#pH# is short for #"pouvoir hydrogen"# i.e. #"the power of hydrogen"# or #"hydrogen potential"#. This is applied to aqueous solutions in the following way.
In water, the following reaction takes place, and this is referred to as #"autoprotolysis"#:
#2H_2O rightleftharpoonsH_3O^+ + HO^-#
As with any equilibrium, we can write the following equilibrium expression:
#([H_3O^+][HO^-])/([H_2O]^2)# #=# #K#.
Because #[H_2O]# is large and effectively constant, this simplifies to:
#[H_3O^+][HO^-]# #=# #K_w#.
This equilibrium has been measured meticulously, and at #298# #K#, we can give a value for #K_w#:
#K_w# #=# #10^(-14)# #=# #[H_3O^+][HO^-]#
We can take #log_10# of both sides to give:
#log_10(10^(-14))# #=# #log_10[H_3O^+]+ log_10[HO^-]#
But #log_10(10^(-14))# #=# #-14# by definition of the log function.
And so, #-14=log_10[H_3O^+]+ log_10[HO^-]# or
#14=-log_10[H_3O^+]- log_10[HO^-]#
Now if we define #-log_10[H_3O^+]=pH# and #- log_10[HO^-]=pOH#, then:
#14=pH+pOH#.
And this given a value of #pH#, I can find #[H_3O^+]# or #[HO^-]# by taking antilogarithms.
What is the #pH# of a solution that is #0.5# #mol*L^-1# with respect to #H_3O^+#? What is the corresponding #pOH#?
