Given.......
#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#
#K_w'=([H_3O^+][HO^-])/([H_2O(l)])#
Because #[H_2O]# is so large, we remove it from the expression to give,,,,,,
#K_w=[H_3O^+][HO^-]=10^-14# under standard conditions of temperature #(298*K)# and pressure #(1*atm)#...
Thus #[HO^-]=10^-14/(5.36xx10^-5)=1.87xx10^-10*mol*L^-1#.
Alternatively, we could take #log_10# of both sides of #K_w=[H_3O^+][HO^-]=10^-14# to give.........
#log_10K_w=log_10[H_3O^+]+log_10[HO^-]#
On rearrangement
#underbrace(-log_10[H_3O^+])_color(red)(pH)underbrace(-log_10[HO^-])_color(blue)(pOH)=underbrace(-log_10K_w)_(pK_w)#
And thus #color(red)(pH)+color(blue)(pOH)=pK_w=14#.
And here, #pH=-log_10(5.36xx10^-5)=4.27#, and clearly, #pOH=9.73#.
#[HO^-]=10^(-9.73)=1.87xx10^-10*mol*L^-1#, as required.......
I don't know about you, but I think the #log# method is better.....
