Hi
I need help. So difficult cant do!!!!
25 cm3 of Phosphoric Acid H3PO3 of initial concentration 0.1M
was titrated against NaOH, of concentration 0.1M.
Given the following equations and corresponding Ka values:-
H3PO3 =
H2PO3- +
H+ Ka1 = 1x 10-2
H2PO3- =
HPO32- +
H+ Ka2 = 2.5 x10-7
Qn: Sketch the shape of the pH curve during this titration,
labelling significant pH values and significant volumes of NaOH on
the axes of the curve.
So far I am only understand that
- 50 cm3 of NaOH is required at end point.
- I am only guessing that there is only one acidic buffer region at
pKa1 at 10cm3 mark?
- I am not sure the pH at first and second equivalence point. So
confusing already.
Thanks for helping me to understand.
First of all, here's some background info
for students :
H3PO3 is more correctly named (for common name; for
systematic name, work out the oxidation state of phosphorus in the
acids) "phosphorous acid" (rather than "phosphoric acid",
H3PO4). Unlike phosphoric acid, a triprotic acid, phosphorous acid
is indeed a diprotic acid (hence Ka1 and Ka2) - the proton
bonded to the phosphorus atom is non-acidic (ie. cannot
be dissociated).
Now, while I'm *not* going to work out for you the exact pH
values and significant NaOH volumes, here are some guidelines
on what the question requires you to do. Based on what you've
written, you've got your fundamentals, and are doing pretty ok.
Just keep at it, and you'll do fine.
[quote]50 cm3 of NaOH is required at end point.[/quote]
Firstly, that's "equivalence point", not end point. Equivalence
point refers to the exact stoichiometric neutralization
point per each acidic proton. "End point" of a titration
occurs when the (relevant) indicator used changes colour.
The end point approximates the equivalence point.
Secondly, there are two equivalence points, and hence two end
points. This is because phosphorous acid is a diprotic acid
(ie. one mole of acid gives two moles of protons; aka dibasic
acid).
[quote]I am only guessing that there is only one acidic buffer
region at pKa1 at 10cm3 mark?[/quote]
Firstly, since phosphorous acid is a diprotic acid,
there will hence be 2 buffer regions (ie. regions in
which any protons or hydroxide ions added will be immediately
removed from solution by the conjugate base and conjugate acid
respectively, hence the 'buffering' effect). As to the pH of these
buffer regions (or to be precise the pH of the exact points of
maximum buffer capacity for the 2 regions), use pKa1 and pKa2 to
determine.
Secondly, how did you obtain the value of "10cm3" mark for NaOH?
Remember that an acid buffer is most effective (ie. at its maximum
buffer capacity) when molarity of acid = molarity of conjugate base
/ salt. Understand that this also means the pH of maximum
buffer capacity occurs midpoint(s) of the neutralization process,
before the equivalence point(s) is/are reached.
Hence you should be able to work out the volumes of NaOH for the
2 points of maximum buffer capacity, which given that the question
is kind enough to have the molarities of phosphorous acid and
sodium hydroxide as being equal, makes your job much
easier.
[quote]I am not sure the pH at first and second equivalence
point. So confusing already.[/quote]
Ahhh, here's the biggie, the main meat of the question. To
determine the pH at 1st and 2nd equivalence point, you need 2
tables :
a) ICF (Initial Change Final) table for 1st equivalence point,
in moles.
b) ICF (Initial Change Final) table for 2nd equivalence
point, in moles.
To determine the volume of NaOH required for both equivalence
points, use stoichiometry (ie. looking at the balanced chemical
equation(s); in this context, for neutralization of the first and
second acidic protons).
The important and tricky PitTrap here (which many students fall
into and lose their precious marks and hence also their chance
for a distinction grade), is when calculating the molarity of
the salt (or acid, or base), you have to be very careful to take
into consideration and correctly add up the total volumes of both
the acid and alkali used (including for 2nd equivalence
point).
Calculating the pH of the 2nd equivalence point.
Notice that for Na2HPO3(aq) of which the anion is the
HPO3 2-(aq) aka monohydrogen phosphite ion (ie. the twice
deprotonated phosphorous acid), because Na+ has no affinity to
covalent bond with (or remove from solution) OH-
ions, while HPO3 2- has significant affinity to
covalent bond with (or remove from solution) H+ ions aka protons,
the salts Na2HPO3 is basic/alkaline upon hydrolysis.
(Most students find it easier to think of it summarily and remember
it as "strong base + weak acid = basic salt").
Hence, using Kb2 expression and value (you are
given Ka2 value; hence using the relationship between Ka, Kb
and Kw, you can determine the Kb2 value), work out the
molarity of hydroxide ions, hence pOH, and therefore
finally pH.
Calculating the pH of the 1st equivalence point.
Now, the first deprotonated phosphorous acid,
NaH2PO3(aq) of which the anion is H2PO3-(aq) aka
dihydrogen phosphite ion, is a little trickier. One the one
hand, it may protonate itself (ie. remove protons from solution to
become phosphorous acid again) and hence be basic. On the other
hand, it may serve as a source of protons itself (since its second
acidic proton may dissociate, to become monohydrogen phosphite ion)
and hence be acidic.
To determine which, compare the relevant Ka value against the
relevant Kb value. You should be able to figure out it's the Ka2
value against the Kb1 value (tip : focus on the protonation and
deprotonation of the ion being discussed, the dihydrogen phosphite
ion).
Because the Ka2 (deprotonation of ion as an acid) value is
significantly greater than the Kb1 (protonation of ion as
a base) value, you can conclude that the dihydrogen phosphite
ion is acidic rather than basic.
Hence, to determine pH at the 1st equivalence point, after
working out the ICF table and upon having determined
the molarity of the salt NaH2PO3(aq), we hence use the Ka2
expression and value to work out the molarity of protons, and
hence pH.
To refresh your memory,
Ka = ( [molarity of protons] x [molarity of conjugate base] ) /
[molarity of conjugate acid]
Kb = ( [molarity of hydroxide ions] x [molarity of conjugate
acid] ) / [molarity of conjugate base]
So to summarize, this is a good discriminating (distinction)
question, in which you have to decide whether the
1st-deprotonated phosphorous acid aka dihydrogen phosphite ion, is
acidic or basic (upon hydrolysis), based on the relevant Ka and Kb
values. Knowing which, will enable you to (decide
whether to use Ka or Kb expression to) calculate the pH at the 1st
equivalence point. The twice-deprotonated phosphorous acid aka
monohydrogen phosphite ion, on the other hand, is very obviously
basic (upon hydrolysis) (since the proton bonded to phosphorus
cannot dissociate and hence isn't acidic), and hence you must use
the Kb expression to calculate OH, then pOH and finally pH, at
the 2nd equivalence point.
[quote]I need help. So difficult cant do!!!! So confusing
already. Thanks for helping me to understand.[/quote]
You're welcome. And you're doing pretty ok,
actually. And remember, the more you've the passion
to understand, enjoy and love Chemistry (for what it
*truly* is, and not just memorizing blindly as instructed
dogmatically by your JC lecturers/tutors, or for the sake of
paper chase exams), the more you will inevitably excel in it!
We All Love Chemistry!