How does the #pH# of an aqueous solution as #[H_3O^+]# is INCREASED?

1 Answer
May 19, 2017

Answer:

#".....decreases"##".........decreases............increases......."#

Explanation:

We assess the autoprotolysis reaction of water:

#H_2O(l) rightleftharpoonsH^(+) +HO^-#

Equivalently we could write this is as...........

#2H_2O(l) rightleftharpoonsH_3O^(+) +HO^-#

And #K_w=[H_3O^+][HO^-]=10^-14# at #298*K#

We can take #log_10# of BOTH sides to give........

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

#log_10[H_3O^+]+log_10[HO^-]=-14#, or, multiplying each side by #-1#

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

But by definition, #-log_10[H_3O^+]=pH# and #-log_10[HO^-]=pOH#. And so after all that..........

#pH+pOH=14#, which is our defining relationship.

At LOW #pH#, #[H_3O^+]# is HIGH and #[HO^-]# is low; and at high #pH#, #[H_3O^+]# is LOW and #[HO^-]# is HIGH.

When #pH=pOH#, #pH=7#. Why?

Remember the definition of the #"log function"#. If #log_(a)b=c#, then #a^c=b#, and thus #log_(10)1=0#, #log_(10)10=1#, #log_(10)10^-1=-1#,#log_(10)100=2#.