What is the value of [h3o] in a solution where [oh]=4.20x10^-6
(H3O^+)(OH^-) = Kw = 1E-14
To find the value of [H3O+] in a solution where [OH-] is given, you can use the equation for water autoprotolysis:
[H3O+][OH-] = Kw
Where Kw is the ion product of water, equal to 1.0 x 10^-14 at 25°C.
In the given solution, [OH-] is given as 4.20 x 10^-6. Plugging this value into the equation, we have:
[H3O+](4.20 x 10^-6) = 1.0 x 10^-14
To solve for [H3O+], divide both sides of the equation by (4.20 x 10^-6):
[H3O+] = (1.0 x 10^-14) / (4.20 x 10^-6)
Calculating this expression, you will find [H3O+] to be approximately 2.38 x 10^-9.