Copyright Sautter 2003
SOLUBILITY EQUILIBRIUM
• Solubility refers to the ability of a substance to dissolve. In
the study of solubility equilibrium we generally deal with
low solubility compounds (those which dissolve poorly)
• Low solubility compounds are classified as precipitates. A
set of solubility rules indicate which combination of ions
generally form precipitates. These rules allow us to
identify low solubility substances and potential
precipitates. The concentrations of the ions present
however are the final determiners of precipitate formation.
SOLUBILITY RULES
• (1) Inorganic acids (those not consisting of carbon, hydrogen and
oxygen) are soluble. Low molecular weight organic acids are soluble.
• (2) Compound containing alkali metals and ammonium (NH4+)
cations are soluble.
• (3) Compounds containing nitrates (NO3-), acetates (C2H3O2-) and
perchlorates (ClO4-) are soluble.
• (4) Compounds containing chlorides (Cl-) are soluble except for AgCl,
PbCl2 and Hg2Cl2. Compounds containing bromides (Br-) and iodides
(I-) have solubility similar to chlorides.
• (5) Sulfate containing compounds are soluble except for PbSO4 and
BaSO4. CaSO4 and Ag2SO4 are slightly soluble.
• (6) Hydroxides are insoluble (precipitates) except Alkali Metal (I) and
heavier Alkaline Earth Metal (II) hydroxides after Mg.
• (7) Carbonates (CO3-2), phosphates (PO4-3) and arsenates (AsO4-3) are
insoluble except for Alkali Metal (I) and Alkaline Earth Metal (II)
compounds
• (8) Sulfides (S-2) are insoluble except for Alkali Metal (I) and Alkaline
Earth Metal (II) compounds.
ALL LOW SOLUBILITY SYSTEMS ARE
EQUILIBRIUM SYSTEMS
RATE OF DISSOLVING = RATE OF CRYSTALIZATION
RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS
The amount of substance dissolved remains constant
The system appears macroscopically static.
Ksp AND SOLUBILITY
• Since solubility is an equilibrium phenomena, equilibrium
concepts can readily be applied.
• If the system is at equilibrium, the solution must be saturated (no
more solid can be dissolved). Equilibrium concepts cannot be
applied to unsaturated solutions (more solid can still be dissolved)
since equilibrium has not yet been reached.
• For a typical saturated solution of silver chloride,
AgCl(s)  Ag+(aq) + Cl-(aq)
the equilibrium expression is:
K = [Ag+] x [Cl-] / 1 or simply [Ag+] x [Cl-]
Since AgCl is and solid it is not included in the equilibrium
expression!
•
This K value is called the Ksp . “sp” stands for solubility
product. The product of the concentrations of the soluble ions.
CALCULATING Ksp
• Problem: A saturated solution of BaSO4 contains 0.0025
grams in one liter. Calculate the Ksp for barium sulfate.
• Solution: Since Ksp requires concentrations is moles per
liter we must first calculate the number of moles present.
Moles = 0.0025 gram / 233 grams per mole = 1.1 x 10-5 M
• BaSO4(s)  Ba+2(aq) + SO4-2(aq)
• Each dissolved BaSO4 forms one Ba+2 ion and one SO4-2
ion, therefore:
• [BaSO4] = 1.1 x 10-5 M, [Ba+2] = [SO4-2] = 1.1 x10-5 M
• Ksp = [Ba+2] x [SO4-2] = (1.1 x 10-5)2 = 1.2 x10 -10
SOLUBILITY
PRODUCTS
OF
SOME
COMMON
SALTS
AT
25 0 C
Calculating Concentrations From Ksp
• Problem: Find the concentration of dissolved Al(OH)3
in a saturated solution.
• Solution: Al(OH)3 (s)  Al+3(aq) + 3 OH-(aq)
• Ksp = [Al+3] x [OH-]3 , Ksp = 1.9 x 10-33
• Each dissolved Al(OH)3 gives one Al+3 and three OHions therefore if [Al+3] = X, [OH-] = 3X and
• Ksp = X (3X)3 = 27 X4 = 3.0 x 10-34
• X = (3.0 x 10-34 / 27)1/4 = 1.8 x 10-9 M
THE COMMON ION EFFECT
• The common ion effect is an application of
LeChatelier’s Principle. Recall that equilibrium
shifts are caused be stresses applied to the system.
• If additional products are added to an equilibrium
system that system will shift towards the reactants.
In a solubility equilibrium adding one of the
dissolved ion components will cause a shift left and
the solubility of the compound will decrease.
• Reducing the concentration of one of the dissolved
ions will cause a shift to the right thereby allowing
the compound to dissolve to a greater extend.
THE COMMON ION EFFECT
Add NaCl to the system
System shifts left!
THE COMMON ION EFFECT
• Problem: Calculate the solubility of silver chloride in pure water and
in a 0.10 M NaCl solution. Compare the results.
• Solution: AgCl(s)  Ag+(aq) + Cl-(aq)
• Ksp = [Ag+] x [Cl-] = 1.8 x 10-10
• Each dissolved AgCl gives one Ag+ and one Cl- ion therefore if
[Ag+] = X, [Cl-] = X
• Ksp = X2 = 1.8 x 10-10, X = (1.8 x 10-10)1/2 = 1.3 x 10-5 M
• In 0.10 M NaCl the concentration of Cl- = 0.10 M and
• Ksp = [Ag+] (0.10)* = 1.8 x 10-10, [Ag+] = 1.8 x 10-10 / 0.10
• ]Ag+] = 1.8 x 10-9 M (the presence of the common Cl- ion in
solution has shifted the equilibrium hard to the left and made the
silver chloride much less soluble than in pure water
• *Note: the Cl- from the AgCl in the common ion system is so small
as compared to the 0.10 M Cl- from the NaCl that is was neglected.
WILL A PRECIPITATE FORM?
• In order to decide whether a precipitate will form
when two solutions are mixed we must first
calculate the value of Q (reaction quotient). Q is
calculated with the same format as Ksp (an ion
concentration product)
• If the value of Q exceeds the Ksp value, a precipitate
will form.
• If the value of Q is less than Ksp, a precipitate will
not be formed.
• If Q = Ksp then the solution is a saturated solution.
WILL A PRECIPITATE FORM?
WILL A PRECIPITATE FORM?
0.0001 M AgNO3
100 ml
0.00002 M NaCl
200 ml
WILL A PRECIPITATE FORM?
• Problem: If 100 ml of 0.0001 M AgNO3 is mixed with 200 ml of
0.0002 M NaCl, will a precipitate form?
• Solution: The potential precipitate is AgCl. (See the solubility
rules).
• Q = [Ag+] x [Cl-], when the two solutions are mixed, the
concentrations of both the Ag+ ion and the Cl- ion will change due
to dilution of one solution by the other.
•
M1V1= M2V2 (the dilution formula)
• For Ag+ , 0.0001 (100) = [Ag+] (100 + 200)
• [Ag+] = 3.33 x 10-5 M
• For Cl- , 0.0002 (100) = [Cl-] (100 + 200)
• [Cl-] = 6.67 x 10-5 M
• Q = (3.33 x 10-5) (6.67 x 10-5) = 2.22 x 10-9, Ksp = 1.8 x 10-10
• Q > Ksp therefore a precipitate will form
DISSOLVING PRECIPITATES
• Several methods of dissolving low solubility materials can be used.
One involves the use of weak acid conjugate bases. Systems that can
be affected by these changes are said to be pH sensitive. Another
method involves the formation of complex ions from one of the
dissolved ions.
• Recall that a weak acid is an equilibrium system in which the acid
dissociates poorly and its conjugate base readily forms the molecular
acid. For example in hydrosulfuric acid H2S(aq)  2 H+(aq) + S-2(aq)
sulfide ion (S-2) will readily be consumed by H+ ion to form molecular
hydrosulfuric acid (H2S).
• Precipitates containing sulfide ion are pH sensitive since added H+ ion
will cause a shift towards the molecular acid thereby consuming S-2
ion and allow further dissolving of the precipitate.
DISSOLVING PRECIPITATES
ADD AN ACID (pH IS LOWERED)
SOLUBILITY OF PbS INCREASES
AS A RESULT OF DECREASED pH
DISSOLVING PRECIPITATES
ADDING A BASE (OH-) RAISES
PH AND EQUILIBRIUM SHIFTS
LEFT MAKING Al(OH)3
LESS SOLUBLE
COMPLEX ION FORMATION & SOLUBILITY
• Complex ions are formed when simple ions are
combined with other ions to form new, more complex
ions. Also simple ions may combine with molecules
in solution which when attached to the ion are called
ligands.
• Examples of complex ion formation include:
• Ag+(aq) + 2 NH3 (aq)  Ag(NH3)2+(aq)
• Fe+3(aq) + SCN-(aq)  FeSCN+2(aq)
• Cu+2(aq) + 4 NH3(aq)  Cu(NH3)4+2(aq)
• Complex ion formation reactions generally have large
K values and occur readily in solution.
COMPLEX ION FORMATION & SOLUBILITY
NH3 is
added
AgCl dissolves well in ammonia (NH3)
Solubility due to equilibrium shift caused
by the complex formation
COMPLEX ION FORMATION & SOLUBILITY
• Problem: What is the solubility of AgCl in 0.10 M NH3 solution?
• Solution: Equ (1) AgCl(s)  Ag+(aq) + Cl-(aq) and
Equ (2) Ag+(aq) + 2 NH3 (aq)  Ag(NH3)2+(aq)
• Adding equ (1) & (2): AgCl(s)+ 2 NH3 (aq)  Ag(NH3)2+(aq) + Cl-(aq)
Ksp AgCl = 1.8 x 10-10, Kf for the complex formation is 1.6 x 107
• K(equ 1 + equ 2) = Ksp x Kf = (1.8 x 10-10)(1.6 x 107) = 2.72 x 10-3
K = ([Ag(NH3)2+] [Cl-]) / [NH3] 2 = 2.72 x 10-3
• Let X = [Ag(NH3)2+]e , and since [Ag(NH3)2+]e = [Cl-]e and two NH3
are used to form one Ag(NH3)2+
[Ag(NH3)2+]e = X, [Cl-]e = X and [NH3]e = (0.10 – 2X)
X2 / (0.10 – 2X)2 = 2.72 x 10-3 and taking the square root of both
sides of the equation we get: X / (0.10 – 2X) = (2.72 x 10-3)1/2
• X = [Ag(NH3)2+]e = 4.7 x 10-3 M (the solubility of AgCl in 0.10 M
NH3)
SOLUBILITY AND LABORATORY ANALYSIS
• A common laboratory analysis involves the identification
of unknown components of a solution. The procedure is
called Qualitative Analysis meaning that just the identity
of the components is sought not the amounts presence.
• An analysis of a solution may often require the
determination of the presence of silver, lead or mercury.
These three ions are referred to a Group I cations. In the
case of Qualitative Analysis, Group I does not pertain to
column I on the periodic table (the alkali metals). It
refers only to Ag+, Pb+2 and Hg2+2 ions.
• The procedures involved in Group I analysis rely on the
solubility characteristic of the three ions of the group.
SOLUBILITY AND LABORATORY ANALYSIS
• A common property of all three Group I ions is that they
form precipitates with chloride ion.
Ag+(aq) + Cl-(aq)  AgCl(s)
Pb+2(aq) + Cl -(aq)  PbCl2(s)
Hg2+2(aq) + Cl- (aq)  Hg2Cl2(s)
• In the second step of the analysis the temperature
dependent properties of the precipitates allows separation
of the three solids. Lead chloride is soluble in hot water
while the solubility of the other solids are independent of
temperature.(Solubility is temperature dependent for many
precipitates. Some are unaffected by temperature, some
dissolve better at higher temperatures and some dissolve
more poorly at elevated temperatures)
GROUP I ANALYSIS
Step I – Chloride Precipitation
Cl- (chloride ion)
Ag+
Pb+2
Hg2+2
PRECIPITATES
AgCl
PbCl2
Hg2Cl2
SOLUBILITY CURVES FOR VARIOUS SOLIDS
GROUP I ANALYSIS
Step II – Precipitate Separation
Hot Water
Lead Chloride Dissolves
PbCl2(s)  Pb+2(aq) + 2 Cl-(aq)
PRECIPITATES
AgCl
PbCl2
Hg2Cl2
SOLUBILITY AND LABORATORY ANALYSIS
• The formation of complex ions is used to separate
AgCl from the remaining Hg2Cl2 precipitate. Recall
that Ag+ ion forms the complex Ag(NH3)2+ and
therefore the addition of aqueous ammonia solution
will dissolve AgCl solid leaving the mercury
precipitate behind.
• A confirmatory test for the Ag+ involves change the
pH of the solution in which the silver complex is
dissolved. Adding nitric acid neutralizes the
ammonia ligands on the complex and the AgCl solid
again precipitates as a white solid.
GROUP I ANALYSIS
Step III – Complex Ion Formation
Ammonia Solution (NH3)
PRECIPITATES
AgCl
Hg2Cl2
AgCl(s)
Silver Chloride Dissolves
+ 2 NH3(aq) Ag(NH3 )2 +(aq) + Cl-(aq)
GROUP I ANALYSIS
Step IIII – Precipitation by pH Change
Nitric Acid (HNO3)
COMPLEX ION
SOLUTION
Ag(NH3)2+
Silver Chloride Precipitates
Ag(NH3 )2+(aq) + Cl-(aq) + H+(aq) AgCl(s) + NH4+(aq)