We know, do we?, that pH+pOH=14pH+pOH=14 for water under standard conditions (see later).
And thus pOH=14-12.52=1.48pOH=14−12.52=1.48...
And so we take antilogarithms,,,[HO^-]=10^(-1.48)*mol*L^-1[HO−]=10−1.48⋅mol⋅L−1
=0.0331*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)Kw=[H3O+][HO−]=10−14...
And we can take log_10log10 of both sides to give....
log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-]log10Kw=log1010−14=log10[H3O+]+log10[HO−].
And thus.... -14=log_(10)[H_3O^+]+log_(10)[HO^-]−14=log10[H3O+]+log10[HO−]
Or.....
14=-log_(10)[H_3O^+]-log_(10)[HO^-]14=−log10[H3O+]−log10[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
