A sample of rain in an area with severe air pollution has a pH OF 3.5. What is the pOH of this rainwater?
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To determine the pOH of the rainwater, we first need to find the concentration of hydroxide ions (OH-) in the water.
pOH is a measure of the concentration of hydroxide ions. It is calculated using the formula: pOH = -log[OH-].
Since we are given a pH value of 3.5, which is a measure of the concentration of hydronium ions (H3O+), we can use the fact that water is neutral (pH = pOH = 7) to find the concentration of hydroxide ions indirectly.
To do this, we use the equation Kw = [H3O+][OH-], where Kw is the ion product of water, which is equal to 1.0 x 10^-14 at 25°C.
Since the solution is acidic (pH < 7), the concentration of H3O+ is greater than the concentration of OH-.
Let's calculate the concentration of H3O+ ions first, using the equation pH = -log[H3O+].
pH = 3.5
[H3O+] = 10^(-pH)
[H3O+] = 10^(-3.5)
[H3O+] = 3.16 x 10^(-4) M (Molar)
Now, using the equation Kw = [H3O+][OH-], we can find the concentration of OH- ions.
Kw = 1.0 x 10^(-14)
[H3O+] = 3.16 x 10^(-4)
[OH-] = Kw / [H3O+]
[OH-] = (1.0 x 10^(-14)) / (3.16 x 10^(-4))
[OH-] = 3.16 x 10^(-11) M (Molar)
Finally, we can calculate the pOH by taking the negative logarithm of the hydroxide ion concentration.
pOH = -log[OH-]
pOH = -log(3.16 x 10^(-11))
pOH = 10.5
Therefore, the pOH of the rainwater is 10.5.
pH + pOH = pKw = 14
You know pH and pKw, solve for pOH.