How do you get pH and pOH given pka or pkb?

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Answer 1 · 2016-10-11T16:07:34

#pH=-log_10[H_3O^+]#, and #pOH=-log_10[HO^-]#,

In water, under standard conditions, we know that the following equilibrium occurs:

#2H_2O rightleftharpoons H_3O^+ + HO^-#

As for any equilibrium, we can measure the equilibrium constant, #K_w#, of this reaction:

#K_w=[H_3O^+][HO^-]=10^-14#.

Given that #K_w=10^-14#, and taking #-log_10# of both sides, we get the defining relationship:

#-log_10K_w# #=# #-log_10[H_3O^+]-log_10[HO^-]# #=# #-log_(10)10^-14#.

Thus #14=pH+pOH#, and this is something that you have to commit to memory.

And thus when quoted #pK_a# for weak acids etc. you have to solve the equilibrium expresssion:

#HA+ H_2O rightleftharpoons H_3O^+ +A^-#.

There should be many model answers on these boards. Here is a start, and here is another attempt.