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

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Answer 1 · 2018-06-07T16:40:10

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

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#