How do you convert pH to pOH?

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Answer 1 · 2016-05-16T14:33:29

#"pOH" = "14.00 - pH"#

#H^+ + OH^-##rightleftharpoons##H_2O#

the ionic product for water,#K_w# = #10^-14#

#=> K_w = ([OH^-][H^+])#

applying log on both sides,

-14= # log[OH^-] +log[H+]#

14= # -log[OH^-] -log[H+]#

therefore 14= pH +pOH

(since #pH= -log[H^+])#

#[H^+]# = concentration of #H^+#