How is pOH determined?

1 Answer
anor277
Jun 4, 2016

#pOH# is detemined equivalently to #pH#, but #pH+pOH=14# under standard conditions.

Explanation:

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

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

Now #K_w# #=# #[H_3O^+][HO^-]# #=# #10^(-14)" at 298K"#.

Thus #-log_10K_w=-log_10[H_3O^+] -log_10[HO^-]#

#=-log_10(10^-14)# #=# #+14#

And thus #pOH+pH=14# under standard conditions.

You should review the logarithmic function if you have any problem with this treatment. You should review it anyway, because I might have made a mistake.

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