Question

What is the [OH-] of a solution whose pH is 2.45?

Answer 1

`[HO^-]=2.82xx10^-12*mol*L^-1`....

Explanation:

See this old answer,

And thus the defining relationship that established `pH` and `pOH` in water under standard conditions.....

`14=pOH+pH`

And so `pOH=14-pH=14-2.45=11.55`

And thus `[HO^-]=10^(-11.55)*mol*L^-1=??*mol*L^-1`