Calculate the p H of the diprotic acid H2 A (such as H2 SO4) where the dissociation of the first H is strong and the dissociation of the second H is weak but the 5% rule does not apply (K2 is too large relative to the initial concentration of H Aminus).
1.Assume the initial concentration of H2A is 0.0140 M.
2.Assume the initial concentration of H2A is 0.0116 M.
Chem help?!?
2020-10-08 2:51 am
回答 (1)
2020-10-08 12:29 pm
In order to calculate the actual pH, you must know a value of Ka2 for the acid. For that ionization:
HA- <--> H+ + A2-
Ka2 = [H+][A2-]/[HA-]
You can set up an ICE table for the second ionization. Since the first ionization is complete, the initial concentrations of H+ and HA- will be 0.0140 M each. Then,
...............HA-..............H+.............A2-
Init........0.0140...........0.0140........0
change....-x..................+x.............+x
Equil.......0.0140x......0.0140+x......x
Ka2 = (0.140+x)(x)/(0.0140-x)
In order to go any farther, you must have a value for Ka2. For HSO4-, Ka2 = 0.012. Using that,
(0.0140+x)(x)/(0.0140-x)= 0.012
1.68X10^-4 - 0.012x = 0.0140x + x^2
x^2 + 0.026x - 1.68X10^-4 = 0
Using the quadratic formula, x = 0.0054
So, [H+] = 0.0140 + 0.0054 = 0.0194 M
pH = 1.71
You can repeat this calculation for the second concentration, but again, you must know a value for Ka2 for the acid in order to do this calculation.
HA- <--> H+ + A2-
Ka2 = [H+][A2-]/[HA-]
You can set up an ICE table for the second ionization. Since the first ionization is complete, the initial concentrations of H+ and HA- will be 0.0140 M each. Then,
...............HA-..............H+.............A2-
Init........0.0140...........0.0140........0
change....-x..................+x.............+x
Equil.......0.0140x......0.0140+x......x
Ka2 = (0.140+x)(x)/(0.0140-x)
In order to go any farther, you must have a value for Ka2. For HSO4-, Ka2 = 0.012. Using that,
(0.0140+x)(x)/(0.0140-x)= 0.012
1.68X10^-4 - 0.012x = 0.0140x + x^2
x^2 + 0.026x - 1.68X10^-4 = 0
Using the quadratic formula, x = 0.0054
So, [H+] = 0.0140 + 0.0054 = 0.0194 M
pH = 1.71
You can repeat this calculation for the second concentration, but again, you must know a value for Ka2 for the acid in order to do this calculation.
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