Attend...we know that water undergoes an acid-base autoprotolysis reaction as shown...
#2H_2O(l) rightleftharpoons H_3O^+ +HO^-#
And under standard conditions of temperature and pressure...
#underbrace([H_3O^+][HO^-]=K_w=10^-14)_"autoprotolysis of water under standard conditions"#
...and so we got #[H_3O^+][HO^-]=K_w=10^-14#...and given this equation, we can manipulate it. One thing we can do is to take #log_10# of BOTH SIDES....
And so...#log_10[H_3O^+]+log_10[HO^-]=log_10K_w=log_(10)10^-14#
But by definition...#log_(10)10^-14=-14#...and if you are not familiar with logarithms, it is time to acquaint yourself with them...
#-14=log_10[H_3O^+]+log_10[HO^-]=log_10K_w#
And on rearrangement...
#+14=-log_10[H_3O^+]-log_10[HO^-]=-log_10K_w#
But by definition...
#+14=underbrace(-log_10[H_3O^+])_"pH"underbrace(-log_10[HO^-])_(+"pOH")=-log_10K_w#
And thus (finally) our working relationship...
#pH+pOH=14#
And so finally to your question...we gots #pH=12.5#..and so necessarily...#pOH=1.5#..and taking antilogarithms...
#[HO^-]=10^(-1.5)*mol*L^-1=0.0316*mol*L^-1#
What is #[H_3O^+]# under these conditions....?
