Question

What is the [OH-] of a solution that has a pOH of 13?

What is the [OH-] of a solution that has a pOH of 13?

Answer 1

On the low side...

Explanation:

By definition, `pOH=-log_10[HO^-]`

And so here we take antilogs...and `[HO^-]=10^(-13)*mol*L^-1`..it might be more useful to quote `pH`...and here `pH=14-pOH=14-13=1`...and thus `[H_3O^+]=0.10*mol*L^-1`.

Answer 2

`10^-13 \ "M"`

Explanation:

The `"pOH"` of a solution is found by:

`"pOH"=-log[OH^-]`

• `[OH^-]` is the hydroxide ion concentration in terms of molarity

And so, we get:

`13=-log[OH^-]`

`[OH^-]=10^-13 \ "M"`