How do you find the value of #[OH^-]# for a solution with a pH of 8.00?

1 Answer
Jun 14, 2017

Answer:

#[HO^-]=10^-6*mol*L^-1#. How does we know.........?

Explanation:

Water undergoes so-called #"autoprotolysis"#......

#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#

At #298*K# we can measure the #"autoprotolysis"# very precisely....

#[HO^-][H_3O^+]=10^-14=K_w#

This is a mathematical equation which we can divide, multiply, or otherwise manipulate provided that we does it to both sides.....One think that we can do is take #log_10# of both sides....

#log_10[HO^-]+log_10[H_3O^+]=log_10(10^-14)#

And thus #log_10[HO^-]+log_10[H_3O^+]=-14#

(Why? Because mathematically, #log_(a)a^y=y# by definition. It is the power to which we raise the #"base a"# to get #a^y#. With me?

And so #14=-log_10[HO^-]-log_10[H_3O^+]#

BUT, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH# BY DEFINITION........

And so (at last!) #14=pH+pOH#

And so (after all that!) if #pH=8#, then CLEARLY, #pOH=6#. And thus #[HO^-]=10^-6*mol*L^-1#.

Claro?

Note the logarithmic function is something that you learn about in mathematics; it is very powerful, and useful. Its application to chemistry is (I think) rather straightforward. If there is still an issue, voice it and someone will help you.