The Solubility Product Constant, K sp

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Solubility
Products
Heterogeneous Equilibria
Slightly Soluble Salts
Prentice-Hall ©2002
Chapter Sixteen
Slide 1 of 32
Solubility of Ionic Salts
The solubility of ionic
salts varies
considerably. Most
nitrates are very
soluble, while
phosphates and may
other salts of
transition metal ions
are often insoluble.
Slide 2 of 36
Solubility and Temperature
The solubility of ionic
salts varies
considerably with the
temperature. Some
salts such as KNO3
have a wide solubility
range. On the other
hand, the solubility of
NaCl in water varies
only slightly.
Slide 3 of 36
The Solubility Product Constant, Ksp


Many important ionic compounds are only slightly
soluble in water and equations are written to
represent the equilibrium between the compound
and the ions present in a saturated aqueous solution.
The solubility product constant, Ksp, is the product
of the concentrations of the ions involved in a
solubility equilibrium, each raised to a power equal to
the stoichiometric coefficient of that ion in the
chemical equation for the equilibrium.
Slide 4 of 36
The Solubility Equilibrium
Equation And Ksp
Example 1
CaF2 (s)  Ca2+ (aq) + 2 F- (aq)
Ksp = [Ca2+][F-]2
Ksp = 5.3x10-9
Example 2
As2S3 (s)  2 As3+ (aq) + 3 S2- (aq)
Ksp = [As3+]2[S2-]3
Ksp = 4.4 x10-27
Slide 5 of 36
Some Values For Solubility Product
Constants (Ksp) At 25 oC
Slide 6 of 36
Ksp And Molar Solubility


The solubility product constant is related to the solubility
of an ionic solute, but Ksp and molar solubility - the
molarity of a solute in a saturated aqueous solution - are
not the same thing.
Calculating solubility equilibria fall into two categories:
determining a value of Ksp from experimental data
 calculating equilibrium concentrations when Ksp is
known.

Slide 7 of 36
Calculating Ksp From Molar Solubility
It is found that 1.2x10-3 mol of lead (II) iodide, PbI2,
dissolves in 1.0 L of aqueous solution at 25 oC. What is the
Ksp at this temperature?
PbI2 (s)  Pb2+ (aq) + 2 I- (aq)
Ksp = [Pb2+] [I-]2
For every lead iodide that dissolves there is one lead ion
and 2 iodide ions. Therefore
[Pb2+] = 1.2x10-3
[I-] = 2.4x10-3
Ksp =(1.2x10-3)(2.4x10-3)2 = 6.9 x 10-9
Slide 8 of 36
Calculating Molar Solubility From Ksp
Calculate the concentration of [Ag+] and [Cl-] in a
saturated solution of silver chloride. The ksp for silver
chloride : Ksp = 1.8x10-10
Ksp = [Ag+] [Cl-] = 1.8x10-10
Let x = [Ag+] = [Cl-] .
Substitute into the Ksp expression
x2 = 1.8x10-10 x = 1.8x10-10 =1.34 x 10-5
[Ag+] = [Cl-] =1.34 x 10-5
Slide 9 of 36
Calculating Molar Solubility From Ksp
Calculate the molar solubility of silver chromate,
Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4 (s)  2 Ag+ + CrO42Ksp = [Ag+]2 [CrO42-] = 1.1x10-12
Let x = [CrO42-] 2x = [Ag+]
Substitute into the Ksp expression
4x3 = 1.1x10-12 x = 3 1.1x10-12 /4
x = [CrO42-] = 6.5 x10-5
2x = [Ag+]
= 1.30 x 10-4
Slide 10 of 36
The Common Ion Effect In Solubility Equilibria




The common ion effect also affects solubility equilibria.
Le Châtelier’s principle is followed for the shift in
concentration of products and reactants upon addition
of either more products or more reactants to a
solution.
The solubility of a slightly soluble ionic compound is
lowered when a second solute that furnishes a
common ion is added to the solution.
The common ion effect is frequently used in analytical
chemistry to determine such things as the percent of
an element present in an unknown sample.
Slide 11 of 36
Solubility Equilibrium Calculation
-The Common Ion Effect
What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?
Ksp = 1.1x10-12 for Ag2CrO4.
Let 2x = Ag+ and CrO42- = 10-1
(4x2)(10-1) = 1.1 x 10-12
4x2 = 1.1 x 10-12/ 10-1 = 1.1 x 10-11
[Ag+] = 2x = 3.32 x 10-6
x = 1.66 x 10-6
Comparison of solubility of Ag2CrO4
In pure water:
6.5 x 10-5 M
In 0.10 M K2CrO4:
1.7 x 10-6 M
The common ion effect!!
Slide 12 of 36
Determining Whether Precipitation
Occurs



Qip is the ion product reaction quotient and is based on
initial conditions of the reaction.
Qip can then be compared to Ksp.
To predict if a precipitation occurs:
- Precipitation should occur if Qip > Ksp.
- Precipitation cannot occur if Qip < Ksp.
- A solution is just saturated if Qip = Ksp.
Slide 13 of 36
Determining Whether Precipitation
Occurs – Example 1
How many grams of KI should be dissolved in 10.0
dm-3 of 0.00200 M Pb(NO3)2 solution so that
precipitation of PbI2 (s) will just begin?
PbI2  Pb2+ (aq) + 2 I- (aq)
Ksp = [Pb2+] [I-]2 = 7.1 x 10-9
[Pb2+] = 2.00 x 10-3 so [I-]2 = (7.1 x 10-9)/(2.0 x 10-3)
[I-]2 = 3.55 x 10-6
[I-] = 1.88 x 10-3 mol dm-3
(1.88 x 10-3 mol dm-3) (10.0 dm3) = 1.88 x10-2 mol
(1.88 x10-2 mol) (166.00 g mol-1KI) = 3.12 g
Slide 14 of 36
Determining Whether Precipitation Occurs
–Example 2
The concentration of calcium ion in blood plasma is 0.0025
M. If the concentration of oxalate ion is 1.0x10-7 M, do you
expect calcium oxalate to precipitate? Ksp = 2.3x10-9.
CaC2O4  Ca2+ (aq) + C2O42- (aq)
[Ca2+ ] [C2O42- ] = 2.3x10-9
Three steps:
(1) Determine the initial concentrations of ions. (Put both in the Ksp
formula)
(2) Evaluate the reaction quotient Qip.
(3) Compare Qip with Ksp. If Qip > Ksp then it will precipitate
Slide 15 of 36
Determining Whether Precipitation
Occurs – Another Example
In applying the precipitation criteria, the effect of
dilution when solutions are mixed must be
considered.
Example:
A 250.0 mL sample of 0.0012 M Pb(NO3)2 (aq) is
mixed with 150.0 mL of 0.0640 M NaI (aq). Should
precipitation of PbI2 (s), Ksp = 7.1x10-9, occur?
Slide 16 of 36
Determining Whether Precipitation Is Complete


A slightly soluble solid never totally precipitates from
solution, but we generally consider precipitation to be
essentially complete if about 99.9% of the target ion is
precipitated and only 0.1% or less is left in solution.
Three conditions that generally favor completeness of
precipitation are:
A very small value of Ksp.
 A high initial concentration of the target ion.
 A concentration of common ion that greatly
exceeds that of the target ion.

Slide 17 of 36
Determining Whether Precipitation Is Complete An Example
A 0.50 L solution of 0.0010 M BaCl2 is added to 0.50 L
solution of 0.00010 M Na2SO4. Will the precipitation of
SO42- as BaSO4 (s) be complete? Ksp = 1.1x10-10.
Slide 18 of 36
Selective Precipitation
a) The first precipitate to
form when AgNO3(aq)
is added to an aqueous
solution containing Cland I- is yellow AgI(s).
b) Essentially all the I- has
precipitated before the
precipitation of white
AgCl(s) begins.
Slide 19 of 36
Selective Precipitation An Example
Example 16.10
An aqueous solution that is 2.00 M in AgNO3 is slowly
added from a buret to an aqueous solution that is
0.0100 M in Cl- and also 0.0100 M in IAgCl (s)  Ag+ (aq) + Cl- (aq)
a.
b.
c.
Ksp = 1.8x10-10
AgI (s)  Ag+ (aq) + I- (aq)
Ksp = 8.5x10-17
Which ion. Cl- or I-, is the first to precipitate from
solution?
When the second ion begins to precipitate, what is the
remaining concentration of the first ion?
Is separation of the two ions by selective precipitation
feasible?
Slide 20 of 36
Effect of pH on Solubility
The solubility of an ionic solute may be greatly
affected by pH if an acid-base reaction also
occurs as the solute dissolves.
 If the anion of the precipitate is that of a weak
acid, the precipitate will dissolve somewhat
when the pH is lowered; if, however, the anion
of the precipitate is that of a strong acid,
lowering the pH will have no effect on the
precipitate.
 Since CO2 is lost from the reaction of an
insoluble carbonate with acid, this concept
does not apply.
Slide 21 of 36

Effect of pH on Solubility - An Example
Determine the molar solubility of Fe(OH)3 in
pH = 2.70 buffer. Ksp = 4.0x10-38 for Fe(OH)3.
Slide 22 of 36
Qualitative Inorganic Analysis



Acid-base chemistry, precipitation reactions, oxidationreduction, and complex-ion formation all come into
sharp focus in an area of analytical chemistry called
classical qualitative inorganic analysis.
“Qualitative” signifies that the interest is in determining
what is present, not how much is present.
Although classical qualitative analysis is not as widely
used today as instrumental methods, it is still a good
vehicle for applying all the basic concepts of equilibria
in aqueous solutions.
Slide 23 of 36
Slide 24 of 36
Cations of Group 1
If aqueous HCl is added to an unknown solution
of cations, and a precipitate forms, then the
unknown contains one or more of these cations:
Pb2+, Hg22+, or Ag+.
 These are the only ions to form insoluble
chlorides.
 If there is no precipitate, then these ions must be
absent from the mixture.
 If there is a precipitate, it is filtered off and saved
for further analysis.
 The supernatant liquid is also saved for further
analysis.
Slide 25 of 36

Cation Group 1 (continued)
Analyzing For Pb2+




Of the three possible ions in solution, PbCl2 is the most
soluble in water.
The precipitate is washed with hot water and the
washings then treated with aqueous K2CrO4.
If Pb2+ is present, chromate ion combines with lead ion
to form a precipitate of yellow lead chromate, which is
less soluble than PbCl2.
If Pb2+ is absent, then the washings just become tinged
yellow but no precipitate is in evidence.
Slide 26 of 36
Cation Group 1 (continued)
Analyzing For Ag+





Next, the undissolved precipitate is treated with
aqueous ammonia.
If AgCl is present, it will dissolve in this solution.
If there is any remaining precipitate, it is separated
from the supernatant liquid and saved for further
analysis.
The supernatant liquid (which contains the Ag+, if
present) is then treated with aqueous nitric acid.
If a precipitate reforms, then Ag+ was present in the
solution, if no precipitate forms, then Ag+ was not
present in the solution.
Slide 27 of 36
Cation Group 1 (continued)
Analyzing For Hg22+
When precipitate was treated with aqueous
ammonia in the previous step, any Hg22+
underwent an oxidation-reduction reaction to
form a dark gray mixture of elemental mercury
and HgNH2Cl that precipitates from the
solution.
 If this dark gray precipitate was observed,
then mercury was present in the original
unknown sample.
 If this dark gray precipitate was not observed,
then mercury must have been absent from the
Slide 28 of 36
original unknown sample.

Group 1 Cation Precipitates
left: cation goup 1 ppt: PbCl2,
PbCl2, AgCl (all white)
middle: product from test for
Hg22+: mix of Hg (black) and
HgNH2Cl (white)
right: product from test for Pb2+:
PbCrO4 (yellow) when
K2CrO4(aq) is reacted with
saturated PbCl2
Slide 29 of 36
Slide 30 of 36
Hydrogen Sulfide In The
Qualitative Analysis Scheme
H2S is a weak diprotic acid; there is very little ionization of
the HS- ion and it is the precipitating agent.
H2S (aq) + H2O  H3O+ (aq) + HS- (aq)
Ka1 = 1.0x10-7
HS- (aq) + H2O  H3O+ (aq) + S2- (aq)
Ka2 = 1.0x10-19
H2S (aq) + H2O  H3O+ (aq) + HS- (aq)
 MS (s) + H3O+ (aq)
M2+ (aq) + H2S (aq) + 2 H2O (l)  MS (s) + 3 H3O+
M2+ (aq) + HS- (aq) + H2O (l)
Overall:
(aq)
Slide 31 of 36
Cation Groups 2, 3, 4 And 5




The concentration of HS- is so low in a strongly acidic
solution, that only the most insoluble sulfides
precipitate.
These include the eight metal sulfides of Group 2.
Of the eight cations in Group 3, five form sulfides that
are soluble in acidic solution but insoluble in an
alkaline ammonia/ammonium chloride buffer solution.
The other three group 3 cation form hydroxide
precipitates in the alkaline solution.
The cations of groups 4 and 5 form soluble sulfides,
even in basic solution.
Slide 32 of 36
Slide 33 of 36
Cation Groups 2, 3, 4 And 5
The hydroxides of Groups 4 and 5, with the
exception of Mg2+, are also moderately or highly
soluble.
 The group 4 cations are precipitated as
carbonates from a buffered alkaline solution.
 The cations of group 5 remain soluble in the
presence of all common reagents.
 Within each of the qualitative analysis groups,
additional reactions to dissolve group
precipitates and to separate and selectively
precipitate individual cations for identification
and confirmation are used.
Slide 34 of 36

Summary
The solubility product constant, Ksp, represents
equilibrium between a slightly soluble ionic
compound and its ions in a saturated aqueous
solution.
 The common ion effect is responsible for the
reduction in solubility of a slightly soluble ionic
compound.
 Precipitation is assumed to be complete if no
more than 0.1% of the target ion remains in
solution.
 The solubilities of some slightly soluble
compounds depends strongly on pH. Slide 35 of 36

Summary

Precipitation, acid-base, and oxidation-reduction
reactions, together with complex-ion formation, are all
used extensively in the classical scheme for the
qualitative analysis of common cations.
Slide 36 of 36
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