Regarding the more accurate way of calculating pH at equivalence
point, for example, the first equivalence point of
HOOCCH2COOH,
I have 2 options:
1) -OOCCH2COOH + H2O --> -OOCCH2COO- + H3O+
2) -OOCCH2COOH + H2O --> HOOCCH2COOH + OH-
How then do I know which one is the correct one?
Here (and in point of fact, for most cases), this (ie. considering
only either the Ka or the Kb alone, instead of considering both
simultaneously vis-a-vis the amphiprotic formula) would be the
less accurate way of determining pH.
As long as the Ka (proton dissociation) and Kb (base hydrolysis) of
the amphiprotic species are comparably large (and in a majority of
cases they will indeed be), then the amphiprotic formula will be
more accurate : pH = 1/2 (pKa1 + pKa2)
But in rare cases, such as in the conjugate acid of hydrazine, ie.
N2H5+, where the Ka value of this (theoretically) amphiprotic
species is much larger than the Kb value, or vice-versa, then you
should (obviously) use whichever value is larger.
Consider the diprotic acid H2A.
If Ka of HA- (ie. Ka2 of H2A) is >>> than Kb of HA- (ie.
Kb2 of A2-), then regard HA- as only acidic and use the Ka of HA-
(ie. Ka2 of H2A) to determine pH.
If Kb of HA- (ie. Kb2 of A2-) is >>> than Ka of HA- (ie.
Ka2 of H2A), then regard HA- as only basic and use the Kb of HA-
(ie. Kb2 of A2-) to determine pH.
But as long as the Ka and Kb of that species (eg. HA-) are
comparably large (ie. less than 1 million times magnitude
difference from each other), then this species is indeed (for all
practical considerations) truly an amphiprotic species, in which
case the amphiprotic formula pH = 1/2 (pKa1 +
pKa2) would obtain a more accurate pH answer, because it
takes into consideration the fact that this amphiprotic species
would actually undergo hydrolysis both ways : as a Bronsted-Lowry
acid and also as a Bronsted-Lowry base, simultaneously.
KickMe remains terribly confused at this point, and no, at this
point he still doesn't see it. At least not yet.
Now I see! We consider the value of the Ka and Kb to determine
if it is effectively an acid or a base. And in the case of
Ka/Kb<10^7, and Kb/Ka<10^7, we use pH=1/2(pKa1+pKa2)
However, what if it is the 2nd equivalence point of a diprotic acid
that I am talking about? I will only have the value of one pKa to
use. Then I will not be able to successfully determine the pH at
the 2nd equivalence point using 1/2(pK1 + pK2)
For example, given HOOCCH2COOH,
pKa1=2.65
pKa2=5.70
For the first equivalence point, since Ka is more than one million
times larger than Kb, we treat it as an acid, hence use -OOCCH2COOH
--> -OOCCH2COO- + H+
For the second equivalence point, Ka/Kb=1000 hence use 1/2(pK1 +
pK2) But theres not pK2 in this case.
And when I attempt to use the approach of "treat it as acid or
base", since Ka is larger than Kb, I treat it as an acid, which in
this case is wrong as I am suppose to treat it as a base.
Once the 1st equivalence point is reached, you have HOOCCH2COO-
(which is the only species present), which is clearly an
amphiprotic species, hence pH = 1/2 (pKa1 + pKa2).
(The only time you only need to compare Ka of a species, with Kb of
that same species (ie. not
Ka of an acid with Kb of that acid's conjugate base), is if the
amphiprotic formula does not appear to be mathematically viable,
ie. if the species' acidic capacity far outweighs that of its basic
capacity, or vice-versa.)
Once the 2nd equivalence point is reached, you have the dinegative
-OOCCH2COO- (which is the only species present), which is clearly
only basic (and not acidic at all, since there are no protons
available to dissociate, obviously) and therefore you have no
choice but to use the Kb (and not Ka, which applies only to acidic
species that have protons available to dissociate) of -OOCCH2COO-,
which could be labeled as Kb1, to calculate the [OH-], hence pOH,
hence pH.
When we define the various Ka and Kb values such that Ka1 > Ka2,
and Kb1 > Kb2, then :
Ka1 x Kb2 = Kw and Ka2 x Kb1 = Kw
For now, the KickMe saga finally comes to a conclusion :
I understood the second equivalence point. The only doubt I
have now is why HOOCCH2COO- is "clearly an amphiprotic species". By
calculating the Ka and Kb, I found the ratio of Ka/Kb=50 million,
shouldn't then I treat it as effectively an acid?
pKa1 = 2.65 hence Ka1 = 2.2387 x 10^-3 hence Kb2 = 4.4669 x
10^-12
pKa2=5.70 hence Ka2 = 1.9953 x 10^-6 hence Kb1 = 5.0118 x
10^-9
For the amphiprotic species, Ka2 and Kb2 are relevant.
Since Ka2 is only 446,669 times larger compared to Kb2, hence this
is indeed an amphiprotic species, and the amphiprotic formula for
pH applies with validity.
