Ryan O. answered • 06/23/22
B.S. Degree in Chemistry with 3+ Years of Tutoring Experience
Hi Denise! So remember that whenever you have an aqueous solution, you can think about the autoionization of water:
2H2O(l) <--> H3O+(aq) + OH-(aq) (note: <--> is an equilibrium arrow)
This is dictated by an equilibrium constant, Kw, which at 25 C is equal to 1.0 x 10^-14. We can thus write the equilibrium constant expression:
Kw = [H3O+][OH-], remember H2O does not get included in the equilibrium constant expression because it is a liquid.
Therefore, we know that [H3O+][OH-] has to be equal to 1.0 x 10^-14. Since [OH-] = [NaOH] for a strong base, we can find the [H3O+]:
[H3O+] = 1.0x10^-14/[OH-] = 1.0 x 10^-14/(1.4 x 10^-2) = 7.1 x 10^-13 M
I hope this helps!
Ryan O.
Sure that is not a problem! To find [OH-] or [H3O+], the one equation that governs this relationship is: 1.0 x 10^-14 = [OH-][H3O+] Since we know the [OH-] concentration given in the question (1.4 x 10^-2 M) we can just plug this number into the equation above: 1.0 x 10^-14 / (1.4 x 10^-2) = [H3O+] [H3O+] = 7.1 x 10^-13 M I hope this helps clear up any issues! :)06/24/22