To calculate the [H3O+], we can use the formula for the ion product of water:
Kw = [H3O+][OH-]
where Kw is the ion product constant of water, equal to 1.0 x 10^-14 at 25°C.
Rearranging the equation, we have:
[H3O+] = Kw / [OH-]
Plugging in the given [OH-] value for urine (1.8 x 10^-9 M) and the value of Kw, we can calculate the [H3O+]:
[H3O+] = (1.0 x 10^-14) / (1.8 x 10^-9) = 5.56 x 10^-6 M
Therefore, the [H3O+] of urine is 5.56 x 10^-6 M.
Calculate the [H3O+] of each aqueous solution with the following [OH-]:
urine, 1.8 x 10^-9 M
Express your answer to two significant figures and include the appropriate units
Final answer
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