Calculate [OH−] in the following aqueous solution at 25 ∘C : [H3O+]= 2.1×10^-8 M

Answer 1

The relationship between [H3O⁺] and [OH^-] in an water is [H3O⁺] x [OH^-] = 10^-14.

To find the [OH^-] when [H3O⁺] is known is to solve the above equation for [OH^-].

[OH^-] = 10^-14 / 2.1 x 10^-8 = 4.8 x 10^-7