At 50.0°C the water ionization constant, Kw, is 5.48 x 10-14. What is the [H3O+] in neutral water at 50.0°C?

Question

PLEASE HELP. I HAVE NO IDEA
a. 5.48 x 10-7
b. 2.34 x 10-7
c. 1.00 x 10-7
d. 5.48 x 10-14
e. 4.27 x 10-13

Answer 1

You should know that

Kw=[H3O+][OH-]

For a neutral solution [H3O+]= [OH-],
thus
Kw=[H3O+]^2 =>
[H3O+]= SquareRoot(Kw) =2.34*10^-7

Answer 2

what there an equation with this, i dont know how water ionized with H30+ + OH -----> 2h2o?

you kind of need the moles of atleast one other product or reactant

[h30][oh]/ =K
[h2o]