In Order to determine equivalence points required looking at graphs and values, all
other calculations involved formulas below.
pK
a
=-log
10
(K)
pK
a
=pH+log
10
([AH]/[A-])
pK
a
=pH+log
10
(conjugated acid/conjugated base)
pK
a1
the point plotted was 1.85
10^-1.85=1.4*10^-2= K
a1

C.) Claim
The exact initial concentration for H
3
PO
4
is .306 this is determined by the change of
the first equivalent point versus the second. In order to determine the pK
a1
and pK
a2
for H
3
PO
4
, you need to analyze the graph and find the equivalent point. Once the
values are found for pK
a1
the point plotted was 1.85. Using the same method to
determine pK
a2
the value is 6.97. To find K
a1
and K
a2
for H
3
PO
4
we take ten raised to
the negative power of the value of pK
a1
and pK
a2
. By doing this we get K
a1
with a value
of 1.4*10^-2 and K
a2
1.07*10^-7. The average H
3
PO
4
concentration in my soft drink is
.00745, the concentration can be determined by the number of moles over the
volume. (Question 2& 3 shown in calculations above)
D.) Evidence
For question one we were able to determine this from the equivalence points on the
graphs. We are able to determine this information for Evidence for research
question number two is seen from the graphs. By understanding where the

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- Fall '11
- U
- Chemistry, pH, Pk, Equivalence point, Phosphoric acid, Equiv. pH