DIPROTIC AND POLYPROTIC ACIDS
Diprotic acids have two ionizable hydrogens and is therefore capable of donating two protons.
Example: 0.10M H2CO3
H2CO3 + H2O === H3O+ + HCO3- Ka1 = [H3O+][HCO3-] = 4.3x10-7
[H2CO3]
HCO3- + H2O === H3O+ + CO32-
Ka2 = [H3O+][CO32-] = 5.6x10-11
[HCO3-]
[H3O+]Total = [H3O+]Ka1 + [H3O+]Ka2
Approximation:
The Ka1>>Ka2 so that the contribution of second dissociation to the concentration of hydronium
ion is negligible. We calculate the [H3O+] only from Ka1. Let x be [H3O+] so that
Ka1 = (x)(x)__ = 4.3x10-7
(0.10-x)
x2 + 4.3x10-7x -4.3x10-8 = 0
x = -4.3x10-7 + [(4.3x10-7)2 -4(1)(-4.3x10-8)]1/2 = -4.3x10-7 + [1.849x10-13 +1.72x10-7]1/2
2(1)
2
x = -4.3x10-7 + 4.15x10-4 = 2.1x10-4
2
pH = -log( 2.1x10-4) = 3.68 – this means that H2CO3 is a weaker acid than HOAc!
Analytical Concentration of H2CO3
CH2CO3 = [H2CO3] + [HCO3-] + [CO32-]
𝛂H2CO3 = ___________[H2CO3]_______
[H2CO3] + [HCO3-] + [CO32-]
Taking the inverse of 𝛂H2CO3,
_1_____ = [H2CO3] + [HCO3-] + [CO32-] = [H2CO3] + [HCO3-] + [CO32-]_
𝛂H2CO3
[H2CO3]. [H2CO3] [H2CO3]
[H2CO3]
Ka1Ka2 = [H3O+][HCO3-]x[H3O+][CO32-] = [H3O+]2[CO32-]
[H2CO3]
[HCO3-]
[H2CO3]
Ka1Ka2 = [CO32-]_
[H3O+]2 [H2CO3]
_1_____ = 1 + Ka1_____ + Ka1Ka2_
𝛂H2CO3
[H3O+] [H3O+]2
Simplifying and taking the inverse of the above expression
𝛂H2CO3 = ___________ [H3O+]2________________
[H3O+]2 + Ka1[H3O+] + Ka1Ka2
Using similar derivations for
HCO3
-
and
, we have
2CO3
𝛂HCO3- = __________Ka1[H3O+]_______________
[H3O+]2 + Ka1[H3O+] + Ka1Ka2
𝛂CO32- = ______________Ka1Ka2______________
[H3O+]2 + Ka1[H3O+] + Ka1Ka2
DISTRIBUTION DIAGRAM FOR CARBONIC ACID AS A FUNCTION OF pH
At pH = pKa1 (6.37), [H2CO3] = [HCO3-] and 𝛂H2CO3 = 𝛂HCO3- = 0.50
At pH = pKa2 (10.25), [HCO3-] = [CO32-] and 𝛂HCO3- = 𝛂CO32- = 0.50
For H3PO4,
H3PO4 + H2O ==== H3O+ + H2PO4-
Ka1 = [H3O+][H2PO4-] = 7.5 x 10-3
[H3PO4]
H2PO4- + H2O ==== H3O+ + HPO42-
Ka2 = [H3O+][HPO42-] = 6.2 x 10-8
[H2PO4-]
HPO42- + H2O ==== H3O+ + PO43-Ka3 = [H3O+][PO43-] = 4.8 x 10-13
[HPO42-]