Transcript Chapter 9

Chapter 9
Systematic Treatment of
Equilibrium
Charge Balance
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Basic concept of electroneutrality
Sum of the positive charges in solution
equals the sum of the negative charges
in solution.
What are charged species?
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Ions in solution.
What species will ionize?
Sodium Chloride (NaCl) will dissolve
into solution to give us Na+ and Cl- in
equal amounts.
Glucose will dissolve in solution but not
generate ions.
What do we get when we put something
like phosphoric acid into solution.
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H3PO4
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It is time to recall some chemistry.
We know that PO43- will result but what else.
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H+ to be sure.
Phosphate goes thru a series of equilibrium steps to give
a range of species
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H2PO4- , HPO42- , PO43- that have charge.
Note: Species with alkali metal ions combined will not
really exist. (There will be no NaHPO4- for example)
What Else
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Since it is a water solution and we know
that water can dissociate then we would
have OH- also.
So we would have H+ for cations.
And OH-, H2PO4- , HPO42- , PO43- for
anions.
The Balanced Equation
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Scations = Sanions
Ions of charge greater than one must
be accounted for. So for example if we
have a calcium ion formed then each
calcium ion has a double charge we
must multiply the concentration of such
an ion by this charge.
Charge Balance
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So for phosphoric acid we have the following charge
balance.
[H+] = [H2PO4-] + 2[HPO4=]+3[PO43-]+[OH-]
The H3PO4 that remains in solution has no charge
and need not be accounted for in the charge balance.
Charge Balance
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What if we put trisodium phosphate
into solution. Na3PO4
This is the salt of a strong base and the
sodium can be assumed to completely
dissociate. (We know what ion pairs are
and we will elect to ignore them unless
directed otherwise.)
What ions will be have?
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Well we will have Na+ and PO43-.
But!!!
Phosphate ion is a fairly strong base. That is
it would love the rob a proton from
somewhere and become HPO42This ion in turn is a relatively strong base too
and will steal another proton (H2PO4-)
This new ion is also capable of the next step
giving us H3PO4.
What ions/species will be have
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So for this solution we will possibly
have.
PO43- , HPO42- , H2PO4- , H3PO4 and Na+
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And the ions from water. H+ and OHSo our charge balance is
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[H+] + [Na+] = [H2PO4-] + 2[HPO4=]+3[PO43-]+[OH-]
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How much of each species will there
be?
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If we were to put K2PO4 into solution
then what would we expect to see for
relative amount of each of these
species.
Warning
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I do not want to see species like
NaPO42- cropping up in charge balances.
They do not exist! This is usually a pit
fall for you all in preparing your charge
balances.
Mass Balance
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All this tells us that what be put into our solution is in
there someplace. This is our statement of
conservation of mass.
So if we put Phosphoric acid into solution it will be in
there as one of the phosphate species. Phosphoric
acid is an ingredient of diet pepsi.
We know that it undergoes these acid base
interactions so our Mass balance accounts for that.
Mass Balance of Phosphoric Acid
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CH3PO4 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
This CH3PO4 is often referred to as the Formal
Concentration. So if we put 1.5 mmoles of
phosphoric acid into 0.50 liters of water then this
formal concentration would be 3.0 mM.
We must account for each species, charged or not,
and since we are looking at concentration and not
charge we should not multiply by the charge in this
case.
Mass Balance of Na3PO4
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If we do not know the concentration we can
still use the mass balance.
When we put this salt into water we know we
get three sodium for each phosphate. To set
this up into an equation we could write.
[Na+] = 3 * phosphate concentration
Have I written this backward?????
Mass Balance
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So we would have a mass balance of
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[Na+] = 3{[H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]}
Steps in solving such systems.
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Write all pertinent reactions (don’t forget
water)
Write the charge balance
Write the mass balances ( there might be
more than one )
Count the equations and unknowns. You will
need to have and equal number here.
Now just solve!
Solve?
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Let us look at the trisodium phosphate.
Unknowns
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Seven Unknown so we need seven equations.
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[Na+] [H3PO4] [H2PO4-] [HPO42-] [PO43-] [H+] [OH-]
Mass balance on sodium ion
Mass balance on phosphate ions
Kw
Charge balance
Three acid equilibria
Full solution will be a seventh order polynomial.
Other systems
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That was just a simple system – these
equilibrium systems can become very
complex and solution depends on our
knowing all the equilibria involved.
Example CaF2 Solubility
CaF2 Crystal Structure
CaF2 Solubility
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Two equilibria involved
Ksp
Kb for F- once it is dissolved
Setting conditions can help solve this
problem
CaF2 Solubility
Acid Rain
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Acid Rain