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
In any aqueous solution, [H3O+] and [OH-] are related by the equation:
[H3O+][OH-] = 1.0 x 10^-14
Therefore, to calculate the [H3O+], we can rearrange the equation as:
[H3O+] = (1.0 x 10^-14) / [OH-]
For the given [OH-] in urine, [OH-] = 1.8 x 10^-9 M. Substituting this value into the equation, we have:
[H3O+] = (1.0 x 10^-14) / (1.8 x 10^-9)
= 5.6 x 10^-6 M
Therefore, the [H3O+] of the aqueous solution with [OH-] = 1.8 x 10^-9 M is 5.6 x 10^-6 M.
To calculate the [H3O+] concentration given [OH-], we can use the equation for the ion product of water (Kw = [H3O+][OH-]) at 25°C, which is 1.0 x 10^-14.
First, we can solve for [H3O+] using the given [OH-] concentration in urine of 1.8 x 10^-9 M:
Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+][1.8 x 10^-9]
[H3O+] = (1.0 x 10^-14) / (1.8 x 10^-9)
[H3O+] ≈ 5.6 x 10^-6 M
Therefore, the [H3O+] concentration in urine with [OH-] of 1.8 x 10^-9 M is approximately 5.6 x 10^-6 M.
To calculate the concentration of H3O+ in each aqueous solution, we can use the equation for the ion product of water (Kw):
Kw = [H3O+][OH-]
Since we are given the concentration of OH- for each solution, we can rearrange the equation and solve for [H3O+]:
[H3O+] = Kw / [OH-]
Now let's calculate the concentration of H3O+ for the urine solution, where [OH-] is 1.8 x 10^-9 M.
Kw, the ion product of water, is equal to 1.0 x 10^-14 at 25 degrees Celsius.
[H3O+] = (1.0 x 10^-14) / (1.8 x 10^-9)
[H3O+] = 5.56 x 10^-6 M
Therefore, the concentration of H3O+ in the urine solution is 5.56 x 10^-6 M, with two significant figures.