If #[H_3O^+]=5.0xx10^-3*mol*L^-1#, what is #pH# and what is #pOH# of this solution?

1 Answer
Apr 17, 2017

Answer:

#pH=2.30#............

Explanation:

.......and #pOH=11.70#

These acid base reactions are governed by the equilibrium:

#2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#.

And at #298*K#, #K_w=[H_3O^+][HO^-]#

#log_10K_w=log_10[H_3O^+]+log_10[HO^-]#

And since we know that #K_w=10^-14#, then..........

#-14=log_10[H_3O^+]+log_10[HO^-]#

OR #14=-log_10[H_3O^+]-log_10[HO^-]#

But, by definition, #-log_10[H_3O^+]=pH#, and #-log_10[H_3O^+]=pOH#

And thus, BY DEFINITON, #pH+pOH=14# under standard conditions. If you can remember this relationship, then these sorts of problems become almost trivial.

And thus, for your problem, #pH=-log_10(0.0050)=-(-2.30)=2.30#.

#pOH=14-2.30=11.70#.

And can you tell me what is #[HO^-]# here? It should be small......