The concentration of HNO3 in a solution is 3.50 10-6 M.
What is the [H3O+] in the solution?
_______M
What is the [OH-] in the solution?
________ M
What is the pH of the solution?
What is the pOH of the solution?
To find the [H3O+] in the solution, we can use the fact that HNO3 is a strong acid and dissociates completely in water. Therefore, the concentration of H3O+ ions is equal to the concentration of HNO3.
[H3O+] = 3.50 x 10^-6 M
To find the [OH-] in the solution, we can use the fact that water is neutral and the concentration of H3O+ and OH- ions in water are equal. Therefore, we can use the equation:
[H3O+] x [OH-] = 1.0 x 10^-14
Using the [H3O+] given above, we can solve for [OH-]:
(3.50 x 10^-6 M) x [OH-] = 1.0 x 10^-14
[OH-] = (1.0 x 10^-14) / (3.50 x 10^-6 M)
To find the pH of the solution, we can use the formula:
pH = -log[H3O+]
Using the [H3O+] given above, we can find the pH:
pH = -log(3.50 x 10^-6 M)
The pOH of the solution can be found using the equation:
pOH = -log[OH-]
Using the [OH-] value we found earlier, we can find the pOH:
pOH = -log([OH-])
Now, let's calculate:
[H3O+] = 3.50 x 10^-6 M
[OH-] = (1.0 x 10^-14) / (3.50 x 10^-6 M)
pH = -log(3.50 x 10^-6 M)
pOH = -log ([OH-])
To find the [H3O+], [OH-], pH, and pOH of a solution, we need to use the concept of acidity and basicity and the dissociation of the acid, HNO3.
HNO3 is a strong acid and will completely dissociate in aqueous solution, giving H+ ions. Therefore, the [H3O+] in the solution will be equal to the concentration of HNO3.
[H3O+] = 3.50 × 10^(-6) M
To find the [OH-] in the solution, we can use the fact that the product of [H3O+] and [OH-] is equal to 1x10^(-14) in water at 25°C.
[H3O+][OH-] = 1x10^(-14)
Since we know the [H3O+] from the previous step, we can rearrange the equation to solve for [OH-]:
[OH-] = 1x10^(-14) / [H3O+]
Now, we can substitute the value of [H3O+] we found earlier into this equation.
[OH-] = 1x10^(-14) / (3.50 × 10^(-6)) M
To find the pH of the solution, we can use the equation:
pH = -log[H3O+]
We can substitute the [H3O+] we found earlier into this equation to get the pH.
pH = -log(3.50 × 10^(-6)) M
To find the pOH of the solution, we can use the equation:
pOH = -log[OH-]
We can substitute the [OH-] we found earlier into this equation to get the pOH.
pOH = -log(1x10^(-14) / (3.50 × 10^(-6))) M
3.5E-6 is quite a dilute solution; however, I don't think that the H3O^+ contribution from H2O will be enough to significantly affect the numbers. Therefore,
HNO3 ==> H^+ + NO3^-
H3O^+ is concn HNO3 and pH -log(H^+).
pH + pOH = pKw = 14.
pOH = -log(OH^-)