In a solution with a [OH] of 1.5 x 10-4 M, what is the [H3O+]?

Question

Hi, I've been stuck on this question for a bit. Help would be greatly appreciated!

Answer 1

At 25^oC,

K_w=[H₃O⁺][OH^-]

K_w=10^-14

10^-14=[x][1.5×10^-4]

x=6.67×10^-11 M

Answer 2

Hello, Kai,

At STP, pure water is in equilibrium with it's component ions:

H₂O = H⁺ + OH^- [Please excuse my laziness for using H⁺ instead of the politically correct H₃O⁺]

The equilibrium coefficient for this is 1x10^-14. So we can write

1x10^-14 =[H⁺]*[OH^-]

Now we can find [H⁺]:

1x10^-14 =[H⁺]*[1.5x10^-4]

[H⁺] = [1x10^-14]/[1.5x10^-4]

[H⁺] = [6.7x10^-11]

If the question wanted to know the pH of this solution, it would be the negative log of 6.7x10^-11, or

pH = 10.2

Bob