constant remains the same, but dissociation % will change, thus one of B/C will be correct. More precisely - diluting solution you are increasing dissociation % - infinitesimally diluted solution is ionized 100%.