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Lecture 10
Analytical Chemistry
November 11, 2002
If river water is saturated with calcite (Box 9-1), the concentration of Ca2+ is governed by
the following equilibria:
CaCO3(s) < == > Ca2+ + CO32-
Ksp = 4.5 x 10-9
CO2(g) < == > CO2(aq)
KCO2 = 0.032
CO2(aq) + H2O < == > HCO3- + H+
K1 = 4.45 x 10-7
HCO3- < == > CO32- + H+
K2 = 4.69 x 10-11
(a) Rearrange these equations to obtain the net reaction below and find the equilibrium
constant for the net reaction.
CaCO3(s) + CO2(aq) + H2O < == > Ca2+ + 2 HCO3The secret to part a is that the 2nd equation is not needed to get the net equation. In or-der
to eliminate the carbonate ion from the net equation we also must reverse the last equation.
CaCO3(s) < == > Ca2+ + CO32-
Ksp = 4.5 x 10-9
CO2(aq) + H2O < == > HCO3- + H+
K1 = 4.45 x 10-7
CO32- + H+ < == > HCO3-
1/K2 = 1/(4.69 x 10-11)
CaCO3(s) + CO2(aq) + H2O < == > Ca2+ + 2 HCO3K = Ksp x K1/K2 = 4.27 x 10-5
b) The mass balance for the net reaction is [HCO3-] = 2[Ca2+]. Find the concentration of
Ca2+ (in mol/L and in mg/L) in equilibrium with atmospheric CO2 (PCO2 = 3.6 x 10-4 bar).
In algebra form the net equation is
K = [Ca2+][HCO3-]2  K x [CO2(aq)] = [Ca2+]{2 [Ca2+]}2 = 4 [Ca2+]3
[CO2(aq)]
The aqueous CO2 concentration is not given. Instead we have CO2(g). So we use the
second equation
KCO2 = [CO2(aq)]/PCO2  [CO2(aq)] = KCO2 x PCO2
2
[Ca2+]3 = ¼ K x KCO2 x PCO2 = ¼ x (4.27 x 10-5)(0.032)( 3.6 x 10-4)
.
.
2+
3
-10
[Ca ] = √ (1.23 x 10 ) = 4.97 x 10-4 M
[Ca2+] = 4.97 x 10-4 M x 40.078 g/mol x 1000 mg/g = 2.0 x 101 mg/L
Why write this in scientific notation?
c) The concentration of Ca2+ in the Don River is 80 mg/L. What effective PCO2 is in
equilibrium with this much Ca2+?
First we convert the calcium concentration to M
[Ca2+] = 8.0 x 101 mg/L x 1 g/103 mg x 1 mol/40.078 g = 2.0 x 10-3 M
Next we rearrange our equation above to solve for PCO2 and get:
PCO2 = 4 [Ca2+]3/ K x KCO2 = 4(2.0 x 10-3)3 / (4.27 x 10-5)(0.032) = 0.022 bar
0.022/0.00036 = 60
The CO2 in the river is not in equilibrium with the atmosphere. There must be a source of
extra carbon dioxide. Possibly from respiration or groundwater high in CO2.
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Now let’s go back to ladder diagrams. We ended with this dual diagram of hydrofluoric
acid and ammonia.
If we mix together solutions of NH3 and HF, the reaction
HF(aq) + NH3(aq) <= => NH4+ + FAnd we noted that the K for this is 1.19 x 106.
Since the equilibrium constant is significantly greater than 1, the reaction’s equilibrium
position lies far to the right.
Here is a problem like the one I started last Wednesday.
What is the pH and composition of a solution prepared by adding 0.090 mol of HF to 0.040
mol of NH3?
We noted that the ladder diagram predicts that these two substances cannot coexist.
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Since HF is present in excess and the reaction between HF and NH3 is favorable, the NH3
will react to form NH4+ consuming HF. At equilibrium, essentially no NH3 remains.
Amount of NH4+ formed is 0.040 mol which also forms 0.040 mol of F-.
This consumes 0.040 mol of HF leaving 0.090 mol – 0.040 mol = 0.050 mol HF.
According to the ladder diagram, for this system, a pH of 3.17 results when there is an
equal amount of HF and F-. Since we have more HF than F-, the pH will be slightly less
than 3.17.
pH = pKa + log [F-]/[HF] = 3.17 + log (0.040/0.050) = 3.17 – 0.097 = 3.07
If the areas of predominance for an acid and a base overlap each other, then practically no
reaction occurs.
For example, of we mix together solutions of NaF and NH4Cl, both substances that
dissociate completely, we would then expect that there would be no significant change in
the moles of F- and NH4+.
Furthermore, the pH of the mixture most be between 3.17 and 9.24. Because F- and NH4+
can coexist over a range of pH values, we cannot be more specific in estimating the
solutions pH from the ladder diagram.
Given specific concentrations of NaF and NH4Cl we can use systematic calculations to get
the real value of the pH.
The ladder diagram for HF/F- also can be used to evaluate the effect of pH on other
equilibria that include either HF or F-. For example, the solubility of CaF2 is affect by pH
because F-is a weak base.
Thus we have two reactions that must come to equilibrium
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CaF2(s) <= => Ca2+ + 2 FHF(aq) <= => H+ + FWhat does Le Chatlier’s principle predict for the solubility of calcium fluoride if we
increase the acidity of the solution?
What does the ladder diagram suggest if we want to minimize the solubility of calcium
fluoride?
Maintain the pH at more than 3.17.
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Now lets do a ladder diagram for the carbonate system that we started with.
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Related to the ladder diagram is the fractional composition diagram. This shows the ratios
of each form of carbonate as a function of pH. Just as the ladder diagrams suggests,
carbonic acid is the predominate species in acidic solutions, hydrogen carbonate
predominates between pH 6 and 10 and carbonate predominates about pH 10.
You can also see that between pH 6 and 10 we have small amounts of both carbonic acid
and carbonate and that there is a point at which the concentration of these two forms are
equal.
The book concludes that a good approximation of this point is when the pH is
pH = ½ ( pK1 + pK2)
This brings us back to chapter 11 and solving diprotic acid H2A systems. There is a
summary on page 211.
Let’s go through the diprotic case using carbonic acid.
Case 1. Suppose we have a solution of carbonic acid. We can treat the system as a
monoprotic acid, i.e. only the first chemical equation applies because the concentration of
carbonate is so low.
We can set up concentration tables for solutions in this pH range.
H2CO3 <= => H+ + HCO3F
F–x
0
x
Ka1 = 4.45 x 10-7
0
x
x2 / (F – x) = Ka1
if you need to know the carbonate concentration, use the Ka2 expression to solve for
[CO32-] using the now know concentration of bicarbonate.
What happens if we are part way between H2CO3 and HCO3-? Like mixing 0.90 mol of
carbonic acid with 0.10 moles of hydrogen carbonate? Then we have a buffer solution and
use the Henderson-Hasselbalch equation.
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Case 2. A solution of NaHCO3. When HA- is the predominate species, in this case
hydrogen carbonate, we need to be aware that both reactions are involved. This means
either using the systematic approach on page 208 (Eq. 11-10), or using the approximation
that [HCO3-] ≈ F you get the equation
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+
[H ] = / Ka1Ka2F + Ka1Kw
√
Ka1 + F
Case 3. A solution of sodium carbonate, Na2CO3. Finally in very basic solutions, the
predominate species is carbonate ion and we can treat the system as a simple monobasic
solution.
CO32- + HOH <= => HCO3- + OHF–x
x
Kb = Kw/Ka2
x
x2 / (F – x) = Kb
Solve for hydroxide and hydrogen carbonate, then get [H+] from Kw and use the Ka2
equilibrium to get carbonic acid.
Polyprotic acids have more than 2 acidic hydrogens. Let’s use phosphoric acid as the
example.
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Here are the chemical equations that are involved:
H3PO4(aq) <= => H+ + H2PO4-
Ka1 = 7.11 x 10-3
pKa1 = 2.148
H2PO4-
<= => H+ + HPO42-
Ka2 = 6.32 x 10-8
pKa2 = 7.199
HPO42-
<= => H+ + PO43-
Ka3 = 7.1 x 10-13
pKa3 = 12.15
PO43- + HOH <= => HPO42- + OH-
Kb1 = 1.4 x 10-2
HPO42- + HOH <= => H2PO4- + OH-
Kb2 = 1.58 x 10-7
H2PO43- + HOH <= => H3PO4 + OH-
Kb3 = 1.41 x 10-12
How would we draw a ladder diagram for phosphoric acid?
Notice in the ladder diagram that the vertical scale is a log scale because it is pH.
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Now how would you deal with this triprotic system?
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Case 1. If we start with H3PO4(aq)?
Treat it as a monprotic weak acid with Ka = Ka1.
Case 2. If we start with H2PO4- ? treat it as an intermediate form of a diprotic acid
between phosphoric acid and hydrogen phosphate
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+
[H ] = / Ka1Ka2F + Ka1Kw
√
Ka1 + F
Case 3. If we start with HPO42- ? treat it as an intermediate form of a diprotic acid, but
now it is intermediate between dihydrogen phosphate and the phosphate ion, so the K’s are
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[H+] = / Ka2Ka3F + Ka2Kw
√
Ka2 + F
Case 4. And finally solutions of PO43- are treated as a monobasic ion with Kb = Kw/Ka3.
Let’s try some calculations.
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1. What is the pH of a 0.10 M H3PO4 solution? F = 0.10 M
Where are we on the ladder diagram? What approach do we use?
H3PO4(aq) <= => H+ + H2PO40.10
0.10 – x
0
x
Ka1 = 7.11 x 10-3
pKa1 = 2.148
0
x
x2 / (0.10 –x ) = 7.11 x 10-3
x2 ≈ 7.11 x 10-4
x1 = 0.0267
Is x small compared to F? No so try again [H3PO4(aq)] = 0.10 – 0.0267 = 0.0733
Carry extra sig figs until the end.
x2 = 7.11 x 10-3 * 0.0733 = 5.21 x 10-4
x2 = 0.0228
try again [H3PO4(aq)] = 0.10 – 0.0228 = 0.0772
x2 = 7.11 x 10-3 * 0.0772 = 5.49 x 10-4
x3 = 0.0234
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try again [H3PO4(aq)] = 0.10 – 0.0234 = 0.0766
x2 = 7.11 x 10-3 * 0.0766 = 5.45 x 10-4
x4 = 0.0233
Look like we can stop. Two sig figs so [H+] = 0.023 M pH = 1.64
Let’s return to the concept of fraction of dissociation introduced in Chapter 10.
α = x/F = 0.0233/0.100 = 0.233
So a 0.1 M solution of phosphoric acid is 23% dissociated. You can see why phosphoric
acid is not on the list of strong acids.
2. What if I have a solution of 0.200 M Na2HPO4. What is its pH?
Where are we on the ladder diagram? Sodium salts are completely soluble so this is a
solution of hydrogen phosphate ion.
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+
[H ] = / Ka2Ka3F + Ka2Kw
√
Ka2 + F
= {( 6.32 x 10-8* 7.1 x 10-13* 0.200 + 6.32 x 10-8* 1.0 x 10-14 )/0.200}1/2
= { (8.974 x 10-21 + 6.32 x 10-22)/ 0.200}1/2
= { 8.342 x 10-21/0.200 }1/2 = { 4.171 x 10-20 }1/2 = 2.04 x 10-10 M
pH = 9.6899 = 9.69
Approximation pH = ½( pKa2 + pKa3) = ½ (7.199 + 12.15) = 9.6745