Calculate the [OH−] in a solution with a pH of 12.52. ?

1 Answer
Jun 7, 2018

[HO^-]=0.0331*mol*L^-1*mol*L^-1

Explanation:

We know, do we?, that pH+pOH=14 for water under standard conditions (see later).

And thus pOH=14-12.52=1.48...

And so we take antilogarithms,,,[HO^-]=10^(-1.48)*mol*L^-1

=0.0331*mol*L^-1

Just to note that in aqueous solution under standard conditions, the ion product...

K_w=[H_3O^+][HO^-]=10^(-14)...

And we can take log_10 of both sides to give....

log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-].

And thus.... -14=log_(10)[H_3O^+]+log_(10)[HO^-]

Or.....

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

14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)

14=pH+pOH
By definition, -log_10[H^+]=pH, -log_10[HO^-]=pOH