In water, the following equilibrium operates under standard conditions of temperature and pressure.......
#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#
And as for any equilibrium, we can define an equilibrium constant. And in this instance, we write #K_w=[H_3O^+][HO^-]=10^(-14)#.
Given #[H_3O^+]=5.4xx10^-9*mol*L^-1#, then..............
#[HO^-]=(10^-14)/(5.4xx10^-9)=1.85xx10^-5*mol*L^-1#.
Since #[HO^-]">>"[H_3O^+]# the solution is BASIC.
Note that we can use #"the log function"# to make this calculation easier.
If we take #log_10# of both sides of #K_w=[H_3O^+][HO^-]=10^(-14)#, we get:
#log_10K_w=log_10[H_3O^+]+log_10[HO^-]#
OR..........
#-log_10K_w=-log_10[H_3O^+]-log_10[HO^-]#
But #-log_10K_w=pK_w#, and #-log_10[H_3O^+]=pH# and #-log_10[HO^-]=pOH# BY DEFINITION.
And also by definition of the #"log function"#, #log_(10)10^-14=-14#.
And thus #14=pH+pOH#. And so (finally!) we can use this relationship in the given problem:
#[H_3O^+]=5.4xx10^-9#, and thus #pH=8.27#, #pOH=5.73#. At neutrality, #pOH=pH=7#. And here the solution is clearly basic.
