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


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 !!!


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.


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.


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.


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]


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


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


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).


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


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


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


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.


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.


...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)


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


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%


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


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


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


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


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


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.


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+


Lab Practical, Qualitative Analysis Qualitative Analysis of Household Chemicals


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


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


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


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


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

14.4 (chp16) solubility and solubility...

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