What is the #[OH^-]# for a solution at 25°C that has a #[H_3O^+]# of #5.36 times 10^-5#?

1 Answer
Jun 22, 2017

Answer:

#[HO^-]=1.87xx10^-10*mol*L^-1#

Explanation:

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.....