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15.4 (Chp17) 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 January 21 When a Substance Dissolve When a reaction occurs in which an insoluble product is produced, the Keq is called a solubility-product. +2 2- BaSO4 (s) D Ba + SO4 (aq) +2 2- Keq = Ksp = [Ba ] [SO4 ]

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 January 21 Solubility Solubility: What is the meaning of solubility ? i. The ability of substance to dissolve in a .

ii. Quantity that dissolves to form a saturated . 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.

3 Solubility and Solubility Product January 21 Factors Affecting Solubility (s).

1 Nature of Solute (). - Like dissolves like (IMF)

2 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 close 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 January 21 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 it is a result of a new specie that is present in the 5 solution. Solubility and Solubility Product January 21 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) g Keq If written in this Write the reverse, (standard convention) form, then mass action is: + - AgCl (s) D Ag (aq) + Cl (aq) g 1/Keq 1 + - Then 1/Keq = Ksp = [Ag ] [Cl ] K eq = + − [Ag ][Cl ]

Ksp - solubility product constant An indicator of the solubility of substance of interest. Yields info. on the amount of allowed in solution. € (Must be determine experimentally)

+ - NaOCH2CH3 (s) + H2O (l) D Na+ (aq) + OH- (aq) + CH3CH2OH (aq) Ksp = [Na ] [OH ] [CH3CH2OH]

6 Solubility and Solubility Product January 21 : Example

What is the solubility product for the following reaction: + 3- -5 Ag3PO4 D 3Ag (aq) + PO4 (aq) s = 4.4•10 M i Excess 0 0 D - 4.4•10-5M 3(4.4•10-5M ) + 4.4•10-5M e Solid 1.32•10-4M 4.4•10-5 M

7 Solubility and Solubility Product January 21 Solubility Equilibrium: Example

What is the solubility product, Ksp, for the following reaction: + 3- -5 Ag3PO4 D 3Ag (aq) + PO4 (aq) [s] = 4.4•10 M i Excess 0 0 D - s +3s +s - 4.4•10-5M 3(4.4•10-5M ) + 4.4•10-5M e Solid 1.32•10-4M 4.4•10-5 M

-4 3 -5 -16 Solubility; Ksp = [1.32•10 ] • [4.4 •10 ] = 1.01 •10

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 January 21 Solubility from

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 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? Assume density of solution = 1.0 g / cc

MWKNO3 = 101.11 g/mol

9 Solubility and Solubility Product January 21 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 temp.

The solubility of KNO3 at 20°C is 30 g per 100ml H2O. What is the Ksp for KNO3 at this temperature? Assume volumes additive of solute and solvent

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 0.297mol [s] = = 2.285M 0.130L

- K = [K+ ][NO ] = [2.285]2 = 5.22 sp 3

10 € Solubility and Solubility Product January 21

€ 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. MW KNO3 = 101.11 g/mol

The solubility of KNO3 at 20°C is 30 g per 100ml H2O. What is the mass % m:m of this solution? Below are density of KNO3 at various % m:m concentration.

Calculate the Ksp of KNO3 using this data.

11 Solubility and Solubility Product January 21 Solubility Vs. Ksp Consider ;

La (IO3)3 Ba(IO3)2 AgIO3

-12 -9 -8 Ksp = 6.2 • 10 < Ksp = 1.5 • 10 < Ksp = 3.1 •10

12 Solubility and Solubility Product January 21 Solubility Vs. Ksp

Consider ;

La (IO3)3 Ba(IO3)2 AgIO3

-12 -9 -8 Ksp = 6.2 • 10 < Ksp = 1.5 • 10 < Ksp = 3.1•10

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).

13 Solubility and Solubility Product January 21 Solubility Rules: What is it used for? How can one predict if a precipitate forms when mixing solutions? Solubility rules -

Soluble Substances Exceptions Insoluble substances Exceptions containing containing

- 2- , (NO3 ) None (CO3 ) Slightly - 3- chlorates (ClO3 ) (PO4 ) soluble - perchlorates (ClO4 ) - (CH3COO )

- - + halogens (X ) Ag, Hg, Pb (OH ) alkali, NH4 , X- = Cl-, Br-, I- Ca, Sr, Ba

2- (SO4 ) 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

14 Solubility and Solubility Product January 21 Precipitation Experiment

Consider the reaction: 2+ 3- 3 Ca + 2PO4 (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 2+ 3- Ca3(PO4)2 (s) D 3 Ca + 2PO4 (aq)

+2 3 3- 2 Q = [Ca ] • [PO4 ] next compare Q to Ksp

This is a solubility product problems, Q is the -product instead of

15 Solubility and Solubility Product January 21 Ion Product; Q

Q - indicator of direction of reaction for reaction not at equilb.

+ - MX (s) D M (aq) + X (aq)

Direction Direction Ksp Reaction Reaction

Q (i) Q Q (h) Unsaturated. Sat. Supersaturated. Q is small, system Point Q is large, system will consist mostly of ion. adjust to reduce high No ppt. forms ion concentration. (s) g (aq) Solid forms (s) f (aq)

16 Solubility and Solubility Product January 21 In Class: Precipitation Determination

Question, (19.88 Sil): Will any solid Ba(IO3)2 form when 6.5 mg of BaCl2 +2 is dissolved in 0.500L of 0.033M NaIO3? ([Ba ] = 6.2e-5 M) -9 Ksp Ba(IO3)2 = 1.5 •10 , BaCl2 MW = 208.23 g/mol

2+ - Ba(IO3)2 = Ba(IO3)2 (s) D Ba (aq) + 2 IO3 (aq)

17 Solubility and Solubility Product January 21 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? -9 Ksp Ba(IO3)2 = 1.5 •10 BaCl2 MW = 208.23 g/mol

2+ - Ba(IO3)2: Ba(IO3)2 (s) D Ba (aq) + 2 IO3 (aq)

[Ba+2] = 6.5 mg .... 6.2•10-5 M

- -2 [IO3 ] = 3.3•10 M

+2 - 2 -5 -2 2 Q = [Ba ] • [IO3 ] = 6.2•10 M • (3.3•10 M)

-8 Q = 6.8 •10 M > Ksp

Precipitate forms.

18 Solubility and Solubility Product January 21 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+ I- Ag I Ag+ I-

AgI(s) AgI(s) AgI(s)

+ - - AgI(s) D Ag (aq) + I (aq) Add I via NaI 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

19 Solubility and Solubility Product January 21 ...according to LeChatelier

AgI(s) Ag+ I- + - AgI(s) D Ag + I Q analogy: + - NaI(s) Q = [Ag ] [I ] AgI(s)

Increase I- Ag+ I- Ksp < Q

AgI(s) Ag+ I-

Ksp Q

More AgI forms and Direction of Rxn Solubility is lowered (ppt occur)

20 Solubility and Solubility Product January 21 Factor Affecting Solubility: Solubility & Common Ion

-12 Calculate the solubility of (I) , CuI (Ksp = 1.1•10 ) a) water b) 0.05 M iodide

+ - CuI (s) D Cu (aq) + I (aq) + - CuI (s) D Cu (aq) + I (aq) a) i Lots 0 0 b) i Lots 0 0.05 D -s +s +s e Solid +s +s D -s +s +s e Solid +s 0.05 + s

21 Solubility and Solubility Product January 21 Factor Affecting Solubility: Solubility & Common Ion

-12 Calculate the solubility of copper(I) iodide, CuI (Ksp = 1.1•10 ) a) water b) 0.05 M

+ - CuI (s) D Cu (aq) + I (aq) + - CuI (s) D Cu (aq) + I (aq) a) i Lots 0 0 b) i Lots 0 0.05 D -s +s +s e Solid +s +s D -s +s +s e Solid +s 0.05 + s

-12 + - 2 a) Ksp = 1.1⋅10 =[Cu ] [I ] = s -6 1.05⋅10 = s

€ Without common ion, solubility is, s = 1.05 •10-6 M

22 Solubility and Solubility Product January 21 Factor Affecting Solubility: Solubility & Common Ion

-12 Calculate the solubility of copper(I) iodide, CuI (Ksp = 1.1•10 ) a) water b) 0.05 M sodium iodide

+ - CuI (s) D Cu (aq) + I (aq) + - CuI (s) D Cu (aq) + I (aq) a) i Lots 0 0 b) i Lots 0 0.05 D -s +s +s e Solid +s +s D -s +s +s e Solid +s 0.05 + s

-12 + - 2 -12 + - a) Ksp = 1.1⋅10 =[Cu ] [I ] = s b) Ksp = 1.1⋅10 =[Cu ] [I ] -6 = s ∗ (0.05M + s) 1.05⋅10 = s = s ∗ (0.05M) -11 2.0 ∗ 10 = s € Without common ion, solubility is, s = 1.05 •10-6 M but with common ion, solubility€ is, s = 2.0 •10-11 M

23 Solubility and Solubility Product January 21 Calculation: Solubility, Iteration Method

Reger Ex 12.19: What is the solubility of in 0.025 M solution? Ksp = 6.2•10-8 M

+2 - CaF2 (s) D Ca (aq) + 2F (aq) i Lots 0 0.025M D -s +s +2s e Solid +s 0.025M +2s

24 Solubility and Solubility Product January 21 Calculation: Solubility, Iteration Method Reger Ex 12.19: What is the solubility of in 0.025 M sodium fluoride solution? Ksp = 6.2•10-8 M

+2 - CaF2 (s) D Ca (aq) + 2F (aq) i Lots 0 0.025M D -s +s +2s e Solid +s 0.025M +2s

-8 +2 - 2 Ksp = 6.2 ⋅ 10 = [Ca ] [F ] = s ∗ (0.025M + 2s)2 assume s << 0.025 6.2 ⋅ 10-8 = s ∗ (0.025M)2 6.2 ⋅ 10-8 1.98⋅10−4 = s = 9.92 ⋅ 10-5 M ∗100 = 0.794% 2 .025 0.025 Check assumption indeed 9.92•10-5 << 0.025 €

25 Solubility and Solubility Product January 21 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 2- 2+ 2+ [SO4 ], [Pb ], and [Sr ] in a solution at equilibrium with both substances ?

2+ 2 - -8 2+ 2- PbSO4 (s) D Pb (aq) + SO4 (aq) ; Ksp = 1.6•10 = [Pb ][SO4 ]

+ 2- -7 2+ 2- SrSO4 (s) D Sr (aq) + SO4 (aq) ; Ksp = 2.8•10 = [Sr ][SO4 ]

2+ 2+ 2- Let x = [Pb ] , y = [Sr ] , x + y = [SO4 ]

26 Solubility and Solubility Product January 21 Calculation: Solubility of Two salts in same solution

-8 At 50°C, the solubility products , Ksp, of PbSO4 and SrSO4 are 1.6•10 M -7 2- 2+ and 2.8•10 M, respectively. What are the values of [SO4 ], [Pb ], and [Sr2+] in a solution at equilibrium with both substances ?

2+ 2 - -8 2+ 2- PbSO4 (s) D Pb (aq) + SO4 (aq) ; Ksp = 1.6•10 = [Pb ][SO4 ]

+ 2- -7 2+ 2- SrSO4 (s) D Sr (aq) + SO4 (aq) ; Ksp = 2.8•10 = [Sr ][SO4 ]

2+ 2+ 2- Let x = [Pb ] , y = [Sr ] , x + y = [SO4 ]

x(x+y) = 1.6•10-8 = x = 0.0571 = 0.057 ; x = 0.057y y(x+y) 2.8•10-7 y

y(x+y) 2.8•10-7 , substitute 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

2+ -5 2+ -4 2- -4 [Pb ] = 2.9•10 M, [Sr ] = 5.1•10 M, [SO4 ]= 5.4•10 M

27 Solubility and Solubility Product January 21 Calculation: Solubility of Two salts in same solution

-8 At 50°C, the solubility products , Ksp, of PbSO4 and SrSO4 are 1.6•10 M and -7 2- 2+ 2+ 2.8•10 M, respectively. What are the values of [SO4 ], [Pb ], and [Sr ] in a solution at equilibrium with both substances ?

2+ 2 - -8 2+ 2- PbSO4 (s) D Pb (aq) + SO4 (aq) ; Ksp = 1.6•10 = [Pb ][SO4 ]

+ 2- -7 2+ 2- SrSO4 (s) D Sr (aq) + SO4 (aq) ; Ksp = 2.8•10 = [Sr ][SO4 ]

2+ 2+ 2- Let x = [Pb ] , y = [Sr ] , x + y = [SO4 ]

y (x+y) = 2.8•10-7 = y = 17.5; y = 17.5x x (x+y) 1.6•10-8 x

x (x+y) = 1.6•10-8 , substitute y = 17.5 x

x (x+(17.5 x)) = 1.6•10-8; x (18.5 x) = 1.6•10-8; x2 = 1.6•10-8 /18.5; x = 2.94•10-5 y = 17.5x; y = 17.5 (2.94•10-5); y = 5.08 •10-4

2+ -5 2+ -4 2- -4 [Pb ] = 2.94•10 M, [Sr ] = 5.08•10 M, [SO4 ]= 5.37•10 M

28 Solubility and Solubility Product January 21 Calculation: Same type problem different data

-6 At 25°C, the solubility product of Zn(IO3)2 and Sr(IO3)2 is 3.9•10 M -7 - +2 and 3.3•10 M, respectively. What are the values of [IO3 ], [Zn ], and [Sr2+] in a solution at equilibrium with both substances ?

2+ - +2 - Zn(IO3)2 (s) D Zn (aq) + 2IO3 (aq) Sr(IO3)2 (s) D Sr (aq) + 2IO3 (aq) i Lots 0 0 i Lots 0 0 D -m +m +2m + 2n D -n +n +2n + 2m e Lots +m +2m + 2n e Lots +n +2n + 2m

- +2 - 2 2+ 2 K sp =[Sr ]∗[IO3 ] Ksp = [Zn ] ∗ [IO3 ]

- 2 K sp - Ksp [IO ] = [IO ]2 = 4 [Sr+2] 3 2+ [Zn ]

+2 - 2 3.9 ⋅10-6 [Zn ][IO ] [m][2m+2n]2 [m] € = 3 = = 3.3€⋅1 0-7 [Sr+2][IO-]2 [n][2m+2n-]2 [n] 3

n ⋅1.1818⋅101 = m Answer: ... [Sr +2] = 7.9•10-4 M [Zn +2] = 9.4•10-3 M - -2 [IO3 ] = 2.0 •10 M +2 - 2 3.9 ⋅10-6 [Zn ][IO ] [m][2m+2n]2 [m] = 3 = = 3.3⋅10-7 [Sr+2][IO-]2 [n][2m+2n-]2 [n] 3

n ⋅1.1818⋅101 = 11.818⋅ n = m

3.3⋅10-7 = [Sr+2][IO-]2= n[2m+2n]2= n[2(11.818n)+2n]2 3 3.3⋅10-7 = n[(23.636n)+2n]2= n[(25.636n]2 = 657.2n3

-7 3.3⋅10 -4 -3 n= 3 = 7.948⋅10 , m = 11.818⋅ n =9.393⋅10 657.2

[Sr+2] = 7.948⋅10-4M, [Zn+2] = 9.393⋅10-3M, [Zn+2] = 2.038⋅10-2M 29 Solubility and Solubility Product January 21 Calculation: Same type problem different data

-17 At 25°C, the solubility product of AgI and PbI2 is 8.3•10 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)

+ - +2 - AgI (s) D Ag (aq) + I (aq) PbI2 (s) D Pb (aq) + 2I (aq) i Lots 0 2t i Lots 0 s D -s +s +s D -t +t +2t e Lots +s +s+2t e Lots +t s+2t

K = [Ag+] ∗[I-] +2 - 2 sp K sp = [Pb ] ∗[I ] K K [I-] = sp [I-]2 = sp + [Pb+2] [Ag ]

-9 +2 - 2 2 7.9 ⋅10 [Pb ][I ] [t] [s+2t] 7 = = = 9.518•10 + - -17 8.3⋅10-17 [Ag+] [I-] [s] [s+2t] € [Ag ] [I ] = 8.3⋅10 = [s] [s+2t] Assume s << t € 8.3⋅10-17 = [s] [2t] = [s] [2.50 •10−3] ; [s] = 3.32•10−14 M [Pb+2][I-]2 = 7.9 ⋅10-9 =[t] [s+2t]2 Assume s << t -9 2 3 −3 +2 7.9 ⋅10 = [t] [2t] = 4[t] ; t =1.25•10 = [Pb ] [Ag+]=3.32•10−14 M, [Pb+2] =1.25•10−3 M, [I-] =2.5•10−3 M

30 Solubility and Solubility Product January 21 Calculation: Same type problem different data

-17 At 25°C, the solubility product of AgI and PbI2 is 8.3•10 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)

+ - +2 - AgI (s) D Ag (aq) + I (aq) PbI2 (s) D Pb (aq) + 2I (aq) i Lots 0 2t i Lots 0 s D -s +s +s D -t +t +2t e Lots +s +s+2t e Lots +t s+2t

K = [Ag+] ∗[I-] +2 - 2 sp K sp = [Pb ] ∗[I ] K K [I-] = sp [I-]2 = sp + [Pb+2] [Ag ]

-9 +2 - 2 2 7.9 ⋅10 [Pb ][I ] [t] [s+2t] 7 = = = 9.518•10 + - -17 8.3⋅10-17 [Ag+] [I-] [s] [s+2t] € [Ag ] [I ] = 8.3⋅10 = [s] [s+2t] Assume s << t € 8.3⋅10-17 = [s] [2t] = [s] [2.50 •10−3] ; [s] = 3.32•10−14 M [Pb+2][I-]2 = 7.9 ⋅10-9 =[t] [s+2t]2 Assume s << t -9 2 3 −3 +2 7.9 ⋅10 = [t] [2t] = 4[t] ; t =1.25•10 = [Pb ] [Ag+]=3.32•10−14 M, [Pb+2] =1.25•10−3 M, [I-] =2.5•10−3 M

31 Solubility and Solubility Product January 21 Calculation: Same type problem different data

-17 At 25°C, the solubility product of AgI and PbI2 is 8.3•10 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)

+ - +2 - AgI (s) D Ag (aq) + I (aq) PbI2 (s) D Pb (aq) + 2I (aq) i Lots 0 2t i Lots 0 s D -s +s +s D -t +t +2t e Lots +s +s+2t e Lots +t s+2t

K =[Ag+] ∗[I-] K =[Pb+2] ∗[I-]2 sp sp

8.3⋅10−17 = s • (2t)2 7.9 ⋅10−9 = t • (2t)2

-9 +2 - 2 2 7.9 ⋅10 [Pb ][I ] [t] [s+2t] 7 = = = 9.518•10 + -14 8.3⋅10-17 [Ag+] [I-] [s] [s+2t] [Ag ] = 3.32⋅10 M [Pb+2] = 1.25•10−3M +2 - 2 -9 2 [Pb ][I ] = 7.9 ⋅10 =[t] [s+2t] Assume s << t [I-] = 2.50•10-3M 7.9 ⋅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 << 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 8.3⋅10-17 [Ag+] [I-] [s] [s+2t]

[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] [Ag+] [I-] = 8.3⋅10-17 = [s] [s+2t] Assume s << t 8.3⋅10-17 = [Ag+] [I-] = [s] [s+2t], Assume s << t 8.3⋅10-17 = [s] [2t] = [s] [2.50 •10−3] ; [s] = 3.32•10−14 M 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

+ −14 +2 −3 - −3 32 s = 3.32⋅10-13 M= [Ag+] , [I-] = s + 2t = 3.32Solubility⋅10-13 +2.50•10-3 = 2.50•10-3M and Solubility Product [Ag ]=3.32•10 M, [Pb ] =1.25•10 M, [I ] =2.5•10 M + -13 January 21 [Ag ] = 3.32⋅10 M [Pb+2] = 1.25•10−3M [I-] = 2.50•10-3M Calculation: Same type problem different data

-17 At 25°C, the solubility product of AgI and Hg2I2 is 8.5•10 M and -29 - + 2+ 5.3•10 M, respectively. What are the values of [I ], [Ag ], and [Hg2 ] in a solution at equilibrium with both substances ?

AgI ! Ag+ + I- Hg I ! Hg+2 + 2I- (s) (aq) (aq) 2 2 (s) (aq) (aq) i Solid 0 2y Solid 0 x Δ -x +x +x -y +y +2y [e] - x 2Y + x - y 2y + x

Assume since K (Hg I ) << K (AgI ) x >>y sp 2 2 (s) sp (s)

-17 2 -9 AgI : 8.5 ⋅10 = x (2y + x) ≈ x ⇒ x = 9.21 ⋅10 (s)

-29 -29 2 2 5.3⋅10 −13 Hg I : 5.3⋅10 = y (2y + x) ≈ y ⋅ x ⇒ y = = 6.25 ⋅10 2 2 (s) 2 (9.21 ⋅10-9 )

+2 −13 + - -9

Hg(aq) = 6.25 ⋅10 , Ag(aq) = I(aq) = 9.21 ⋅10

33 Solubility and Solubility Product January 21 Common Ion Effect -10 B&L 17.44 (7th Ed.): The Ksp for cerium iodate, Ce(IO3)3, is 3.2 •10

a) Calculate the molar solubility of Ce(IO3)3 in pure water.

b) What concentration of NaIO3 in solution would be necessary to 3+ reduce the Ce concentration in a saturated solution of Ce(IO3)3 by a factor of 10 below that calculation in part (a).

+3 - Ce(IO3)3 (s) D Ce (aq) + 3 IO3 (aq) i Lots 0 0 D -s +s +3s e Lots +s +3s

34 Solubility and Solubility Product January 21 Common Ion Effect

-10 B&L 17.44 (7th Ed.): The Ksp for cerium iodate, Ce(IO3)3, is 3.2 •10

a) Calculate the molar solubility of Ce(IO3)3 in pure water. 3+ b) What concentration of NaIO3 in solution would be necessary to reduce the Ce concentration

in a saturated solution of Ce(IO3)3 by a factor of 10 below that calculation in part (a).

+3 - Ce(IO3)3 (s) D Ce (aq) + 3 IO3 (aq) i Lots 0 0 D -s +s +3s e Lots +s +3s

3+ - 3+ [Ce ] a) K = [Ce+3] ∗[IO ]3 b) reduce Ce by factor of 10: sp 3 10 -10 3 4 3.2⋅10-10 =1.86⋅10-4 •[IO ]3 3.2 ⋅ 10 = s ∗ 27s = 27s 3 Total -11 4 1.7204 ⋅10-6 = [IO ] 1.185 ⋅ 10 = s 3 Total −3 1.198⋅10−2M = [IO ] 1.86 ⋅ 10 M = s 3 Total [NaIO ] = 1.198⋅10−2M - [IO-] 3 3 (from part previous) [NaIO ] = 1.198⋅10−2M - (3 x [1.86⋅10-4 ]) € 3 **(same as Ce+3) [NaIO ] = 1.14 ⋅10-2 M 3

+3 ** The IO3- in solution must be the same as Ce for this condition

35 Solubility and Solubility Product January 21 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 involves adding precipitating ions to separate the unknown ions into ion groups. The groups are then analyzed further through precipitation and complex ion formation.

36 Solubility and Solubility Product January 21 Qualitative Analysis Q < Ksp, solid dissolve until Q=Ksp

Q = Ksp, equilib

Q > Ksp, ppt until Q=Ksp Grp1: Insoluble + 2+ 2+ Ag , Pb , Hg2 Grp2: Acid-Insoluble 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

37 Solubility and Solubility Product January 21 Selective Precipitation Qualitative Analysis

Q < Ksp, solid dissolve until Q=Ksp

Q = Ksp, equilib

Q > Ksp, ppt until Q=Ksp

Grp1: Insoluble Chlorides

+ 2+ 2+ Ag , Pb , Hg2 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

38 Solubility and Solubility Product January 21 Lab Practical, Qualitative Analysis Qualitative Analysis of Household Chemicals

39 Solubility and Solubility Product January 21 Selective Precipitation(2)

A solution contains 2.0 •10-4 M Ag+ and 1.5 •10-3 M Pb2+. If NaI is added, will -17 -9 AgI (Ksp = 8.51 •10 ) or PbI2 (Ksp = 9.8 •10 ) precipitate first? Specify the concentration of I- needed to begin precipitation.

+ - AgI ! Ag + I 2+ - (s) (aq) (aq ) PbI ! Pb + 2I 2 (s) (aq) (aq ) -4 i Excess 2.0 •10 2x i Excess 1.5•10-3 y Δ -y y y Δ -x x 2x -4 [e] Excess 2.0 •10 + y 2x+y -3 [e] Excess 1.5•10 + x y + 2x

40 Solubility and Solubility Product January 21 Selective Precipitation(2)

A solution contains 2.0 •10-4 M Ag+ and 1.5 •10-3 M Pb2+. If NaI is added, will -17 -9 AgI (Ksp = 8.51 •10 ) or PbI2 (Ksp = 9.8 •10 ) precipitate first? Specify the concentration of I- needed to begin precipitation. This is a Q problem PbI ! Pb2+ + 2I- 2 (s) (aq) (aq ) AgI ! Ag+ + I- i Excess 1.5•10-3 y (s) (aq) (aq ) i Excess ! 2.0 •10-4 2x Δ -x x 2x [e] Excess 1.5•10-3 +x y + 2x Δ -y y y [e] Excess 2.0 •10-4 + y 2x+y Assume y << x, furthermore assume x << 1.5•10-3

Assume y << x and 2.0 •10-4 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 8.51•10-17 < [Ag+] [I-] = [2.0 •10-4 + y] [2x + y] -17 -4 8.51•10 < [2.0 •10 ] [2x] -9 9.8•10 x = = 1.278•10-3M = x, 4•1.5•10-3 -17 8.51•10 -3 -3 2x > = 4.256•10-13, x > 2.13•10-13 M Note x = 1.5•10 therefore need second iteration for 1.5•10 +x . 2.0 •10-4 [Pb2+] = 1.5•10-3 + 1.278•10-3 = 2.778•10-3 -9 -3 -3 2 Second it−e9ration, 9.8•10 = [1.5•10 + 1.278•10 ] [2x] 9.8•10 2 -4 - = x ; x =9.39•10 -3 [I ] = 2x+y = 2x, y << x, −3 ( 4 ) 2 . 7 8 • 1 0 2x = 1.88•10 M [I-] = 4.256•10-13 M - -3 [I- ] = 2x = 1.88•10 M-3 for PbI to precipitate [I ] = y + 2x = 1.88•10 2 (s)

AgI will precipitate first. The [I-] concentration needs to be 4.26•10-13M, - -3 PbI2 will not precipitate until the [I ] concentration reaches 1.88•10 M

41 Solubility and Solubility Product January 21 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 + [Ag(NH3) ] K = 2 = 1.7 • 107 Formation Constant, Kf f + 2 [Ag ]∗[NH3] + + Ag (aq) + 2NH3(aq) D Ag(NH3)2 (aq)

- - Solubility of metal salts affected presence of Lewis Base: e.g., NH3, CN , OH € 42 Solubility and Solubility Product January 21 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

43 Solubility and Solubility Product January 21 Calculations: Formation of Complex Ion

nd 2+ Tro, End of chapter Q 109 (2 Edition): Kf Zn(NH3)4 = 2.8e9

A solution is made that is 1.1e-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 2+ 2+ Zn (aq) + 4NH3 (aq) D Zn(NH3)4 (aq)

Zn2+ + NH ! Zn(NH )2+ (aq) 3 (aq) 3 4 (aq) i 1.1•10-3 0.150 0 Δ -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(NH )2+ = 1.1•10-3M 3 4

2+ [Zn(NH ) ] x 1.1•10-3 2.8•109 = 3 4 = = [Zn+2] [NH ]4 [1.1•10-3-x] [0.150 − 1.1•10-3]4 [1.1•10-3-x] [0.1456]4 3

1.1•10-3 1.1•10-3-x = = 8.74•10-10, [Zn2+] = 8.74•10-10M 2.8•109[0.1456]4 eq

44 Solubility and Solubility Product January 21 Calculations: Formation of Complex Ion

- By using the values of Ksp for AgI and Kf for Ag(CN)2 , calculate the for the following reaction: - - - AgI(s) + 2CN (aq) D Ag(CN)2 (aq) + I (aq)

Note that this equation is the sum of

+ - -17 + - AgI (s) D Ag (aq) + I (aq) ; Ksp = 8.3•10 = [Ag ][I ]

+ - - 21 - + - 2 Ag (aq) + 2CN (aq) D Ag(CN)2 (aq) ; Kf = 1.0•10 = [Ag(CN)2 ]/[Ag ][CN ]

45 Solubility and Solubility Product January 21 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

+ - -17 + - AgI (s) D Ag (aq) + I (aq) ; Ksp = 8.3•10 = [Ag ][I ]

+ - - 21 - + - 2 Ag (aq) + 2CN (aq) D Ag(CN)2 (aq) ; Kf = 1.0•10 = [Ag(CN)2 ]/[Ag ][CN ]

- - - AgI(s) + 2CN (aq) D Ag(CN)2 (aq) + I (aq)

+ - + - 2 K = Ksp• Kf = ( [Ag ][I ] ) • ( [Ag(CN)2]/[Ag ][CN ] )

-17 21 K = Ksp• Kf = 8.3•10 • 1.0•10 K = 8.3•104

46 Solubility and Solubility Product January 21