Question

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

Answer 1

`[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.