In a NEUTRAL solution, how does #[HO^-]# compare to #[H_3O^+]#?

1 Answer
Apr 10, 2017

Answer:

How do you think it compares...........?

Explanation:

In aqueous solution under standard conditions, the following equilibrium operates:

#2H_2OrightleftharpoonsH_3O^+ + HO^-#

At #298*K#, the ion-product, #[H_3O^+][HO^-]=10^-14#,

we know that #[H_3O^+]=[HO^-]# by specification (we have a neutral solution), and thus,

#[H_3O^+]=[HO^-]=sqrt(10^-14)*mol*L^-1=10^-7*mol*L^-1#.