What is #K_a# for #HOCl#, given that initially #[HOCl]=0.08*mol*dm^3#, and #pH=2.85#?

1 Answer
Apr 23, 2017

Answer:

#K_a=([H_3O^+][""^(-)OCl])/([HOCl])=2.54xx10^-5#.........

Explanation:

The #"acid dissociation constant,"# #K_a#, measures the particular acid dissociation............

#HOCl(aq) + H_2O rightleftharpoonsH_3O^(+) + ""^(-)OCl#

Where #K_a=([H_3O^+][""^(-)OCl])/([HOCl])#

We are given that the initial concentration was #0.08*mol*dm^-3#, and that #pH=2.85#. But we know that #pH=-log_(10)[H_3O^+]#.

And thus #[H_3O^+]=10^(-2.85)*mol*L^-1#.

But, by stoichiometry #[""^(-)OCl]=10^(-2.85)*mol*L^-1#.

And, thus......................

#[HOCl]=(0.08-10^(-2.85))*mol*L^-1=...............#

#=0.0786*mol*L^-1#.

And thus we can calculate #K_a# sans approximation, i.e.

#K_a=((10^(-2.85))^2)/(0.0786)=2.54xx10^-5#.