How do you calculate the hydrogen ion concentration of an aqueous solution, given the pOH of the solution is..?

To find Ph, you do 14-pOH= 9.5

10^-ph= Hydrogen ion concentration
so 10^-9.5= 3.16 x10^-10

To start, the value of Kw is 1.00 x 10^-14, not 1.00 x 10^-16. And you cannot alter this constant, because its a constant.

But to answer your question... we can use the following steps.

-log(pOH) = OH^-1

So we should start by saying:

a) -log(4.50) = OH^-1

So the answer to this is

a) OH^-1 = -0.65

Now that we have OH^-1, we can use our formula.

Kw = [H3O+][OH-]

We know Kw = 1.00 x 10^-14 and OH^-1 = -0.65

So..

a) 1.00 x 10^-14 = [H3O+][-0.65]

I don't have a calculator on hand, but this is what you would plug in:

(1.00 x 10^-14)/(-0.65) = Your answer

If you persist that Kw = 1.00 x 10^-16, this is what you'd plug in:

(1.00 x 10^-16)/(-0.65) = Your answer

Hope this helps.

so to get the H concentration you need to have pH, which you can get from the pOH by using pH+pOH= 14 so 14-4.5=9.5. then you can get the concentration by taking 10^-(pH) *make sure you put in the negative

I get a 3.16x10-10M

pH + pOH= 14

pOH= -log[OH-]

kw= [H3O+][OH-]=1*10^-14