What is the hydrogen ion concentration of a 0.22 M hypochlorous acid solution with Ka = 3.5E-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq) = H+(aq) + OCl-(aq).
I keep getting 3.60 as the concentration, but that is the wrong answer.
Using ICE:
Eq: HOCl -> H+ + OCl-
I: 0.22
C: -x +x +x
E: (0.22-x) x x
kb = x^2/(0.22-x)
kb = 1.01E-14/3.5E-8 = 2.89E-7
x^2 = 2.89E-7 - (0.22-x)
x = square root of 2.89E-7 *(0.22-x)
x'= 2.5196E-4
x'' = 2.5182E-4
x'' = 2.5182E-4; x= 2.5182E-4 = [H+]
-log(2.5182E-4) = 3.5989, rounded to 3.60
I am not sure where I went wrong
Please help
Lynn
1 answer