Transcript No Slide Title

CHAPTER 17
Acid-Base Equilibriu and
Solubility Equilibria
Neutralization Reaction
A neutralization reaction is the reaction of an acid with a base,
usually giving salt plus water as products.
For three types of neutralization reactions the reaction goes
essentially to completion:
strong acid + strong base
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O()
strong acid + weak base
HCl(aq) + NH3(aq)  NH4Cl(aq)
weak acid + strong base
HF(aq) + NaOH(aq)  NaF(aq) + H2O()
For weak acid + weak base an equilibrium is established between
reactants and products. This case is mathematically difficult.
Buffers
A buffer solution is a solution whose pH does not change
significantly upon the addition of a small amount of a strong acid or
strong base.
Buffer solutions contain a substantial amount of a weak acid and
a weak base, usually a weak acid-weak base conjugate pair.
Buffer solutions are important for systems where a particular
value for pH is required, such as in biological systems.
Examples:
CH3COOH + CH3COO-
acetate buffer
H2CO3 + HCO3-
carbonate buffer
NH3 + NH4+
ammonia buffer
How Buffers Work
Consider a buffer solution containing a weak acid “HA” and the
conjugate base of the weak acid “A-”. What will happen if we add a
small amount of a strong acid (HCl) or strong base (NaOH)?
strong acid
HCl(aq) + A-(aq)  HA(aq) + Cl-(aq)
strong base
NaOH(aq) + HA(aq)  Na+(aq) + A-(aq) + H2O()
Buffer solution converts the strong acid or base into a weak acid
or base. By doing so, they minimize the effect of these additions on pH.
Example of Buffering
Consider 1.000 L of each of the following two solutions.
Solution A. 0.100 M CH3COOH and 0.200 M CH3COONa.
Solution B. Pure water
Find the initial pH and the pH after addition of 0.0010 moles HCl
to each of the above solutions, at T = 25. C. The value for Ka for
CH3COOH at this temperature is Ka = 1.8 x 10-5.
Solution A. 0.100 M CH3COOH and 0.200 M CH3COONa.
CH3COONa(aq)  Na+(aq) + CH3COO-(aq)
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
Ka = [H3O+] [CH3COO-] = 1.8 x 10-5
[CH3COOH]
Substance
Initial
Change
Equilibrium
H3O+
0.000
x
x
CH3COO- 0.200
x
0.200 + x
CH3COOH 0.100
-x
0.100 - x
Ka = [H3O+] [CH3COO-] = (x) (0.200 + x) = 1.8 x 10-5
[CH3COOH]
(0.100 - x)
Assume x << 0.100
(x) (0.200) = 1.8 x 10-5
(0.100)
x = (1.8 x 10-5) (0.100) = 9.0 x 10-6
(0.200)
pH = - log10(9.0 x 10-6) = 5.05
Is 9.0 x 10-6 << 0.100? Yes (at least 10 times smaller).
Now, consider the effect of adding 0.0010 moles of HCl (equivalent to 0.0010 M HCl, since 1.00 L solution).
HCl(aq) + CH3COO-(aq)  CH3COOH(aq) + Cl-(aq)
initial
after rxn.
0.0010 M
0.0000 M
0.2000 M
0.1990 M
0.1000 M
0.1010 M
0.0010 M
CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq)
Ka = [H3O+] [CH3COO-] = 1.8 x 10-5
[CH3COOH]
Substance
H3O+
CH3COOCH3COOH
Initial After HCl
0.000 0.000
0.200 0.199
0.100 0.101
Change
x
x
-x
Equilibrium
x
0.199 + x
0.101 – x
Ka = (x) (0.199 + x) = 1.8 x 10-5
(0.101 – x)
Assume x << 0.100, then
(x) (0.199) = 1.8 x 10-5
(0.101)
x = (0.101) (1.8 x 10-5) = 9.1 x 10-6
(0.199)
pH = - log10(9.1 x 10-6) = 5.04
The pH decreased from 5.05 to 5.04, a change of 0.01 pH unit.
Solution B (pure water)
pure water. pH = 7.00
After addition of 0.0010 moles of HCl we simply have a 0.001 M
solution of HCl, a strong acid.
So pH = - log10(0.0010) = 3.00
We may summarize the results as follows:
solution A
solution B
initial pH
5.05
7.00
pH after addn. of
5.04
3.00
0.0010 mol HCl
The pH of solution A (the buffer solution) changed by only a
small amount after addition of the HCl. The pH of solution B
(unbuffered water) changed by 4 pH units (a change in H3O+
concentration of a factor of 10,000).
Henderson Equation
The Henderson (or Henderson-Hasselbalch) equation applies for
solutions of a weak acid/conjugate base pair. We may develop the
equation as follows.
Consider a weak acid HA
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
Ka = [H3O+] [A-] so [H3O+] = Ka [HA]
[HA]
[A-]
Therefore
- log10[H3O+] = - log10Ka - log10([HA]/[A-])
pH = pKa - log10([HA]/[A-]) , or
pH = pKa + log10([A-]/[HA]) , the Henderson equation.
Properties of the Henderson Equation
1) The Henderson equation applies for solutions of a weak
acid/conjugate base pair.
2) There must be appreciable amounts (at least 0.001 M) of both
the weak acid and the conjugate base present in the system.
3) If the weak acid and conjugate base concentrations are the
same ([HA] = [A-]), then
pH = pKa + log10([A-]/[HA]) = pKa + log10(1) = pKa
This suggests that the best buffers are those where pH ~ pKa. In
fact, buffers work best if the pKa of the weak acid is within about 1 pH
unit of the desired buffer pH.
4) While buffers are often prepared by combining a weak acid
and conjugate base pair, they can also be prepared by adding a small
amount of strong base to a weak acid solution, or a small amount of
strong acid to a weak base solution.
Example: Consider 1.000 L of a 0.100 M solution of acetic acid
(CH3COOH). If we add 0.050 moles of NaOH, then
CH3COOH(aq) + NaOH(s)  CH3COO-(aq) + Na+(aq) + H2O()
initial
(.100 mol)
(.050 mol)
(.000 mol)
final
(.050 mol)
(.000 mol)
(.050 mol)
The neutralization reaction is used to produce the conjugate base
needed for the buffer. Note that we require that there be a larger number
of moles of weak acid than strong base.
Buffer Calculation (Example)
We have 0.500 M stock solutions of the following:
CH3COOH
HF
HOCl
NaCH3COO
NaF
NaOCl
Ka = 1.8 x 10-5
Ka = 3.5 x 10-4
Ka = 3.5 x 10-8
pKa = 4.74
pKa = 3.46
pKa = 7.46
How would you prepare 1.000 L of a pH = 3.75 buffer solution?
Example: We have 0.500 M stock solutions of the following:
CH3COOH
HF
HOCl
NaCH3COO
NaF
NaOCl
Ka = 1.8 x 10-5
Ka = 3.5 x 10-4
Ka = 3.5 x 10-8
pKa = 4.74
pKa = 3.46
pKa = 7.46
How would you prepare 1.000 L of a pH = 3.75 buffer solution?
Which acid/conjugate base pair should we use?
We should use the one where the pKa is as close as possible to the
pH of the buffer we wish to make. That would be the HF/NaF system.
Now, let
x = mL of the HF stock solution
(1000 - x) = mL of the NaF stock solution
Then, in the 1.000 L buffer solution (using MiVi = MfVf)
[HF] = (0.500 M) x
[F-] = (0.500 M) (1000 - x)
1000
1000
pH = pKa + log10([A-]/[HA])
3.75 = 3.46 + log10 (0.500)[(1000 - x)/1000]
(0.500) (x/1000)
3.75 = 3.46 + log10[(1000 - x)/x]
3.75 - 3.46 = 0.29 = log10[(1000 - x)/x]
100.29 = 1.95 = (1000 - x)/x
1.95 x = 1000 - x
2.95 x = 1000 ; x = 1000/2.95 = 339.
So 339 mL of the HF solution and 661 mL of the NaF solution.
Titration
By titration, we mean the reaction of a fixed amount of one substance by the slow addition of a second substance.
In an acid-base titration we slowly add an acid to a fixed amount
of base, or we slowly add a base to a fixed amount of acid.
We use titrations to determine the
concentration of acid or base in a stock
solution. We may also be interested in the
titration curve, a plot of pH vs mL of added
titrant (solution being slowly added).
Equivalence point.
The equivalence point in a titration is the point
where sufficient titrant has been added to
have a complete reaction with no excess
acid or base.
Indicator
An indicator is a substance (usually a weak acid) that changes
colors at a particular pH.
HInd(aq) + H2O(l)  H3O+(aq) + Ind-(aq)
color 1
color 2
We often use indicators in titrations. We choose an indicator so that the
end point (point where color change occurs) is as close as possible to the
equivalence point for the titration. Based on the Henderson equation
pH = pKa + log10([A-]/[HA])
we expect the color change will occur at pH  pKInd.
In choosing an indicator we would like the color change to be as
close as possible to the equivalence point for the titration.
Phenolphthalein is a suitable indicator for both the strong acid+strong
base titration (blue line) and weak acid+strong base titration (red line).
Methyl red is a suitable indicator only for the strong acid+strong base
titration. Other indicators would also be suitable choices.
Titration of a Strong Base With a Strong Acid
Consider the titration of 25.00 mL of a 0.0100 M solution of
NaOH (a strong base) with a 0.0100 M solution of HCl (a strong acid).
The reactions that will occur are as follows:
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
neutralization
NaOH(aq)  Na+(aq) + OH-(aq)
if excess base
HCl(aq)  H+(aq) + Cl-(aq)
if excess acid
If there is excess NaOH the pH will be greater than 7.0. If there
is excess HCl the pH will be less than 7.0. At the equivalence point the
moles of NaOH and HCl will be equal, and the solution will be neutral
(pH = 7.0).
The equivalence point occurs at pH = 7.0. A good indicator
would be one where the color change occurs near pH = 7.0 (such as
bromthymol blue). Notice that the titration curve is steep (nearly
vertical) near the equivalence point.
Note that we can calculate the pH at every point in the titration
(which was done in generating the above titration curve).
pH Calculation During Titration
Consider the titration of 25.00 mL of a 0.0100 M solution of
NaOH (a strong base) with a 0.0100 M solution of HCl (a strong acid).
The reactions that will occur are as follows:
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
neutralization
NaOH(aq)  Na+(aq) + OH-(aq)
if excess base
HCl(aq)  H+(aq) + Cl-(aq)
if excess acid
What is the pH after addition of 10.00 mL of HCl solution?
Strategy:
1) Find the moles of acid and base present, and the total volume.
2) Find the species present after the neutralization reaction.
3) Find the pH (as in a regular acid/base calculation)
mol NaOH = 0.02500 L (0.0100 M) = 2.5 x 10-4 mol NaOH
mol HCl = 0.0100 L (0.0100 M) = 1.0 x 10-4 mol HCl
Volume - 0.0250 L + 0.0100L = 0.0350 L
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
neutralization
After neutralization, excess NaOH is
Excess NaOH = 2.5 x 10-4 mol - 1.0 x 10-4 mol = 1.5 x 10-4 mol
[NaOH]excess = (1.5 x 10-4 mol)/(0.0350 L) = 0.00429 M
NaOH(aq)  Na+(aq) + OH-(aq)
pOH = - log10(.00429) = 2.37
pH = 14.00 - 2.47 = 11.63
if excess base
Titration of a Strong Acid With a Strong Base
Consider the same system, but the case where 25.00 mL of a
0.0100 M solution of HCl is titrated with a 0.0100 M solution of NaOH.
The reactions taking place are the same, but now the solution is initially
acidic.
Titration of a Weak Acid With a Strong Base
Consider the titration of 25.00 mL of a 0.0100 M solution of
CH3COOH (a weak acid; Ka = 1.8 x 10-5) with a 0.0100 M solution of
NaOH (a strong base).
The reactions that will occur are as follows:
CH3COOH(aq) + NaOH(aq)  H2O(l) + Na+(aq) + CH3COO-(aq)
neutralization
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
if excess weak acid
NaOH(aq)  Na+(aq) + OH-(aq)
if excess base
If there is excess CH3COOH then there will usually be a buffer
solution with both CH3COOH and CH3COO- present.
CH3COOH(aq) + NaOH(aq)  H2O(l) + Na+(aq) + CH3COO-(aq)
If there is excess base then the pH will be controlled by the concentration
of excess strong base.
NaOH(aq)  Na+(aq) + OH-(aq)
Note that at the half-equivalence point pH  pKa, since
pH = pKa + log10([A-]/[HA])
At the half-equivalence point (point where we have added half
the titrant needed to reach the equivalence point) pH  pKa for the weak
acid. Notice that there will be a region where the solution forms a buffer
solution. Note that the pH at the equivalence point is greater than 7.0.
Indicator - phenolphthalein
Example of Weak Acid + Strong Base
Calculation
25.00 mL of a 0.100 M solution of formic acid (HCOOH, Ka =
1.7 x 10-4) is titrated with a 0.100 M solution of KOH, a strong base.
What is the pH after 10.00 mL of KOH solution has been added?
25.00 mL of a 0.100 M solution of formic acid (HCOOH, Ka =
1.7 x 10-4) is titrated with a 0.100 M solution of KOH, a strong base.
What is the pH after 10.00 mL of KOH solution has been added?
Moles HCOOH = 0.02500 L (0.100 mol/L) = 2.50 x 10-3 mol
Moles NaOH = 0.01000 L (0.100 mol/L) 1.00 x 10-3 mol
Volume = 25.00 mL + 10.00 mL = 35.00 mL
HCOOH(aq) + NaOH(aq)  Na+(aq) + HCOO-(aq) + H2O(l)
Initial moles
Moles after
neutralization
Concentration after
neutralization
HCOOH
2.50 x 10-3
1.50 x 10-3
0.04286
HCOO-
0
1.00 x 10-3
0.02857
NaOH
1.00 x 10-3
0
0
We can now use the reaction for a weak acid in equilibrium with
its conjugate base.
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO-(aq)
Ka = [H3O+][HCOO-] = 1.7 x 10-4
[HCOOH]
After neutralization
Change
Equilibrium
HCOOH
0.04286
-x
0.04286 - x
COOH-
0.02857
x
0.02857 + x
x
x
H3O+
0
(0.02857 + x)(x) = 1.7 x 10-4 If we assume x << 0.028, then
(0.04286 - x)
x = (0.04286)(1.7 x 10-4) = 2.6 x 10-4
(0.02857)
pH = 3.59
Titration of a Weak Base With a Strong Acid
Consider the titration of 25.00 mL of a 0.0100 M solution of NH3
(a weak base; Kb = 1.8 x 10-5) with a 0.0100 M solution of HCl (a strong
acid).
The reactions that will occur are as follows:
NH3(aq) + HCl(aq)  NH4+(aq) + Cl-(aq) neutralization
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
if excess weak base
HCl(aq)  H+(aq) + Cl-(aq)
if excess acid
If there is excess NH3 then there will usually be a buffer solution
with both NH3 and NH4+ present. If there is excess acid then the pH will
be controlled by the concentration of excess strong acid.
As was the case for the titration of a weak acid with a strong
base, at the half equivalence point pH  pKa (this will be for NH4+, the
conjugate acid of NH3).
At the half-equivalence point (point where we have added half
the titrant needed to reach the equivalence point) pH  pKa for the
conjugate acid of the weak base. Notice that there will be a region where
the solution forms a buffer solution. Note that the pH at the equivalence
point is less than 7.0.
Indicator - chlorophenol red
Titration of a Polyprotic Acid With a Strong Base
Consider the titration of 1.000 L of a 1.000 M solution of a
diprotic acid (H2A) with NaOH, a strong base.
Solubility Product
We previously divided ionic compounds into two general
categories:
insoluble - does not dissolve in water
soluble - dissolves in water
However, “insoluble” ionic compounds usually dissolve in water
to a very small extent.
The solubility product, Ksp, is defined as the equilibrium constant
for the solubility reaction.
Examples:
AgCl AgCl(s)  Ag+(aq) + Cl-(aq)
CaF2
CaF2(s)  Ca2+(aq) + 2 F-(aq)
Ksp = [Ag+] [Cl-]
Ksp = [Ca2+] [F-]2
Since our reactant is a solid, equilibrium requires only that there
be some solid present in the system.
Ksp(Al(OH)3) = [Al3+][OH-]3 = 1.8 x 10-33
Ksp(CaF) = [Ca2+][F-]2 = 4.0 x 10-11
Ksp(AgBr) = [Ag+][Br-] = 7.7 x 10-13
Ksp(AgCl) = [Ag+][Cl-] = 1.6 x 10-6
Ksp(ZnS) = [Zn2+][S2-] = 3.0 x 10-23
The solid compound does not appear in the expression for Ksp, but must
be present in some amount for equilibrium for a saturated solution.
Solubility and Molar Solubility
There are two terms that are related to the solubility product.
Solubility (solubility by mass) - The number of grams of a
compound that will dissolve per 1.00 L of water.
Molar solubility - The number of moles of a compound that will
dissolve per 1.00 L of water.
Of course, our previous methods for indicating solubility (such as
grams per 100 grams solvent) are still valid.
Example: The solubility product for lead II chloride (PbCl2, MW
= 278.1 g/mol) is Ksp = 1.2 x 10-5 at T = 25. C. What are the solubility
by mass and the molar solubility of lead II chloride?
Example: The solubility product for lead II chloride (PbCl2, MW
= 278.1 g/mol) is Ksp = 1.2 x 10-5 at T = 25. C. What are the solubility
by mass and the molar solubility of lead II chloride?
PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
Ksp = [Pb2+] [Cl-]2 = 1.2 x 10-5
Substance
Initial
Change
Equilibrium
Pb2+
0.00
+x
x
Cl-
0.00
+ 2x
2x
(x) (2x)2 = 4x3 = 1.2 x 10-5
x3 = 3.0 x 10-6 ; x = (3.0 x 10-6)1/3 = 1.44 x 10-2
molar solubility = 1.44 x 10-2 moles/L
solubility = 1.44 x 10-2 mol 278.1 g = 4.01 g
1L
1 mol
L
Solubility and the Reaction Quotient
Consider a solution where only ions are present (no solid initially
present). As a specific example consider a 1:1 ionic compound MX
MX(s)  M+(aq) + X-(aq)
Qsp = [M+][X-]
There are three possibilities for Qsp, the reaction quotient
If Qsp < Ksp the solution is unsaturated (more solid could
dissolve). If there is no solid present, the solution is stable, but not at
equilibrium.
If Qsp = Ksp the solution is saturated (no more solid could
dissolve).
If Qsp > Ksp the solution is supersaturated (a precipitate will form
as the system goes to equilibrium).
Consider the solubility of lead II bromide, PbBr2.
PbBr2(s)  Pb2+(aq) + 2 Br-(aq)
Qsp = [Pb] [Br]2
Qsp < Ksp
Qsp = Ksp
Qsp > Ksp
unsaturated
saturated
supersaturated
“stable”
equilibrium
“unstable”
Factors Affecting Solubility
1) Temperature. By Le Chatlier’s principle, we may make the
following general statements
If the solubility reaction is exothermic (H < 0) then solubility
decreases as temperature increases.
AB(s)  A+(aq) + B-(aq) + “heat”
If the solubility reaction is endothermic (H > 0) then solubility
increases as temperature increases.
AB(s) + “heat”  A+(aq) + B-(aq)
2) pH. This will be true for hydroxide compounds (those
containing OH- ion) or compounds containing weak base anions (F-, CN-,
CO32-, S2-, PO33-, and other conjugate bases of weak acids).
Example: What is the maximum concentration of Ni2+ ion in a buffer
solution with pH = 10.00? Ksp(Ni(OH)2) = 5.5 x 10-16.
Example: What is the maximum concentration of Ni2+ ion in a buffer
solution with pH = 10.00? Ksp(Ni(OH)2) = 5.5 x 10-16.
Ni(OH)2  Ni2+(aq) + 2 OH-(aq)
Ksp = [Ni2+] [OH-]2 = 5.5 x 10-16
Since pH = 10.00, pOH = 14.00 - 10.00 = 4.00.
So [OH-] = 10-4.00 = 1.0 x 10-4 M
Ksp = [Ni2+] [OH-]2
[Ni2+] = Ksp
= 5.5 x 10-16 = 5.5 x 10-8 M
[OH-]2
(1.0 x 10-4)2
Note that it is possible for [Ni2+] to be  5.5 x 10-8 M.
3) Common ion effect. The common ion effect refers to the fact
that in equilibrium reactions all that matters are the concentrations of the
ions involved in the equilibrium. The source of the ions does not matter.
If a common ion is already present in solution this can affect the
solubility of a compound.
Example: What is the molar solubility of AgCl in a) pure water
and b) in a 0.100 M solution of NaCl. Ksp(AgCl) = 1.8 x 10-10.
Example: What is the molar solubility of AgCl in a) pure water
and b) in a 0.100 M solution of NaCl. Ksp(AgCl) = 1.8 x 10-10.
AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp = [Ag+] [Cl-] = 1.8 x 10-10
a) pure water
compound
initial
change
equilibrium
Ag+
0.00
+x
x
Cl-
0.00
+x
x
(x) (x) = x2 = 1.8 x 10-10
x = (1.8 x 10-10)1/2 = 1.34 x 10-5
molar solubility = 1.34 x 10-5 mol/L.
b) 0.100 M solution of NaCl
NaCl(s)  Na+(aq) + Cl-(aq)
compound
initial
change
equilibrium
Ag+
0.00
+x
x
Cl-
0.100
+x
0.100 + x
(x) (0.100 + x) = 1.8 x 10-10 Assume x << 0.100
(x) (0.100) = 1.8 x 10-10
x= (1.8 x 10-10) = 1.8 x 10-9
(0.100)
molar solubility = 1.8 x 10-9 mol/L.
This is much smaller than the molar solubility we found for silver
chloride in pure water (1.34 x 10-5 mol/L).
Selective Precipitation
We may use differences in solubility to remove specific cations
from a mixture by selective precipitation, the formation of a precipitate
with one ion in the mixture.
Example: Consider a solution containing 0.0100 M Mg2+ ion and
0.0100 M Cu2+ ion. Can we separate these ions?
Example: Consider a solution containing 0.0100 M Mg2+ ion and
0.0100 M Cu2+ ion. Can we separate these ions?
Consider adding OH- ion.
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
Ksp = [Mg2+] [OH-]2 = 5.6 x 10-12
Cu(OH)2(s)  Cu2+(aq) + 2 OH-(aq)
Ksp = [Cu2+] [OH-]2 = 1.6 x 10-19
Mg
[OH-] = {Ksp/[Mg2+]}1/2 = {5.6 x 10-12/0.01}1/2 = 2.4 x 10-5 M
Cu
[OH-] = {Ksp/[Cu2+]}1/2 = {1.6 x 10-19/0.01}1/2 = 4.0 x 10-9 M
So if OH- is added to the solution Cu(OH)2 will begin to
precipitate when [OH-] reaches 4.0 x 10-9. No Mg(OH)2 will precipitate
until [OH-] reaches 2.4 x 10-5 M. So we can separate most of the Cu2+
ion out of solution.
Complex Ion Formation
Some metal cations will react with small molecules or ions (NH3,
CN-, OH-) to form complex ions. A complex ion is defined as a metal
cation bonded to one or more small molecules or ions.
For example
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
The equilibrium constant for the formation of a complex ion is
called the formation constant (or stability constant), Kf
Kf = [Ag(NH3)2+]
[Ag+] [NH3]2
Formation constants are usually large numbers (meaning complex
ion formation is favored when the reactants are present). We may do
equilibrium calculations in the same way we do with other systems at
equilibrium.
Kf(Cu(CN)42-) = [Cu(CN4)2-]
[Cu2+][CN-]4
Kf(CO(NH3)63+) = [CO(NH3)63+]
[Co3+][NH3]6
End of Chapter 17
Tony Cortino: Where did you learn to dance like that?
Pepper: Julliard. I wanted to be a research chemist, but my legs were too
long.
- dialogue from the movie Mafia!
17.4
Solubility Equilibria
Quantitative predictions about how much of a given ionic compound will dissolve
in water is possible with the solubility product constant, Ksp.
AgCl(s) ⇌ Ag+(aq) +
Cl–(aq)
Ksp = [Ag+][Cl–]
Compound
Dissolution Equilibrium
Ksp
Al(OH)3(s) ⇌ Al3+(aq) + 3OH–(aq)
1.8 x 10–33
Calcium fluoride
CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
4.0 x 10–11
Silver bromide
AgBr(s) ⇌ Ag+(aq) + Br–(aq)
7.7 x 10–13
Silver chloride
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
1.6 x 10–6
Zinc sulfide
ZnS(s) ⇌ Zn2+(aq) + S2–(aq)
3.0 x 10–23
Aluminum hydroxide