14.4 (Chp16) Solubility and Solubility...
Transcript of 14.4 (Chp16) Solubility and Solubility...
March 14 1 Solubility and Solubility Product
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
March 14 2 Solubility and Solubility Product
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 + SO4
2- (aq) Keq = Ksp = [Ba+2] [SO4
2-]
How water dissolves ionic compounds
Do not confuse solubility (s) with solubility product Ksp.
Solubility depends on Temp., conc., and pH
Solubility Product depends on Temp. Only !!!
March 14 3 Solubility and Solubility Product
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 g per 100 mL or g /L (molar solubility, grams per liter saturated solution.)
g • mol g mol g Molarity g
The greater the solubility, the smaller the amount of precipitation.
March 14 4 Solubility and Solubility Product
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.
March 14 5 Solubility and Solubility Product
In a Chemical Reaction - For a reaction to take place -
Reactants should be soluble in solvent- This ensures that reactants combine (come in contact) with each other efficiently.
When reaction proceeds- Products 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.
March 14 6 Solubility and Solubility Product
Precipitation Reaction
Consider the reaction - i AgNO 3 (aq) + HCl (aq) D AgCl(s) + HNO3(aq)
Net ionic equation: Ag+
(aq) + Cl- (aq) D AgCl (s) Write the reverse, (standard convention) AgCl (s) D Ag+
(aq) + Cl- (aq)
Then 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)
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Keq =1
[Ag+][Cl−]
If written in this form, then mass action is:
NaOCH2CH3 (s) + H2O (l) D Na+ (aq) + OH- (aq) + CH3CH2OH (aq) Ksp = [Na+] [OH-] [CH3CH2OH]
March 14 8 Solubility and Solubility Product
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 0 0 Δ - 4.4•10-5M 3(4.4•10-5M ) + 4.4•10-5M
e Solid 1.32•10-4M 4.4•10-5 M
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).
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
March 14 10 Solubility and Solubility Product
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
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mol KNO3 = 30.0g ∗ mol101.11g
= 0.297mol
Assume solution has ρ = 1.00 g/ccMass solution = 100g H2O + 30g KNO3 = 130 g Solution130 g solution = 0.130 L since ρ = 1.00 g/cc
[s] = 0.297mol0.130L
= 2.285M
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Ksp = [K+ ][NO3
-] = [2.285]2 = 5.22
March 14 12 Solubility and Solubility Product
Solubility Vs. Ksp Consider ;
La (IO3)3 Ba(IO3)2 AgIO3
Ksp = 6.2 • 10-12 < Ksp = 1.5 • 10-9 < Ksp = 3.1 •10-8
s = 6.9•10-4 < s = 7.2•10-4 > 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).
March 14 13 Solubility and Solubility Product
Solubility Rules: What is it used for? How can one predict if a precipitate forms when mixing solutions?
Solubility rules -
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
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
March 14 14 Solubility and Solubility Product
Precipitation Experiment
Consider the reaction: 3 Ca2+ + 2PO4
3- (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+ + 2PO4
3- (aq)
Q = [Ca+2]3 • [PO43-]2 next compare Q to Ksp
In solubility product problems, Q is the ion-product instead of reaction quotient
March 14 15 Solubility and Solubility Product
Ion Product; Q
Q - indicator of direction of reaction for reaction not at equilb.
MX (s) D M+(aq) + X-
(aq)
Direction Reaction
Q (i) Unsaturated. Q is small, system consist mostly of ion. No ppt. forms (s) g (aq)
Direction Reaction
Q (h) Supersaturated.
Q is large, system will adjust to reduce high
ion concentration. Solid forms (s) f (aq)
Ksp Q
Sat. Point
March 14 17 Solubility and Solubility Product
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 (s) D 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.
March 14 18 Solubility and Solubility Product
Common Ion Effect
Ag+ I-
AgI(s) AgI(s)
Ag+ I-
Add I- via NaI
Ag+ I-
AgI(s)
AgI(s) D Ag+(aq) + I-
(aq)
add common ion i.e., NaI to solution
Reduce solubility i.e., less AgI will dissolve in soln’ which means more ppt. forms in solution
Direction of Reaction
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.
March 14 19 Solubility and Solubility Product
...according to LeChatelier
AgI(s) D Ag+ + I-
Q analogy:
Q = [Ag+] [I-]
Ksp < Q
AgI(s) Ag+ I-
AgI(s) NaI(s)
Ag+ I-
AgI(s) Ag+ I-
More AgI forms and Solubility is lowered
Increase I-
Ksp Q
Direction of Rxn (ppt occur)
March 14 22 Solubility and Solubility Product
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
CuI (s) D Cu+(aq) + I- (aq)
a) i Lots 0 0 Δ -s +s +s e Solid +s +s
CuI (s) D Cu+(aq) + I- (aq)
b) i Lots 0 0.05 Δ -s +s +s e Solid +s 0.05 + s
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a) Ksp = 1.1 ⋅10-12 =[Cu+] [I-] = s2
1.05 ⋅10-6 = s
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b) Ksp = 1.1 ⋅10-12 =[Cu+] [I-]
= s ∗ (0.05M +s) = s ∗ (0.05M)
2.0 ∗ 10-11 = s
Without common ion, solubility is, s = 1.05 •10-6 M
but with common ion, solubility is, s = 2.0 •10-11 M
March 14 24 Solubility and Solubility Product
Check assumption indeed 9.92•10-5 << 0.025
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
CaF2 (s) D Ca+2
(aq) + 2F- (aq) i Lots 0 0.025M Δ -s +s +2s e Solid +s 0.025M +2s
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Ksp = 6.2 ⋅ 10-8 = [Ca+2] [F-]2
= s ∗ (0.025M + 2s)2
6.2 ⋅ 10-8 = s ∗ (0.025M)2
6.2 ⋅ 10-8
0.0252= s = 9.92 ⋅ 10-5M
assume s << 0.025
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9.92 ⋅10-5
0.025• 100 = 0.4%
March 14 26 Solubility and Solubility Product
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 [SO4
2-], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances ?
PbSO4 (s) D Pb2+(aq) + SO4
2 -(aq) ; Ksp = 1.6•10-8 = [Pb2+][SO4
2-]
SrSO4 (s) D Sr+ (aq) + SO4
2-(aq) ; Ksp = 2.8•10-7 = [Sr2+][SO4
2-]
Let x = [Pb2+] , y = [Sr2+] , x + y = [SO42-]
x(x+y) = 1.6•10-8 = x = 0.0571 = 0.057 ; x = 0.057y y(x+y) 2.8•10-7 y
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
March 14 27 Solubility and Solubility Product
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 (s) D Zn2+ (aq) + 2IO3
-2(aq)
i Lots 0 0 Δ -m +m +2m + 2n e Lots +m +2m + 2n
Sr(IO3)2 (s) D Sr +2(aq) + 2IO3
- (aq)
i Lots 0 0 Δ -n +n +2n + 2m e Lots +n +2n + 2m
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Ksp = [Zn2+] ∗ [IO3
-]2
[IO3
-]2 =
Ksp
[Zn2+]
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Ksp =[Sr+2]∗[IO3-]2
[IO4-]2 =
Ksp
[Sr+2]
Answer: [Sr +2] = 7.9•10-4 M [Zn +2] = 9.4•10-3 M [IO3
- ] = 2.0 •10-2 M
...
3.9 ⋅10-6
3.3 ⋅10-7=
[Zn+2][IO3-]2
[Sr+2][IO3-]2
= [m][2m+2n]2
[n][2m+2n-]2=[m]
[n]
n ⋅1.1818 ⋅101 = m
3.9 ⋅10-6
3.3 ⋅10-7=
[Zn+2][IO3-]2
[Sr+2][IO3-]2
= [m][2m+2n]2
[n][2m+2n-]2=[m]
[n]
n ⋅1.1818 ⋅101 = 11.818 ⋅n = m
3.3 ⋅10-7 = [Sr+2][IO3-]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= 3.3 ⋅10-7
657.23 = 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
March 14 28 Solubility and Solubility Product
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 (s) D Ag + (aq) + I-
(aq) i Lots 0 2t Δ -s +s +s e Lots +s +s+2t
PbI2 (s) D Pb +2(aq) + 2I-
(aq) i Lots 0 s Δ -t +t +2t e Lots +t s+2t
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Ksp = [Ag+] ∗ [I-]
[I-] =Ksp
[Ag+]
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Ksp = [Pb+2] ∗ [I-]2
[I-]2 =Ksp
[Pb+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
[Pb+2][I-]2 = 7.9 ⋅10-9=[t] [s+2t]2 Assume s << t7.9 ⋅10-9 = [t] [2t]2 = 4[t]3 ; t =1.25•10−3 = [Pb+2]
[Ag+] [I-] = 8.3 ⋅10-17= [s] [s+2t] Assume s << t8.3 ⋅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 29 Solubility and Solubility Product
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 (s) D Ag + (aq) + I-
(aq) i Lots 0 2t Δ -s +s +s e Lots +s +s+2t
PbI2 (s) D Pb +2(aq) + 2I-
(aq) i Lots 0 s Δ -t +t +2t e Lots +t s+2t
Ksp
=[Ag+] ∗[I-]
8.3 ⋅10−17= s • (2t)2
Ksp
=[Pb+2] ∗[I-]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
[Pb+2][I-]2 = 7.9 ⋅10-9=[t] [s+2t]2 Assume s << t7.9 ⋅10-9 = [t] [2t]2 = 4[t]3 ; t = 1.25•10−3 = [Pb+2]
8.3 ⋅10-17 = [Ag+] [I-] = [s] [s+2t], Assume s << t8.3 ⋅10-17 = [s] [2t] but t = 1.25•10−3
8.3 ⋅10-17 = [s] [2(1.25•10−3)] = 2.50 •10−3s
s = 3.32 ⋅10-14 M= [Ag+] , [I-] = s + 2t = 3.32 ⋅10-13+2.50•10-3 = 2.50•10-3M[Ag+] [I-] = 8.3 ⋅10-17= [s] [s+2t] Assume s << t8.3 ⋅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
7.9 ⋅10-9
8.3 ⋅10-17=[Pb+2][I-]2
[Ag+] [I-]=
[t] [s+2t]2
[s] [s+2t]= 9.518•107
[Pb+2][I-]2 = 7.9 ⋅10-9=[t] [s+2t]2 Assume s << t7.9 ⋅10-9 = [t] [2t]2 = 4[t]3 ; t = 1.25•10−3 = [Pb+2]
8.3 ⋅10-17 = [Ag+] [I-] = [s] [s+2t], Assume s << t8.3 ⋅10-17 = [s] [2t] but t = 1.25•10−3
8.3 ⋅10-17 = [s] [2(1.25•10−3)] = 2.50 •10−3s
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•10−3M [I-] = 2.50•10-3M
[Ag+] = 3.32 ⋅10-14 M [Pb+2] = 1.25•10−3M [I-] = 2.50•10-3M
March 14 32 Solubility and Solubility Product
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). Ce(IO3)3 (s) D Ce+3
(aq) + 3 IO3-
(aq) i Lots 0 0 Δ -s +s +3s e Lots +s +3s
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a) Ksp = [Ce+3] ∗[IO3-]3
3.2 ⋅ 10-10 = s ∗27s3 = 27s4
1.185 ⋅ 10-11 = s4
1.86 ⋅ 10−3M = s
b) reduce Ce3+ by factor of 10: [Ce3+]10
3.2 ⋅10-10=1.86 ⋅10-4 •[IO3]Total3
1.7204 ⋅10-6 = [IO3]Total
1.198 ⋅10−2M = [IO3]
Total
[NaIO3] = 1.198 ⋅10−2M - [IO
3-]
(from part previous)
[NaIO3] = 1.198 ⋅10−2M - (3 x [1.86 ⋅10-4
**(same as Ce+3)])
[NaIO3] = 1.14 ⋅10-2 M
** The IO3- in solution must be the same as Ce+3 for this condition
March 14 33 Solubility and Solubility Product
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.
March 14 35 Solubility and Solubility Product
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+
March 14 36 Solubility and Solubility Product
Lab Practical, Qualitative Analysis Qualitative Analysis of Household Chemicals
March 14 38 Solubility and Solubility Product
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
AgI(s)
! Ag+(aq)
+ I-(aq )
i Excess ! 2.0 •10-4 2xΔ -y y y[e] Excess 2.0 •10-4 + y 2x+y
Assume y << x and 2.0 •10-4
8.51•10-17 < [Ag+] [I-] = [2.0 •10-4 + y] [2x + y] 8.51•10-17 < [2.0 •10-4] [2x]
2x > 8.51•10-17
2.0 •10-4= 4.256•10-13, x > 2.13•10-13 M
[I-] = 2x+y = 2x, y << x, [I-] = 4.256•10-13 M
PbI2 (s)
! Pb2+(aq)
+ 2I-(aq )
i Excess 1.5•10-3 yΔ -x x 2x[e] Excess 1.5•10-3+x y + 2x
Assume y << x, furthermore assume x << 1.5•10-3
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 = 9.8•10-9
4•1.5•10-3= 1.278•10-3M = x,
Note x = 1.5•10-3 therefore need second iteration for 1.5•10-3+x .[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 PbI2 (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
March 14 39 Solubility and Solubility Product
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
Formation Constant, Kf Ag+
(aq) + 2NH3(aq) D Ag(NH3)2+ (aq)
Solubility of metal salts affected presence of Lewis Base: e.g., NH3, CN-, OH-
€
Kf =[Ag(NH3)2
+ ]
[Ag+]∗[NH3]2 = 1.7 • 107
March 14 40 Solubility and Solubility Product
Formation Constants Table
Complex Kf Complex Kf Complex Kf [Ag(CN)2]– 5.6e18 [Cr(OH)4]– 8e29 [HgI4]2– 6.8e29
[Ag(EDTA)]3– 2.1e7 [CuCl3]2– 5e5 [Hg(ox)2]2– 9.5e6
[Ag(en)2]+ 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
March 14 41 Solubility and Solubility Product
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)4
2+ (aq)
Zn2+(aq)
+ NH3 (aq)
! Zn(NH3)
42+
(aq)
i 1.1•10-3 0.150 0Δ -x -1.1•10-3 +1.1•10-3 [e] 1.1•10-3-x 0.150 − 1.1•10-3 1.1•10-3
Assume reaction goes to completion Zn(NH3)
42+ = 1.1•10-3M
2.8•109 = [Zn(NH
3)
42+]
[Zn+2] [NH3]4
= x [1.1•10-3-x] [0.150 − 1.1•10-3]4
= 1.1•10-3 [1.1•10-3-x] [0.1456]4
1.1•10-3-x = 1.1•10-3 2.8•109[0.1456]4
= 8.74•10-10, [Zn2+]eq
= 8.74•10-10M
March 14 43 Solubility and Solubility Product
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