Question #f3621

1 Answer
anor277
May 21, 2016

#pH# and #pOH# depend on the concentrations of #[H_3O^+]# or #[HO^-]# respectively.

Explanation:

#pH# #=# #-log_10[H_3O^+]#

#pOH# #=# #-log_10[HO^-]#

For strong acids, for which protonation of the water molecule is almost quantitative, #pH# #=# #-log_10[H_3O^+]#. Likewise for strong bases, #pOH# #=# #-log_10[HO^-]#.

For water at #298# #K#, the following equilibrium applies:

#K_w# #=# #10^-14# #=# #[H_3O^+][HO^-]#. And, after taking negative #log_10# of both sides:

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

And thus,

#14=pH+pOH#, under standard conditions.

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