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
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.
To calculate the [H3O+] of a solution given the [OH-], we can use the Kw expression, which is the ion product of water. Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C.
Given that the [OH-] in urine is 1.8 x 10^-9 M, we can substitute this value into the Kw expression:
[H3O+][OH-] = 1.0 x 10^-14
[H3O+] = 1.0 x 10^-14 / [OH-]
[H3O+] = 1.0 x 10^-14 / 1.8 x 10^-9
[H3O+] ≈ 5.6 x 10^-6 M
Therefore, the approximate value of [H3O+] in urine is 5.6 x 10^-6 M.
To calculate the concentration of [H3O+] (hydronium ion), we can use the relationship with [OH-] (hydroxide ion) in water, which is given by the equation:
Kw = [H3O+][OH-]
Where Kw is the ion product constant for water and has a value of 1.0 x 10^-14 at 25 degrees Celsius.
Rearranging the equation, we get:
[H3O+] = Kw / [OH-]
Now, let's substitute the given [OH-] value into the equation:
[H3O+] = (1.0 x 10^-14) / (1.8 x 10^-9)
Calculating this, we get:
[H3O+] ≈ 5.6 x 10^-6 M
Therefore, the concentration of [H3O+] in urine with [OH-] of 1.8 x 10^-9 M is approximately 5.6 x 10^-6 M. Remember to express your answer to two significant figures and include the appropriate units, which in this case is M (molarity).