A solution has [H+] = 4.0 x 10-8 M. The pOH of this solution is

Answer 1

The pOH of a solution can be calculated using the formula:

pOH = -log[OH-]

Since we know the concentration of [H+], we can use the relationship between [H+] and [OH-] in a neutral solution, which is given by the equation:

[H+][OH-] = 1.0 x 10^-14 M^2

Rearranging this equation, we get:

[OH-] = 1.0 x 10^-14 M^2 / [H+]

Substituting the given [H+] value of 4.0 x 10^-8 M into the equation, we have:

[OH-] = 1.0 x 10^-14 M^2 / (4.0 x 10^-8 M)

Simplifying the equation, we get:

[OH-] = 2.5 x 10^-7 M

Now we can calculate the pOH:

pOH = -log[OH-]
pOH = -log(2.5 x 10^-7)
pOH ≈ 6.60

Therefore, the pOH of this solution is approximately 6.60.