chapter 11: solving equilibrium problems for complex systems 1/30

Chapter 11: Solving Equilibrium Problems for

Complex Systems

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For simultaneous equilibria in aqueous solutions, BaSO4(s) in water for example, there are three equilibria:

BaSO4(s) Ba2+ + SO42- (1)

SO42- + H3O+ HSO4

- +H2O (2)2H2O H3O+ + OH- (3)

The addition of H3O+ causes: (2) shift right and (1) shift right.

since Ba2+ + OAc- BaOAc+(4)The addition of OAc- causes:(1) shift right*The introduction of a new equilibrium system into a solution does not change the equilibrium constants for any existing equilibria.

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11 A Solving multiple-equilibrium problems using A systematic methodThree types of algebraic equations are used to solve multiple-equilibrium problems:(1) equilibrium-constant expressions(2) mass-balance equations(3) a single charge-balance equation

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11 A-1 Mass-Balance EquationsMass-balance equations: The expression that relate the equilibrium concentrations of various species in a solution to one another and to the analytical concentrations of the various solutes. These equations are a direct result of the conservation of mass and moles.

A weak acid HA dissolved in water for example:HA+ H2O H3O+ + A- (1)2H2O H3O+ + OH- (2)

mass equation 1: cHA = [HA] + [A-]cHA is analytical concentration, [HA] and [A-] are

equilibrium concentration.mass equation 2: [H3O+] = [A-] + [OH-]since [H3O+] = [H3O+]from HA + [H3O+]from H2O ,where [H3O+]from HA = [A-] , [H3O+]from H2O = [OH-]

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*

*

5/30: conc. of H3O+ at equilibrium

*

*

*

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11 A-2 Charge-Balance EquationCharge-Balance Equation: An expression relating the concentrations of anions and cations based on charge neutrality in any solution.Charge balance equation: n1[C1

+n1] + n2[C2+n2] + ..... = m1[A1

-m1] + m2[A2-m2] + .....

Example: A solution contains H+, OH–, K+, H2PO4–, HPO4

2– ， , and PO4

3– ， what is the charge balance equation?Solution:[H+] + [K+] = [H2PO4

–] + 2[HPO42–] + 3[PO4

3–] + [OH–]

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+ [Cl-]

Figure 11-1 A systematic method for solving multiple-equilibrium problems.

11A-3 Steps for solving problems with several equilibria

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11A-4 Using Approximations to Solve Equilibrium CalculationsApproximations can be made only in charge-balance and mass-balance equations, never in equilibrium-constant expressions.If the assumption leads to an intolerable error, recalculate without the faulty approximation to arrive at a tentative answer.

11A-5 Use of Computer Programs to Solve Multiple-Equilibrium ProblemsSeveral software packages are available for solving multiple nonlinear simultaneous equations include Mathcad, Mathematica, Solver, MATLAB, TK, and Excel.

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A simple example of systematic calculationsQ Calculate [H3O+] and [OH-] in pure water

A Step 1: 2H2O H3O+ + OH-

Step 2: [H3O+]=? and [OH-]=? 2 unknowns

Step 3: [H3O+][OH-] = 1x10-14 (1)

Step 4: mass-balance equation:[H3O+]=[OH-] (2)

Step 5: charge-balance equation:[H3O+]=[OH-] (3)

Step 6: equations (2) and (3) are identical, omit equation (3)two unknowns two different equations (1) and (2), OK

Step 7: Approximation, omitStep 8: equation (2) substitute into equation (1)

[H3O+] [OH-] = [H3O+]2 = 1x10-14

∴ [H3O+] = 1x10-7 and [OH-] = 1x10-7

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11B Calculating solubilities by the systematic method11B-1 Solubility of metal hydroxides for High Ksp value, pH controlled by the solubility

Example 11-5Calculate the molar solubility of Mg(OH)2 in water.

Solution

2+ -2(s) (aq) (aq)

+ -2 (l) 3 (aq) (aq)

2+ + -3

Step 1. Balanced equations

Mg(OH) Mg +2OH

2H O H O +OH

Step 2. Unknown(s)?

s=[Mg ]=? [H O ]=? [OH ]=?

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2+ - 2 -12sp

+ - 14w 3

-

Step 3. Equilibrium-constant expressions

K =[Mg ][OH ] =7.1x10 (1)

K =[H O ][OH ]=1x10 (2)

Step 4. Mass-balance equation(s)

[OH

2+ +3

- 2+ +3

]=2[Mg ] [H O ] (3)

Step 5. Charge-balance equation

[OH ]=2[Mg ] [H O ] (4)

identical to equation (3)

Step 6. Independent equations and unkn

owns

3 unknowns and 3 independent equations, OK

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+ 2+3

- 2+

Step 7. Make approximations

assume [H O ] <<2[Mg ], equation (3) simplifies to

[OH ] 2[Mg ] (5)

Step 8. Solve the equations

subst

2+ - 2 2+ 2+ 2 -12sp

2+ -4

- -4

itute (5) into (1)

K =[Mg ][OH ] =[Mg ](2[Mg ]) =7.1x10

[Mg ] s=1.21x10 M

substitute into (5) [OH ]=2.42x10 M

substitute into (2)

+ -113

+ 2+3

[H O ]=4.1x10 M

Step 9. Check

[H O ] << 2[Mg ] OK

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for Low Ksp value, pH ≈7, controlled by autoprotolysis of water

Example 11-6Calculate the molar solubility of Fe(OH)3 in water.

Solution

3+ -3(s) (aq) (aq)

+ -2 (l) 3 (aq) (aq)

3+ + -3


Fe(OH) Fe +3OH

2H O H O +OH

Step 2. Unknown(s)?

s=[Fe ]=? [H O ]=? [OH ]=?

Step 3. Equilibrium-con

3+ - 3 -39sp

+ - 14w 3

stant expressions

K =[Fe ][OH ] =2x10 (1)

K =[H O ][OH ]=1x10 (2)

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- 3+ +3

- 3+ +3


[OH ]=3[Fe ] [H O ] (3)


[OH ]=3[Fe ] [H O ] (4)

identical to

3+ +3

equation (3)

Step 6. Independent equations and unknowns



assume 3[Fe ]<<[H O ], equation (3) simplifi- +

3

es to

[OH ] [H O ] (5)

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- + -73

3+ - 3 3+ -7 3 -39sp


substitute (5) into (2)

[OH ]=[H O ]=1.0x10 M

substitute into (1)

K =[Fe ][OH ] =[Fe ](1.0x10 ) =2x10

3+ -18

3+ +3

[Fe ] s=2x10 M

Step 9. Check

3[Fe ] [H O ] OK

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11B-2 The Effect of pH on Solubility*The solubility of precipitates containing an anion with basic

properties, a cation with acidic properties, will depend on pH. (simultaneous equilibria)

Solubility Calculations When the pH Is Constant ([OH-] and [H3O+] are known)

Example 11-7Calculate the molar solubility of calcium oxalate in a solution

that has been buffered so that its pH is constant and equal to 4.00.

Solution

2+ 2-2 4(s) (aq) 2 4(aq)

+ -2 2 4(aq) 2 3 (aq) 2 4(aq)

- + 2-2 4(aq) 2 3 (aq) 2 4(aq)


CaC O Ca +C O

H C O +H O H O +HC O

HC O +H O H O +C O

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2+ 2- -2 4 2 4 2 2 4

2 2 192 4 sp

Step 2. Unknown(s)

s=[Ca ]=? [C O ]=? [HC O ]=? [H C O ]=?

Step 3. Equilibrium-constant expressions

[Ca ][C O ] K 1.7 10 (1)

[H

23 2 41

2 2 4

253 2 4

22 4

2+ 2- -2 4 2 4 2 2

O ][HC O ]K 5.60 10 (2)

[H C O ]

[H O ][C O ] K 5.42 10 (3)

[HC O ]


[Ca ] = s = [C O ]+[HC O ]+[H C

4O ] (4)


buffer is not specified, omit

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Step 6. Independent equations and unknowns



omit


from equat2- -4 2-

- 2-3 2 4 2 42 4 2 4-5

2

- -4 2-3 2 4 2 4

2 2 4 -21

ion (3)

[ ][ ] 1.00 10 [ ] [ ] 1.85[ ] (5)

5.42 10

substitute (5) into (2)

[ ][ ] 1.00 10 1.85[ ] [ ]

5.60 10

H O C O C OHC O C O

K

H O HC O C OH C O

K

-3 2-

2 4 3.30 10 [ ] (6)C O

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2 2- 2- -3 2- 2-2 4 2 4 2 4 2 4

22 2- 2 9

2 4

2

substitute (5) and (6) into (4)

[ ] [ ] 1.85[ ] 3.30 10 [ ] 2.85[ ]

[Ca ] [Ca ][ ] [Ca ] =1.7 10

2.85

[Ca ] 7.0

sp

Ca C O C O C O C O

C O K

s

5

2- 5 - 52 4 2 4

82 2 4

10

Also

[C O ]=2.46x10 M [HC O ]=4.54x10 M

[H C O ]=8.11x10 M

M

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Solubility Calculations When the pH Is Variable ([OH-] and [H3O+] are unknown)

for High Ksp value (pH controlled by the solubility)omit

for Low Ksp value (pH≈7, controlled by autoprotolysis of water)omit

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11B-3 The Effect of Undissociated Solutes on Precipitation Calculations

-7(s) (aq) s

+ - -4(aq) (aq) (aq) d

+ - -10(s) (aq) (aq) sp s d

AgCl AgCl K =3.6x10

AgCl Ag + Cl K =5.0x10

AgCl Ag + Cl K =K K 1.8x10

For example, a saturated solution of AgCl(s) contains significant amounts of undissociated silver chloride molecules, AgCl(aq) complexs:

Example 11-8Calculate the solubility of AgCl in distilled water.Solution

+(aq) (aq)

-7 -7 5 5

(aq)

solubility = s = [AgCl ] + [Ag ]

= 3.6x10 = = 3.6x10 1.34x10 = 1.38x10 M

neglecting [AgCl ] leads to -3% error.

spK

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11B-4 The Solubility of Precipitates in the Presence of Complexing Agents

The solubility increase in the presence of reagents that form complexes with the anion or the cation of the precipitate.Ex. F- prevent the precipitation of Al(OH)3

3+ -3(s) (aq) (aq)

3+ - 3-(aq) (aq) 6(aq)

Al(OH) Al + 3OH

Al + 6F AlF

for High stability constantomit

for Low High stability constantomit

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• Complex formation with a common ion to the precipitate may increase in solubility by large excesses of a common ion.

Example

(s) (aq)

+ -(aq) (aq) (aq)

+ -(s) (aq) (aq)

-(s) (aq) 2(aq)

- 22(aq) (aq) 3(aq)

Given:

AgCl AgCl

AgCl Ag + Cl

AgCl Ag + Cl

AgCl + Cl AgCl

AgCl + Cl AgCl

What is the the concentration of KCl at which the solubility of AgCl is a minimum?Solutionomit

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Figure 11-2 The effect of chloride ion concentration on the solubility of AgCl. The solid curve sows the total concentration of dissolved AgCl. The broken lines show the concentrations of the various silver-containing species.

CKCl = 0.003 M26/30

11C Separation of ions by control of the concentration of the precipitating agent

11C-1 Calculation of the feasibility of separationsGenerally, complete precipitation is considered as 99.9% of the target ion is precipitated, i.e., 0.1% left.

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[Fe3+][OH-]3 = 2x10-39

[Mg2+][OH-]2 = 7.1x10-12

Fe3+ 會先沈澱設剩餘 0.1% 之 Fe3+ 為 complete precipitation for Fe(OH)3 ，則 [Fe3+] = 0.1x0.1% = 1x 10-4 M1 x 10-4 x [OH-]3 = 2x10-39

[OH-] = 3 x 10-12 M 完全沈澱 Fe3+所需之 [OH-]

Mg(OH)2 開始沉澱之 [OH-] ：0.1 x [OH-]2 = 7.1x10-12

[OH-] = 8.4 x 10-6 M 　 Mg2+開始沈澱之 [OH-]

控制水溶液之 [OH-] = 3 x 10-12 ~ 8.4 x 10-6 M ，可將 0.1 M 的 Fe3+ 與 0.1 M 的 Mg2+ 分離。

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11C-2 Sulfide Separations

832 2 3 1

2

22 143

2 3 2 -

2 23

1 22

2 2 83 2 1 2

[H O ][HS ]H S H O H O + HS K 9.6 10 (1)

[H S]

[H O ][S ]HS H O H O + S K 1.3 10 (2)

[HS ]

[H O ] [S ](1) (2) =K K

[H S]

[H O ] [S ] [H S]K K (0.10)(9.6 10 )(1.3 10

2

14 2

22

23

2

1.2 10[S ]

) 1.2 10

[H O ]

x

x

Saturated H2S, [H2S](aq) = 0.1 M

[S2-] in H2S saturated solution depend on the pH

2sp

2+ 2- 2 2(s) (aq) (

3222

aq) sp

sp2-

K [H O ][M

MS M + S K =[M ][S ]

K =] = solubility =

[S . 0]

1 2 1

Metal sulfide solubility for (M2+S2-) in saturated H2S

* The solubility of MS in H2S saturated solution depend on the pH29/30

Homework (Due 2014/11/27)

Skoog 9th edition, Chapter 11, Questions and Problems11-5 (e) (g)11-6 (e) (g)11-7 (a)11-14

End of Chapter 11

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chapter 11: solving equilibrium problems for complex systems 1/30

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