HELP WITH BOTH QUESTIONS #1) The K a values of ortho-phthalic acid, HOOCC 6 H 4
ID: 1014965 • Letter: H
Question
HELP WITH BOTH QUESTIONS
#1) The Ka values of ortho-phthalic acid, HOOCC6H4COOH are shown below.
What ratio of [-OOCC6H4COO-]/[HOOCC6H4COO-] is required to produce a buffer with a pH of 5.00?
#2)
+
-
due to the dissolving of a compound
due to the precipitation of a compound
due to an increase in the number of moles of gas
due to a decrease in the number of moles of gas
due to the formation of a more ordered physical state
due to the formation of a less ordered physical state
Ka1 1.3 10-3 Ka2 3.9 10-6Explanation / Answer
1)
Ka1 for HOOCC6H4COOH and HOOCC6H4COO- combination
Ka2 for HOOCC6H4COO- and -OOCC6H4COO- combination
in this case Ka2 should be used
we know that
for buffers
pH = pKa + log [conjugate base / acid ]
also
pKa = -log Ka
so
pH = -log Ka2 + log [-OOCC6H4COO-]/[HOOCC6H4COO-]
5 = -log 3.9 x 10-6 + log [-OOCC6H4COO-]/[HOOCC6H4COO-]
[-OOCC6H4COO-]/[HOOCC6H4COO-] = 0.39
so
the ratio should be 0.39
2)
consider the given reaction
C4H10 (l) ---> C4H10 (g)
we know that
the entropy order is given by
gases > liquids > solids
it is because
gases are less ordered then liquids which in turn are less ordered than solids
so
gaseous C4H10 has higher entropy than liquid C4H10
so
there is an increase in entropy
so
the sign of dSo is +ve
and it is due to formation of less ordered physical state
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