I'm reviewing my notes for a test tomorrow, and I found that I have a question concerning acids that under go multiple protonizations. I know the molarity [or moles, if that's being calculated] of H3O+ of the first protonization can be found using the Ka, etc. I have in my notes that for the next protonization, the molarity of H3O+ on the products side is always the molarity/moles of the H3O+ found in the first prot. + x.
Ex. in case my explanation is confusing:
Prot 1:
H3PO4 + H20 <==> H3O+ + H2PO4-
[H3PO4]= 5.00 - x
[H3O+]= x
Ka= x2/5.00 = .193
Prot 2:
H2PO4- + H2O <==> H3O+ + HPO4-2
[H2PO4-]= .193-x
[H3O+]= .193+x
If I continued this to the third protonization of H3PO4 [HPO4-2 + H20 <==> H+ + PO4-3], would the concentration of H3O+ be [H3O+ in prot 2] + x?
I wonder what you mean by..
Ka= x2/5.00 = .193
You need to solve for x from Ka, so I assume that last equal sign means..
Ka= x2/5.00 => x= .193
Then..
Go to step two..
H2PO4- + H2O <==> H3O+ + HPO4-2
[H2PO4-]= .193-x
[H3O+]= .193+x and so on.
In the example you provided, you are correct that the concentration of H3O+ on the products side in the second protonization is [H3O+] in the first protonization + x. This is based on the principle of conservation of mass and the fact that one proton from H3O+ is transferred to the H2PO4- ion to form the next product.
To calculate the value of x, you start with the expression for the equilibrium constant (Ka) for the first protonization:
Ka = [H3O+][H2PO4-]/[H3PO4]
In this equation, [H3O+] represents the concentration of H3O+ after the first protonization reaction, [H2PO4-] represents the concentration of H2PO4-, and [H3PO4] represents the initial concentration of H3PO4.
Since the concentration of H3O+ is x and the concentration of H2PO4- is 0.193 - x (as you correctly stated), we can substitute these values into the equation and solve for x.
x^2 / (5.00 - x) = 0.193
To solve this quadratic equation, we rearrange it to:
x^2 - 0.193(5.00 - x) = 0
x^2 - (0.965 - 0.193x) = 0
x^2 - 0.965 + 0.193x = 0
x^2 + 0.193x - 0.965 = 0
Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 1, b = 0.193, and c = -0.965, and x should be positive.
Once you solve for x and find its value, you can proceed to the second protonization and use the same concept to calculate the concentration of H3O+ on the products side.
Continuing this process for the third protonization, the concentration of H3O+ would be [H3O+] in the second protonization + x, just as you mentioned.
It's important to note that these calculations assume that the concentrations of H2O and HPO4-2 remain constant throughout the protonization reactions.