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chapter
Salts
>> Parts of this equation/concept include:
Soluble salts completely dissolve into their component ions in
water. The concentrations of the ions are determined from the stoichiometry.
Solubility can be determined based on the solubility rules in Chapter
5. Once a salt is dissolved in water, the ions may (or may not)
react with water as an acid or a base.
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Cations normally act as acids. The cations associated with the
strong bases (group I ions, Ca2+, Sr2+, Ba2+)
do not act as either an acid or a base. The other cations tend to
fall into the following categories.
>> Cations of Nitrogen Bases
Cations that are conjugate acids of the nitrogen bases donate the
H+ bonded to the nitrogen to the water and return to
their base form. The Ka for this reaction is usually
not found in tables. Instead, the Kb for the base
form is listed. The Ka can be determined from
the equation
KaKb = Kw
= 1.0 x 10–14 (Equation
16.17)
The most common mistake with this type of problem is to use the
Kb rather than the Ka to solve
the equilibrium expression. Always use Ka with
acid reactions (H+ produced) and Kb
with base reactions (OH– produced).
The general formula for the equilibrium reaction is
BH+ + H2O 
B + H3O+
>> Example 1
What is the pH of a 0.029 M (NH4)2SO4 solution?
Solution:
This is an ammonium salt. All ammonium salts are soluble. Consequently,
in water it reacts as
(NH4)2 SO4 
2 NH4+ + SO42–
Therefore the ion concentrations are
[NH4+] = 0.058 M
[SO42–] = 0.029 M
The sulfate does not react further in water (see the section
on basic ions). Ammonium ion is the conjugate acid of ammonia.
Its reaction in water is
NH4+ + H2O
NH3 + H3O+
The value for Ka is determined from the Kb
for ammonia. (Since Ka and Kb
must be a conjugate acid–base pair, the substance to look
for in the table will be a product of the equilibrium reaction.)
Kb of NH3 = 1.8 x 10–5
KaKb = Kw
Ka(1.8 x 10–5) = 1.0 x 10–14
Ka = 5.6 x 10–10
Using the ICE table to determine equilibrium concentrations,
the initial values are determined from the stoichiometry of the
ionization. (Always do complete reactions before equilibrium reactions.)
| |
NH4+ |
H3O+ |
NH3 |
| Initial |
0.058 |
0 |
0 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.058 – x |
x |
x |
Assume x is small.
5.7 x 10–6 = x
Checking the assumption, 0.057 – 5.6 x 10–6
= 0.057, we find the assumption is good.
[H+] = x = 5.7 x 10–6 M
pH = 5.24
>> Example 2
What is the pH of 0.19 M CH3CH2NH3Br?
Solution:
CH3CH2NH3Br
CH3CH2NH3+ + Br–
Some of the nitrogen base cations are not obvious. However, the
anions associated with them usually are. Recognizing that there
are normally only two ions in a salt is usually sufficient to
be able to write the reaction.
[CH3CH2NH3+] =
[Br–] = 0.19 M
Bromide ion does not react further with water (see the section
on basic ions). CH3CH2NH3+
is the conjugate acid of ethyl amine, CH3CH2NH2.
So its reaction with water is
CH3CH2NH3+ + H2O
CH3CH2NH2 + H3O+
| Ka |
= |
| [CH3CH2NH2][H3O+] |
|
| [ CH3CH2NH3+] |
|
The Kb for CH3CH2NH2
is 6.4 x 10–4. Therefore the Ka
is
KaKb = Kw
= 1.0 x 10–14
Ka(6.4 x 10–4) = 1.0 x 10–14
Ka = 1.6 x 10–11
Using the ICE table,
| |
CH3CH2NH3+ |
CH3CH2NH2 |
H3O+ |
| Initial |
0.19 |
0 |
0 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.19 – x |
x |
x |
Using these values in the Ka expression
Assuming x is small,
1.7 x 10–6 = x
Checking the assumption, 0.19 – 1.7 x 10–6
= 0.19. The assumption is good.
[H3O+] = x = 1.7 x 10–6
M
pH = 5.76
>> Hydrated Metal Ions
Small, highly charged metal ions are acidic in aqueous solution.
The H+ ions come from the waters coordinated, as ligands
or Lewis bases, to the metal ion. Normally, six waters are coordinated
with each metal ion. (The exceptions are rarely worth learning.)
The general formula for the equilibrium is
M(H2O)6x+ + H2O
H3O+ + M(H2O)5(OH)(x–1)+
Because it is the lone pair of the oxygen in water that forms the
bond with the metal, some prefer to write the formula as
M(OH2)6x+ + H2O
H3O+ + M(OH2)5(OH)(x–1)+
The Ka values for acidic, metal cations are listed
in Table 16.2. Cations associated with strong bases (group I ions,
Ca2+, Sr2+, and Ba2+) do not react
with water.
>> Example 3
What is the pH of 0.41 M AlCl3?
Solution:
AlCl3 is a soluble salt, so
AlCl3
Al3+ + 3 Cl–
[Cl–] = 1.23 M
[Al3+] = [Al(H2O)63+]
= 0.41 M
Chloride ion does not react with water. However, aluminum ion
is acidic, with Ka = 1.4 x 10–5.
The reaction of the hydrated ion is
Al(H2O)63+ + H2O
H3O+ + Al(H2O)5(OH)2+
| Ka |
= |
| [H3O+][Al(H2O)5(OH)2+] |
|
| [ Al(H2O)63+] |
|
= |
1.4 x 10–5 |
Using the ICE table
| |
Al(H2O)63+ |
Al(H2O)5(OH)2+ |
H3O+ |
| Initial |
0.41 |
0 |
0 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.41 – x |
x |
x |
Assuming x is small,
s2.4 x 10–3 = x
Checking the assumption that x is small, 0.41 – 0.0024
= 0.4076 = 0.41. The assumption is good.
x = [H3O+] = 2.4 x 10–3
M
pH = 2.62
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Most anions are basic in water. The anions react by accepting an
H+ from water, leaving the water as OH–.
These anions are conjugate bases of weak acids. The anions associated
with strong acids are so weak that they do not react with water.
These anions are Cl–, Br–, I–,
NO3–, ClO3–,
ClO4–, and SO42–.
Another exception is HSO4– which reacts
as an acid in water. (See the section on sulfuric acid.)
The general equilibrium reaction for a basic anion is
A– + H2O
HA + OH–
>> Example 4
What is the pH of 0.36 M NaF?
Solution:
Since sodium salts are always soluble,
NaF
Na+ + F–
and [Na+] = [F–] = 0.36 M.
Sodium, as the cation of a strong base, does not react further
in water. Fluoride ion is the conjugate base of HF. It reacts
in water as
F– + H2O
HF + OH–
The Kb (as with most ions) will not be listed
in the Appendix. It can be determined from the Ka
of the conjugate acid, HF
KaKb = Kw
= 1.0 x 10–14
(3.5 x 10–4)Kb = 1.0 x 10–14
Kb = 2.9 x 10–11
The equilibrium concentrations are determined from the ICE table.
| |
F– |
HF |
OH– |
| Initial |
0.36 |
0 |
0 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.36 – x |
x |
x |
Assuming x is small,
3.2 x 10–6 = x
Checking the assumption, 0.36 – 3.2 x 10–6
= 0.36. The assumption is good.
[OH–] = x = 3.2 x 10–6
pOH = 5.49
pH = 8.51
>> Example 5
What is the pH of 0.15 M K2CO3?
Solution:
K2CO3
2 K+ + CO32–
Therefore, [K+] = 0.30 M, [CO32–]
= 0.15 M. Potassium, the cation of a strong base, does
not react with water. Carbonate does.
CO32– + H2O
HCO3– + OH–
As an ion, the Kb for carbonate is not listed.
Therefore the Kb is determined from the Ka
of the conjugate acid, HCO3–. This
is the Ka2 for carbonic acid, H2CO3.
(Always use the Ka of conjugate acid, which
will also be the product in the equilibrium reaction. It can get
confusing with polyprotics—which Ka?)
For H2CO3, Ka1 = 4.3
x 10–7, Ka2 = 5.6 x 10–11.
Ka2 refers to HCO3–.
KaKb = Kw
= 1.0 x 10–14
(5.6 x 10–11)Kb = 1.0 x 10–14
Kb = 1.8 x 10–4
Using the ICE table to determine equilibrium concentrations,
| |
CO32– |
HCO3– |
OH– |
| Initial |
0.15 |
0 |
0 |
| Change |
–x |
+x |
+x |
| Equilibrium |
0.15– x |
x |
x |
Assuming x is small,
5.2 x 10–3 = x
Checking the assumption, 0.15 – 0.0052 = 0.144. Not quite
true; so, using the quadratic,
2.7 x 10–5 – 1.8 x 10–4x
= x2
0 = x2 + 1.8 x 10–4x
– 2.7 x 10–5
| x |
= |
| –1.8 x 10–4±
[(1.8 x 10–4)2 – 4(1)(–2.7
x 10–5)]1/2 |
|
| 2 |
|
| x |
= |
| –1.8 x 10–4±
[3.24 x 10–8 + 1.08 x 10–4]1/2 |
|
| 2 |
|
Ignoring the negative answer,
x = 0.0051 = [OH–]
pOH = 2.29
pH = 11.71
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Insoluble salts do react slightly in water. They react in the same
way as soluble salts, with the solid forming its component ions.
However, it is an equilibrium reaction rather than a complete reaction.
For a solution to be in equilibrium, both products and reactants
must be present. Recall that a saturated solution has the maximum
amount of solute dissolved in it. If solid is still present, the
solution is saturated.
Solubility is the amount of solid that dissolves in a solution.
If the units are expressed as moles of solid dissolved in a liter
solution, it is molar solubility. In the change line of the following
tables, it is represented as x.
>> Example 6
What is the molar solubility and concentration of each ion in a saturated solution of CaF2?
Solution:
CaF2
Ca2+ + 2 F–
Ksp = [Ca2+][F–]2
= 1.5 x 10–10
Recall that the concentration of a solid is included in the constant,
not in the concentration part of the mass action expression.
Using the ICE table,
| |
Ca2+ |
F– |
| Initial |
0 |
0 |
| Change |
+x |
+2x |
| Equilibrium |
x |
2x |
1.5 x 10–10 = (x)(2x)2
1.5 x 10–10 = 4x3
To solve this equation with a nonprogrammable scientific calculator
you will need the root button, the x1/y
key, usually the shift of the xy key.
This example is solved on the calculator as 1.5 x 10–10
( ) 4 = 3.75 x 10–11
(x1/y) 3 = 3.3 x 10–4
3.3 x 10–4 = x = molar solubility (units
are mol/L)
[Ca2+] = 3.3 x 10–4 M
[F–] = 6.6 x 10–4 M
>> Example 7
What is the molar solubility and concentration of each ion in
a saturated solution of Al(OH)3?
Solution:
Al(OH)3
Al3+ + 3 OH–
Ksp = [Al3+][OH–]3
= 1.9 x 10–33
With the ICE table,
| |
Al3+ |
OH– |
| Initial |
0 |
0 |
| Change |
+x |
+3x |
| Equilibrium |
x |
3x |
1.9 x 10–33 = (x)(3x)3
1.9 x 10–33 = 27x4
For some reason, everyone's first instinct is that 33
= 9, but it's 27. Really!
2.9 x 10–9 mol/L = x = molar solubility
[Al3+] = 2.9 x 10–9 M
[OH–] = 8.7 x 10–9 M
>> Example 8
What is the molar solubility and concentration of each ion in
a saturated solution of Cu3(PO4)2?
Solution:
Cu3(PO4)2
3 Cu2+ + 2 PO43–
Ksp = [Cu2+]3[PO43–]2
= 1.4 x 10–37
| |
Cu2+ |
PO43– |
| Initial |
0 |
0 |
| Change |
+3x |
+2x |
| Equilibrium |
3x |
2x |
1.4 x 10–37 = (3x)3(2x)2
1.4 x 10–37 = (27x3)(4x2)
1.4 x 10–37 = 108x5
1.7 x 10–8 mol/L = x = molar solubility
[Cu2+] = 5.0 x 10–8 M
[PO43–] = 3.4 x 10–8
M
>> Common Ion Effect
The common ion effect refers to a solution that contains both an
insoluble salt and a soluble salt with one ion in common. The molar
solubility decreases with the addition of a common ion.
>> Example 9
What is the concentration of calcium ion in a solution of 0.12
M NaF and saturated with CaF2?
Solution:
Solubility rules and the concentration tip you off that this
is a complete reaction. The term "saturated" lets you know that
CaF2 is in equilibrium with its ions.
NaF
Na+ + F–
[F–] = 0.12 M
CaF2
Ca2+ + 2 F–
Ksp = [Ca2+][F–]2
= 1.5 x 10–10
Using the ICE table,
| |
Ca2+ |
F– |
| Initial |
0 |
0.12 |
| Change |
+x |
+2x |
| Equilibrium |
x |
0.12 + 2x |
1.5 x 10–10 = (x)(0.12 + 2x)2
Assume 2x is small. It wasn't very large in Example 6
and the common ion will make it smaller. Besides, if you don't
assume it, the math can get really ugly.
1.5 x 10–10 = (x)(0.12)2
1.5 x 10–10 = x(0.0144)
1.0 x10–8 = x = [Ca2+]
>> Example 10
What is the molar solubility and concentration of aluminum ion
in a saturated solution of aluminum hydroxide and 0.0032 M
calcium hydroxide?
Solution:
Ca(OH)2
Ca2+ + 2 OH–
[OH–] = 0.0064 M
Al(OH)3
Al3+ + 3 OH–
Ksp = [Al3+][OH–]3
= 1.9 x 10–33
| |
Al3+ |
OH– |
| Initial |
0 |
0.0064 |
| Change |
+x |
+3x |
| Equilibrium |
x |
0.0064 + 3x |
1.9 x 10–33 = (x)(0.0064 + 3x)3
Assume 3x is small.
1.9 x 10–33 = x(0.0064)3
7.2 x 10–27 = x = molar solubility =
[Al3+]
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