14.4 (Chp16) Solubility and
Solubility Product
What drives substances to dissolve and others to
remain as a precipitate?
Dr. Fred Omega Garces
Chemistry 201
Miramar College
1
Solubility and Solubility Product
March 14
When a Substance Dissolve
When a reaction occurs in which an insoluble product is
produced, the Keq is called a solubility-product.
BaSO4
(s)
D
Ba+2 + SO42-
Keq = Ksp = [Ba+2] [SO42-]
(aq)
Do not confuse
solubility (s)
with solubility
product Ksp.
How water dissolves ionic compounds
Solubility depends on Temp., conc., and pH
Solubility Product depends on Temp. Only !!!
2
Solubility and Solubility Product
March 14
Solubility
Solubility: What is the meaning of solubility ?
i.
The ability of substance to dissolve in a solvent.
ii. Quantity that dissolves to form a saturated solution.
s
g per 100 mL or g /L
g
(molar solubility, grams per liter saturated solution.)
g • mol
g
g
mol g Molarity
The greater the
solubility, the smaller
the amount of
precipitation.
3
Solubility and Solubility Product
March 14
Factors Affecting Solubility (s).
1 Nature of Solute (Concentration). - Like dissolves like (IMF)
2 Temperature Factor i) Solids/Liquids- Solubility increases with Temperature (mostly)
Increase K.E. increases motion and collision between solute / solvent.
ii) gas - Solubility decreases with Temperature
Increase K.E. result in gas escaping to atmosphere.
3 Pressure Factor i) Solids/Liquids - Very little effect
Solids and Liquids are already lose together, extra pressure will not
increase solubility.
ii) Gas- Solubility increases with Pressure.
Increase pressure squeezes gas solute into solvent.
4
Solubility and Solubility Product
March 14
In a Chemical Reaction For a reaction to take place Reactants should be soluble in solventThis ensures that reactants combine (come in contact) with each
other efficiently.
When reaction proceedsProducts which are formed have different properties than that of
the reactants. One such property is the solubility.
When a precipitate forms upon mixing two solution it
is a result of a new specie that is present in the
solution.
5
Solubility and Solubility Product
March 14
Precipitation Reaction
Consider the reaction i AgNO 3 (aq) + HCl
(aq)
D
AgCl(s) + HNO3(aq)
(aq)
D
AgCl
Net ionic equation:
Ag+(aq) + Cl-
If written in this
form, then mass
action is:
(s)
Write the reverse, (standard convention)
AgCl
Then
(s)
D
Ag+(aq) + Cl-
1
K eq =
[Ag + ][Cl− ]
(aq)
Keq = Ksp = [Ag+] [Cl-]
Ksp - solubility product constant
An indicator of the solubility of substance of interest.
€
Yields info. on the amount of ions allowed in solution.
(Must be determine experimentally)
NaOCH2CH3 (s) + H2O (l) D Na+ (aq) + OH- (aq) + CH3CH2OH (aq)
6
Solubility and Solubility Product
Ksp = [Na+] [OH-] [CH3CH2OH]
March 14
Solubility Equilibrium: Example
What is the solubility product for the following reaction:
Ag3PO4
D 3Ag+ (aq) + PO43- (aq)
[s] = 4.4•10-5M
i
Excess
Δ
- 4.4•10-5M
3(4.4•10-5M )
Solid
1.32•10-4M
e
0
0
+ 4.4•10-5M
4.4•10-5 M
Solubility; Ksp = [1.32•10-4]3 • [4.4 •10-5] = 1.01 •10 -16
No direct proportionality of Ksp to solubility [s]
Ksp e
# ions in solution
but in general the greater the solubility (s), the less the amount of precipitate (solid) in solution
Unlike in homogeneous equilibrium, Ksp isn’t a good indicator of
a substance solubility. The solubility is ultimately determine by
the solving the equilibrium problem for solubility (s).
8
Solubility and Solubility Product
March 14
Solubility from Solubility Chart
The diagram below shows the solubility of some salts as a function of temperature.
This graph can be used to determine the Ksp for any one of the salts shown. Consider
KNO3, the solubility of KNO3 at 20°C according to the graph is approximately 30 g per
100ml H2O. The strategy is to find the molar concentration and then use this
information to determine the Ksp for the salt at the specified temperature.
The solubility of KNO3 at 20°C is 30 g per
100ml H2O. What is the Ksp for KNO3 at
this temperature?
MW KNO3 = 101.11 g/mol
mol
mol KNO3 = 30.0g ∗
= 0.297mol
101.11g
Assume solution has ρ = 1.00 g/cc
Mass solution = 100g H2O + 30g KNO3
= 130 g Solution
130 g solution = 0.130 L since ρ = 1.00 g/cc
[s] =
0.297mol
0.130L
= 2.285M
-
K sp = [K ][NO3 ] = [2.285]2 = 5.22
+
€
10
Solubility and Solubility Product
March 14
Solubility Vs. Ksp
Consider ;
La (IO3)3
Ksp = 6.2 • 10-12
<
s = 6.9•10-4
<
Ba(IO3)2
Ksp = 1.5 • 10-9
<
s = 7.2•10-4 >
AgIO3
Ksp = 3.1 •10-8
s
= 1.8•10-4
In general there is no direct proportionality of Ksp to
solubility because Ksp depends on the number of ions in
solution.
But if same number of ions form between two chemicals,
then greater the solubility, the greater the degree that a
substance dissolves in solution (less precipitate).
12
Solubility and Solubility Product
March 14
Solubility Rules: What is it used for?
How can one predict if a precipitate forms when mixing solutions?
Solubility rules Soluble Substances
containing
Exceptions
Insoluble substances
containing
Exceptions
nitrates, (NO3-)
chlorates (ClO3-)
perchlorates (ClO4-)
acetates (CH3COO-)
None
carbonates (CO32-)
phosphates (PO43-)
Slightly soluble
halogens (X-)
X- = Cl-, Br-, I-
Ag, Hg, Pb
hydroxides (OH-)
alkali, NH4+, Ca,
Sr, Ba
sulfates (SO42-)
Ba, Hg, Pb
alkali & NH4+
None
Solubility rule provides information on the specie that will form a precipitate.
In general, a molar solubility of 0.01M or greater for a substance is
considered soluble.
http://www.ausetute.com.au/solrules.html
13
Solubility and Solubility Product
March 14
Precipitation Experiment
Consider the reaction:
3 Ca2+ + 2PO43-
(aq)
D Ca3(PO4)2
(s)
According to the solubility rules, this should form a precipitate (ppt) .
The question is what concentration is required before a precipitate forms ?
To solve,
It is always convenient to write the equation as a dissolution of solid
Ca3(PO4)2
(s)
D
3 Ca2+ + 2PO43-
Q = [Ca+2]3 • [PO43-]2
(aq)
next compare Q to Ksp
In solubility product problems,
Q is the ion-product instead of reaction quotient
14
Solubility and Solubility Product
March 14
Ion Product; Q
Q - indicator of direction of reaction for reaction not at equilb.
MX (s) D M+(aq) + X-(aq)
Direction
Ksp
Reaction
Q (i)
Unsaturated.
Q is small, system
consist mostly of ion.
No ppt. forms
(s) g (aq)
15
Q
Sat.
Point
Solubility and Solubility Product
Direction
Reaction
Q (h)
Supersaturated.
Q is large, system will
adjust to reduce high
ion concentration.
Solid forms
(s) f (aq)
March 14
In Class: Precipitation Determination
Question, (19.88 Sil): Will any solid Ba(IO3)2 form when 6.5 mg of BaCl2 is
dissolved in 0.500L of 0.033M NaIO3?
Ksp Ba(IO3)2 = 1.5 •10-9 , BaCl2 MW = 208.23 g/mol
Ba(IO3)2:
Ba(IO3)2
D
(s)
Ba2+(aq) + 2 IO3-
(aq)
[Ba+2] = 6.5 mg .... 6.2•10-5 M
[IO3-] = 3.3•10-2 M
Q = [Ba+2] • [IO3-]2 = 6.2•10-5 M • (3.3•10-2 M)2
Q = 6.8 •10-8 M
> Ksp
Precipitate forms.
17
Solubility and Solubility Product
March 14
Common Ion Effect
For a system containing a precipitate in equilibrium with its ions, solubility
of solid can be reduced if an ionic compound is dissolved in the solution.
Ag+
Ag+ I-
I-
AgI(s)
AgI(s)
Ag+ I-
Add Ivia NaI
AgI(s)
AgI(s) D Ag+(aq) + I-(aq)
add common ion i.e.,
NaI to solution
Direction of Reaction
Reduce solubility
i.e., less AgI will dissolve in soln’ which
means more ppt. forms in solution
18
Solubility and Solubility Product
March 14
...according to LeChatelier
AgI(s)
Ag+
I-
AgI(s) D Ag+ + IQ analogy:
AgI(s)
Ag+
AgI(s)
Q = [Ag+] [I-]
NaI(s)
Ag+
I-
Ksp < Q
I-
Ksp
More AgI forms and
Solubility is lowered
19
Increase I-
Q
Direction of Rxn
(ppt occur)
Solubility and Solubility Product
March 14
Factor Affecting Solubility: Solubility & Common Ion
Calculate the solubility of copper(I) iodide, CuI (Ksp = 1.1•10-12)
a) water
b) 0.05 M sodium iodide
a)
i
Δ
e
CuI (s) D
Lots
-s
Solid
Cu+(aq) + I- (aq)
0
0
+s
+s
+s
+s
a) Ksp = 1.1⋅10 -12 =[Cu+ ] [I - ] = s2
b) i
Δ
e
CuI (s) D Cu+(aq) + I- (aq)
Lots
-s
Solid
0
+s
+s
0.05
+s
0.05 + s
b) Ksp = 1.1⋅10 -12 =[Cu+ ] [I -]
= s ∗ (0.05M + s)
1.05⋅10 -6 = s
= s ∗ (0.05M)
2.0 ∗ 10 -11 = s
Without common ion, solubility is, s = 1.05 •10-6 M
€ is, s = 2.0 •10-11 M
but with common ion, solubility
22
Solubility and Solubility Product
March 14
Calculation: Solubility, Iteration Method
Reger Ex 12.19: What is the solubility of calcium fluoride in 0.025 M
sodium fluoride solution? Ksp = 6.2•10-8 M
i
Δ
e
CaF2 (s) D Ca+2(aq) + 2F- (aq)
Lots
0
0.025M
-s
+s
+2s
Solid
+s
0.025M +2s
Ksp = 6.2 ⋅ 10 -8 = [Ca +2 ] [F - ]2
assume s << 0.025
= s ∗ (0.025M + 2s)2
6.2 ⋅ 10 -8 = s ∗ (0.025M)2
6.2 ⋅ 10 -8
0.0252
9.92 ⋅10 -5
• 100 = 0.4%
0.025
= s = 9.92 ⋅ 10 -5 M
€
Check assumption
indeed 9.92•10-5 << 0.025
€
24
Solubility and Solubility Product
March 14
Calculation: Solubility of Two salts in same solution
At 50°C, the solubility products , Ksp, of PbSO4 and SrSO4 are
1.6•10-8 M and 2.8•10-7M, respectively. What are the values of
[SO42-], [Pb2+], and [Sr2+] in a solution at equilibrium with both
substances ?
PbSO4 (s) D Pb2+(aq) + SO42 -(aq) ; Ksp = 1.6•10-8 = [Pb2+][SO42-]
SrSO4 (s) D Sr+ (aq) + SO4 2-(aq) ; Ksp = 2.8•10-7 = [Sr2+][SO42-]
Let x = [Pb2+] , y = [Sr2+] , x + y = [SO42-]
x(x+y) = 1.6•10-8 = x
y(x+y) 2.8•10-7
y
= 0.0571 = 0.057 ; x = 0.057y
y(0.057y+y) = 2.8 •10-7 ; 1.057 y2~ = 2.8•10-7; y= 5.147•10-4 = 5.1•10-4
x = 0.057 y; x = 0.057 (5.1•10 -4 ) = x = 2.9 •10-5
[Pb2+] = 2.9•10-5 M, [Sr2+] = 5.1•10 -4 M, [SO42-]= 5.4•10 -4 M
26
Solubility and Solubility Product
March 14
Calculation: Same type problem different data
At 25°C, the solubility product of Zn(IO3)2 and Sr(IO3)2 is 3.9•10-6 M and
3.3•10-7M, respectively. What are the values of [IO3-], [Zn+2], and [Sr2+]
in a solution at equilibrium with both substances ?
Zn(IO3)2
i Lots
Δ -m
e Lots
(s)
D Zn2+ (aq)
0
+m
+m
+ 2IO3-2(aq)
0
+2m + 2n
+2m + 2n
Sr(IO3)2
i Lots
Δ -n
e Lots
-
[IO3 ] =
2
-
K sp
-
[IO4 ]2 =
[Zn2+ ]
+2
- 2
3.9 ⋅ 10-6 [Zn ][IO 3] [m][2m+2n]2 [m]
=
=
=
3.3 €
⋅ 10-7 [Sr +2][IO-3]2 [n][2m+2n- ]2 [n]
n ⋅ 1.1818 ⋅ 101 = m
+2
- 2
3.9 ⋅ 10-6 [Zn ][IO 3] [m][2m+2n]2 [m]
=
=
=
3.3 ⋅ 10-7 [Sr +2][IO-3]2 [n][2m+2n- ]2 [n]
D Sr +2(aq) + 2IO3- (aq)
0
0
+n
+2n + 2m
+n
+2n + 2m
K sp = [Sr+2 ] ∗ [IO3 ]2
K sp = [Zn2+ ] ∗ [IO3 ]2
-
(s)
...
K sp
[Sr+2 ]
€
Answer:
[Sr +2] = 7.9•10-4 M [Zn
[IO3- ] = 2.0 •10-2 M
+2]
= 9.4•10-3 M
n ⋅ 1.1818 ⋅ 101 = 11.818 ⋅ n = m
3.3 ⋅ 10-7 = [Sr +2][IO-3]2 = n[2m+2n]2 = n[2(11.818n)+2n]2
3.3 ⋅ 10-7 = n[(23.636n)+2n]2 = n[(25.636n]2 = 657.2n3
n=
27
3
3.3 ⋅ 10-7
657.2
= 7.948 ⋅ 10-4 , m = 11.818 ⋅ n =9.393 ⋅ 10-3
[Sr +2] = 7.948 ⋅ 10-4M, [Zn+2] = 9.393 ⋅ 10-3M, [Zn+2] = 2.038 ⋅ 10-2M
Solubility and Solubility Product
March 14
Calculation: Same type problem different data
At 25°C, the solubility product of AgI and PbI2 is 8.3•10-17 M and
7.9•10-9M, respectively. What are the values of [I-], [Ag+], and [Pb2+] in a
solution at equilibrium with both substances ? (Answer behind)
AgI
i Lots
Δ -s
e Lots
(s)
D Ag + (aq) +
0
+s
+s
I-(aq)
PbI2
i Lots
Δ -t
e Lots
2t
+s
+s+2t
K sp = [Ag+ ] ∗ [I -]
[I ] =
-
7.9 ⋅ 10-9
8.3 ⋅ 10
€
-17
=
[Pb+2][I- ]2
[Ag ] [I ]
+
-
=
[I -]2 =
[Ag+ ]
[s] [s+2t]
= 9.518 •107
€
[Pb+2][I- ]2 = 7.9 ⋅ 10-9 =[t] [s+2t]2 Assume s << t
7.9 ⋅ 10-9 = [t] [2t]2 = 4[t]3 ; t =1.25 •10−3 = [Pb+2]
28
Pb
D
0
+t
+t
+2
(aq)
+
2I- (aq)
s
+2t
s+2t
K sp = [Pb+2 ] ∗ [I -]2
K sp
[t] [s+2t]2
(s)
K sp
[Pb+2 ]
[Ag+ ] [I- ] = 8.3 ⋅ 10-17 = [s] [s+2t] Assume s << t
8.3 ⋅ 10-17 = [s] [2t] = [s] [2.50 •10−3] ; [s] = 3.32 •10−14 M
[Ag+ ]=3.32 •10−14 M, [Pb+2] =1.25 •10−3 M, [I- ] =2.5 •10−3 M
Solubility and Solubility Product
March 14
Calculation: Same type problem different data
At 25°C, the solubility product of AgI and PbI2 is 8.3•10-17 M and
7.9•10-9M, respectively. What are the values of [I-], [Ag+], and [Pb2+] in a
solution at equilibrium with both substances ? (Answer behind)
AgI
i Lots
Δ -s
e Lots
(s)
D Ag + (aq) +
0
+s
+s
I-(aq)
2t
+s
+s+2t
PbI2
i Lots
Δ -t
e Lots
(s)
D
K sp =[Ag+ ] ∗ [I- ]
K sp =[Pb+2 ] ∗ [I- ]2
8.3 ⋅ 10−17 = s • (2t)2
7.9 ⋅ 10−9 = t • (2t)2
7.9 ⋅ 10-9
8.3 ⋅ 10
-17
=
[Pb+2][I- ]2
[Ag ] [I ]
+
-
=
[t] [s+2t]2
[s] [s+2t]
= 9.518 •107
7.9 ⋅ 10 = [t] [2t] = 4[t]
2
3
; t = 1.25 •10
−3
0
+t
+t
+2
(aq)
+
2I- (aq)
s
+2t
s+2t
[Ag+ ] = 3.32 ⋅ 10-14 M
[Pb+2] = 1.25 •10−3 M
[Pb+2][I- ]2 = 7.9 ⋅ 10-9 =[t] [s+2t]2 Assume s << t
-9
Pb
= [Pb ]
+2
[I- ] = 2.50•10-3M
8.3 ⋅ 10-17 = [Ag+ ] [I- ] = [s] [s+2t], Assume s << t
8.3 ⋅ 10-17 = [s] [2t] but t = 1.25 •10−3
8.3 ⋅ 10-17 = [s] [2(1.25 •10−3 )] = 2.50 •10−3 s
s = 3.32 ⋅ 10-14 M= [Ag+ ] , [I- ] = s + 2t = 3.32 ⋅ 10-13 +2.50•10-3 = 2.50•10-3M
7.9 ⋅ 10-9 [Pb+2][I- ]2 [t] [s+2t]2
=
=
= 9.518 •107
[Ag+ ] [I- ] [s] [s+2t]
8.3 ⋅ 10-17
[Pb+2][I- ]2 = 7.9 ⋅ 10-9 =[t] [s+2t]2 Assume s << t
7.9 ⋅ 10-9 = [t] [2t]2 = 4[t]3 ; t = 1.25 •10−3 = [Pb+2]
29
8.3 ⋅ 10-17 = [Ag+ ] [I- ] = [s] [s+2t], Assume s << t
8.3 ⋅ 10-17 = [s] [2t] but t = 1.25 •10−3
8.3 ⋅ 10-17 = [s] [2(1.25 •10−3 )] = 2.50 •10−3 s
s = 3.32 ⋅ 10-13 M= [Ag+ ] , [I- ] = s + 2t = 3.32 ⋅ 10-13 +2.50•10-3 = 2.50•10-3M
[Ag+ ] = 3.32 ⋅ 10-13 M
[Pb+2] = 1.25 •10−3 M
[I- ] = 2.50•10-3M
Solubility and Solubility Product
[Ag+ ] [I- ] = 8.3 ⋅ 10-17 = [s] [s+2t] Assume s << t
8.3 ⋅ 10-17 = [s] [2t] = [s] [2.50 •10−3] ; [s] = 3.32 •10−14 M
[Ag+ ]=3.32 •10−14 M, [Pb+2] =1.25 •10−3 M, [I- ] =2.5 •10−3 M
March 14
Common Ion Effect
B&L 17.44 (7th Ed.): The Ksp for cerium iodate, Ce(IO3)3, is 3.2 •10-10
a) Calculate the molar solubility of Ce(IO3)3 in pure water.
b) What concentration of NaIO3 in solution would be necessary to reduce the Ce3+ concentration
in a saturated solution of Ce(IO3)3 by a factor of 10 below that calculation in part (a).
i
Δ
e
Ce(IO3)3
Lots
-s
Lots
(s)
+
D Ce+3(aq)
0
+s
+s
- 3
b) reduce Ce
a) Ksp = [Ce+3 ] ∗ [IO3 ]
3.2 ⋅ 10
-10
3
= s ∗ 27s = 27s
1.185 ⋅ 10 -11
= s4
1.86 ⋅ 10−3 M = s
3 IO3- (aq)
0
+3s
+3s
4
3+
by factor of 10:
3.2 ⋅ 10 =1.86 ⋅ 10 •[IO 3]Total
-10
1.7204 ⋅ 10-6
-4
3
[Ce3+ ]
10
= [IO 3]Total
1.198 ⋅ 10−2M = [IO 3] Total
[NaIO 3] = 1.198 ⋅ 10−2M - [IO-3] (from part previous)
[NaIO 3] = 1.198 ⋅ 10−2M - (3 x [1.86 ⋅ 10-4
**(same as Ce +3 )
€
])
[NaIO 3] = 1.14 ⋅ 10-2 M
** The IO3- in solution must be the same as Ce+3 for this condition
32
Solubility and Solubility Product
March 14
Qualitative Analysis
Ions are precipitated selectively by adding a precipitation ion until the Ksp of one
compound is exceeded without exceeding the Ksp of the others. An extension of this
approach is to control the equilibrium of the slightly soluble compound by
simultaneously controlling an equilibrium system that contains the precipitating ion.
Qualitative analysis of ion mixtures involves adding precipitating ions to separate the
unknown ions into ion groups. The groups are then analyzed further through
precipitation and complex ion formation.
33
Solubility and Solubility Product
March 14
Selective Precipitation
Qualitative Analysis
Q < Ksp, solid dissolve until Q=Ksp
Q = Ksp, equilib
Q > Ksp, ppt until Q=Ksp
Grp1: Insoluble Chlorides
Ag+, Pb2+, Hg22+
Grp2: Acid-Insoluble sulfides
Cu2+, Bi3+, Cd2+, Pb2+, Hg2+, As3+, Sb3+, Sn4+
Grp3: Base-Insoluble sulfides
Cu2+, Al3+, Fe2+, Fe3+, Co2+, Ni2+, Cr3+, Zn2+, Mn2+
Grp4: Insoluble Phosphates
Ba2+, Ca2+, Mg2+
Grp5: Alkali Metals and NH4+
Na+, K+, NH4+
35
Solubility and Solubility Product
March 14
Lab Practical, Qualitative Analysis
Qualitative Analysis of Household Chemicals
36
Solubility and Solubility Product
March 14
Selective Precipitation(2)
A solution contains 2.0 •10-4 M Ag+ and 1.5 •10-3 M Pb2+. If NaI is added, will
AgI (Ksp = 8.51 •10-17) or PbI2 (Ksp = 9.8 •10-9) precipitate first? Specify the
concentration of I- needed to begin precipitation. This is a Q problem
PbI 2 (s) !
AgI (s) !
i
Excess
Δ
-y
[e] Excess
+
Ag(aq)
+
2.0 •10-4
y
2.0 •10-4 + y
!
I-(aq )
i
Δ
2x
y
2x+y
[Ag+ ] [I- ] = [2.0 •10-4 + y] [2x + y]
8.51 •10-17 <
[2.0 •10-4 ] [2x]
2x >
2.0 •10-4
y
2x
y + 2x
9.8•10-9 < [Pb2+ ] [I- ]2 = [1.5•10-3 +x ] [y + 2x]2
9.8 •10-9 = [1.5 •10-3] [2x]2
x=
= 4.256 •10 , x > 2.13•10
[I- ] = 2x+y = 2x, y << x,
[I- ] = 4.256•10-13 M
1.5•10-3 +x
2I-(aq )
Assume y << x, furthermore assume x << 1.5•10-3
8.51 •10-17 <
-13
+
1.5•10-3
x
[e] Excess
Assume y << x and 2.0 •10-4
8.51 •10-17
Excess
-x
Pb2+
(aq)
-13
M
9.8•10-9
= 1.278•10-3M = x,
4•1.5•10
Note x = 1.5 •10-3 therefore need second iteration for 1.5•10-3 +x .
-3
[Pb2+ ] = 1.5•10-3 + 1.278•10-3 = 2.778•10-3
Second iteration, 9.8 •10-9 = [1.5 •10-3 + 1.278•10-3] [2x]2
2x = 1.88•10-3 M
[I- ] = 2x = 1.88•10-3 M for PbI 2 (s) to precipitate
AgI will precipitate first. The [I-] concentration needs to be 4.26•10-13M,
PbI2 will not precipitate until the [I-] concentration reaches 1.88•10-3 M
38
Solubility and Solubility Product
March 14
Complex Ion Process
Metals forming Complexes: Metals by virtue of electron pairs will act as Lewis
acids and bind a substrate to form a Complex Consider the following:
Ag(NH3)2+ (aq) Complex ion
AgCl(s) D Ag+(aq) + Cl- (aq)
Ag+(aq) + 2NH3(aq) D Ag(NH3)2+ (aq)
AgCl(s) + 2NH3(aq) D Ag(NH3)2+ (aq) + Cl- (aq)
Normally insoluble AgCl can be made soluble
By the addition of NH3. The presence of NH3
drives the top reaction to the right and
increase the solubility of AgCl
An assembly of metal ion and the Lewis base (NH3)
is called a complex ion.
The formation of this complex is describe by
Kf =
Formation Constant, Kf
Ag+(aq) + 2NH3(aq) D Ag(NH3)2+ (aq)
[Ag(NH3 )2+ ]
[Ag+ ] ∗[NH3 ]2
= 1.7 • 10 7
Solubility of metal salts affected presence of Lewis Base: e.g., NH3, CN-, OH39
€
Solubility and Solubility Product
March 14
Formation Constants Table
40
Complex
Kf
[Ag(CN)2]–
Complex
Kf
5.6e18
[Cr(OH)4]–
[Ag(EDTA)]3–
2.1e7
[Ag(en)2]+
Complex
Kf
8e29
[HgI4]2–
6.8e29
[CuCl3]2–
5e5
[Hg(ox)2]2–
9.5e6
5.0e7
[Cu(CN)2]–
1.0e16
[Ni(CN)4]2–
2e31
[Ag(NH3)2]+
1.6e7
[Cu(CN)4]3–
2.0e30
[Ni(EDTA)]2–
3.6e18
[Ag(SCN)4]3–
1.2e10
[Cu(EDTA)]2–
5e18
[Ni(en)3]2+
2.1e18
[Ag(S2O3)2]3–
1.7e13
[Cu(en)2]2+
1e20
[Ni(NH3)6]2+
5.5e8
[Al(EDTA)]–
1.3e16
[Cu(CN)4]2–
1e25
[Ni(ox)3]4–
3e8
[Al(OH)4]–
1.1e33
[Cu(NH3)4]2+
1.1e13
[PbCl3]–
2.4e1
[Al(Ox)3]3–
2e16
[Cu(ox)2]2–
3e8
[Pb(EDTA)]2–
2e18
[Cd(CN)4]2–
6.0e18
[Fe(CN)6]4–
1e37
[PbI4]2–
3.0e4
[Cd(en)3]2+
1.2e12
[Fe(EDTA)]2–
2.1e14
[Pb(OH)3]–
3.8e14
[Cd(NH3)4]2+
1.3e7
[Fe(en)3]2+
5.0e9
[Pb(ox)2]2–
3.5e6
[Co(EDTA)]2–
2.0e16
[Fe(ox)3]4–
1.7e5
[Pb(S2O3)3]4–
2.2e6
[Co(en)3]2+
8.7e13
[Fe(CN)6]3–
1e42
[PtCl4]2–
1e16
[Co(NH3)6]2+
1.3e5
[Fe(EDTA)]–
1.7e24
[Pt(NH3)6]2+
2e35
[Co(ox)3]4–
5e9
[Fe(ox)3]3–
2e20
[Zn(CN)4]2–
1e18
[Co(SCN)4]2–
1.0e3
[Fe(SCN)]2+
8.9e2
[Zn(EDTA)]2–
3e16
[Co(EDTA)]–
1e36
[HgCl4]2–
1.2e15
[Zn(en)3]2+
1.3e14
[Co(en)3]3+
4.9e48
[Hg(CN)4]2–
3e41
[Zn(NH3)4]2+
2.8e9
[Co(NH3)6]3+
4.5e33
[Hg(EDTA)]2–
6.3e21
[Zn(OH)4]2–
4.6e17
[Co(ox)3]3–
1e20
[Hg(en)2]2+
2e23
[Zn(ox)3]4–
1.4e8
[Cr(EDTA)]–
1e23
Solubility and Solubility Product
March 14
Calculations: Formation of Complex Ion
Tro, End of chapter Q 109 (2nd Edition)
A solution is made that is 1.1•10-3M Zn(NO3)2 and 0.150 M NH3. After the
solution reaches equilibrium, what concentration of Zn2+ (aq) remains.
Answer = 8.7•10-10M
Zn2+(aq) + 4NH3 (aq) D Zn(NH3)42+ (aq)
Zn2+
(aq)
i
1.1•10
-3
+
NH 3 (aq)
! Zn(NH 3 )2+
4
0.150
(aq)
0
Δ
-x
-1.1•10-3
[e] 1.1•10-3 -x 0.150 − 1.1•10-3
+1.1•10-3
1.1•10-3
-3
Assume reaction goes to completion Zn(NH 3 )2+
=
1.1•10
M
4
2.8 •10
9
=
1.1•10 -x =
-3
41
[Zn(NH 3 )2+
]
4
[Zn+2] [NH 3] 4
=
x
[1.1•10-3 -x] [0.150 − 1.1•10-3] 4
1.1•10-3
2.8 •109 [0.1456] 4
=
1.1•10-3
[1.1•10-3 -x] [0.1456] 4
= 8.74•10-10 , [Zn2+ ] eq = 8.74•10-10M
Solubility and Solubility Product
March 14
Calculations: Formation of Complex Ion
By using the values of Ksp for AgI and Kf for Ag(CN)2-, calculate the
equilibrium constant for the following reaction:
AgI(s) + 2CN-(aq) D Ag(CN)2- (aq) + I-(aq)
Note that this equation is the sum of
AgI (s) D Ag+(aq) + I -(aq) ;
Ksp = 8.3•10-17 = [Ag+][I -]
Ag+(aq) + 2CN-(aq) D Ag(CN)2-(aq) ;
Kf = 1.0•1021 = [Ag(CN)2-]/[Ag+][CN -]2
AgI(s) + 2CN-(aq) D Ag(CN)2- (aq) + I-(aq)
K = Ksp• Kf = ( [Ag+][I -] ) • ( [Ag(CN)2]/[Ag+][CN -]2 )
K = Ksp• Kf = 8.3•10-17 • 1.0•1021
K = 8.3•104
43
Solubility and Solubility Product
March 14