Version 2012 Updated on 030212 Copyright © All rights reserved

Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University

Chapter 10

Solving Equilibrium Problems for Complex Systems

Erosion of carbonate stone by acid rain. The column at the left is exposed only to air inside a monument. The column at the right is exposed to acidic rainfall that is slowly dissolving the marble.

CaCO3 Ca2+ + CO32–

CO32– + H+ HCO3

–

Mass balance ; material balance

The sum of the amounts of all species containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution. The mass balance is a statement of the conservation of matter. It really refers to conservation of atoms, not to mass.

Ex. 1) 0.050 mol of acetic acid in water 1L solution

CH3COOH = CH3COO– + H+

mass balance : 0.050M = [CH3COOH] + [CH3COO– ]

2) Aqueous solution of La(IO3) 3

La(IO3) 3 = La3 + + 3 IO3–

initial solid 0 0

final solid x 3x

mass balance : [IO3–] = 3 [La3 +]

Ex. 3) 0.0100 M HCl + excess solid BaSO4

Three equilibria :

BaSO4(s) Ba2+ + SO42–

SO42– + H3O+ HSO4

– + H2O

2H2O H3O+ + OH–

Because the only source for the two sulfate species is the dissolved BaSO4, the barium ion concentration must equal the total concentration of sulfate containing species.

Mass-balance equation:

[Ba2+] = [SO42–] + [HSO4

–]

The hydronium ions in the solution can exist either as free H3O+ ions or combined with SO4

2– to form HSO4–. These hydronium ions have two

sources: HCl and the dissociation of water.

[H3O+] + [HSO4–] = cHCl + [OH–] = 0.0100 + [OH–]

Charge balance

The charge balance is an algebraic statement of electroneutrality of the solution. That is the sum of the positive charges in an electrolyte solution equals the sum of the negative charges in solution.

Solution must have zero total charge.

no. mol/L positive charge = no. mol/L negative charge

[positive charges] = [negative charges]

The general form of the charge balance for any solution:

n1[C1] + n2[C2] + … = m1[A1] + m2[A2] + ...

where ni = charge of the ith cation

[Ci] = concentration of the ith cation

mi = charge of the ith anion

[Ai] = concentration of the ith anion

Ex. 1. Charge balance of a solution containing

H+, OH–, K+, H2PO4–, HPO4

2–, PO43–

[H+] + [K+] = [OH–] + [H2PO4–] + 2[HPO4

2–] + 3[PO4

3–]

Ex. 2. Charge balance of a solution containing

H2O, H+, OH–, ClO4–, Fe(CN)6

3–, CN–, Fe3+, Mg2+, CH3OH, HCN, NH3, NH4+

[H+] + 3[Fe3+] + 2[Mg2+] + [NH4+] = [OH–] + [ClO4

–] + 3[Fe(CN)63–] + [CN–]

Neutral species (H2O, CH3OH, HCN, and NH3) do not appear in the charge balance.

Ex. 0.010 M NH3 solution saturated with AgBr

Pertinent equilibria:

AgBr(solid) Ag+ (aq)

+ Br– (aq)

Ag+ (aq)

+ 2NH3 (aq) Ag(NH3)+

(aq)

Ag(NH3)+ + NH3 Ag(NH3)2

+

NH3 + H2O NH4+ + OH–

2H2O H3O+ + OH–

Mass balance equations:

[Ag+] + [Ag(NH3)+] + [Ag(NH3)2

+] = [Br–]

cNH3 = [NH3] + [NH4+] + [Ag(NH3)

+] + [Ag(NH3)2+] = 0.010

[OH–] = [NH4+] + [H3O+]

Charge balance :

[Ag+] + [Ag(NH3)+] + [Ag(NH3)2

+] + [NH4+] + [H3O+] = [OH–] + [Br–]

Systematic steps for solving problems involving several equilibria

1.Write balanced chemical equations for all pertinent equilibria.

2.Define the unknown.

3.Write all equilibrium-constant expressions.

4.Write mass balance expressions

5.Write the charge balance expression

6.Count the number of independent equations and unknowns

7.Make approximations

8.Solve the equations

9.Check the assumptions.

Ex. Calculate the molar solubility of Mg(OH)2 in water.

1. Mg(OH)2 (solid)

Mg+2 (aq)

+ 2OH– (aq)

2H2O H3O+ + OH–

2. Solubility Mg = [Mg+2]

3. Ksp = [Mg+2][OH–]2 = 7.1 10–12

Kw = [H3O+][OH–] = 1.0 10–14

4. Mass balance : [OH–] = 2[Mg+2] + [H3O+]

5. Charge balance: [OH–] = 2[Mg+2] + [H3O+]

6. Three independent algebraic equations and three unknowns

7. Ksp of Mg(OH)2 is relatively large: [H3O+] << [OH–] 2[Mg+2] [OH–]

8. Ksp = [Mg+2][OH–]2 = [Mg+2](2[Mg+2])2 = 7.1 10–12

[Mg+2] = {(7.1 10–12) /4}1/3= 1.21 10–4

9. Validation: [OH–] 2[Mg+2] = 2 1.21 10–4

[H3O+] =(1.0 10–14) / (2 1.21 10–4) = 4.1 10–11

Systematic treatment of equilibrium

General prescription :

Step 1. Write all the pertinent

chemical reactions.

Step 2. Write the charge balance.

Step 3. Write the mass balance.

Step 4. Write the equilibrium

constant for each chemical

reaction.

Step 5. Count the equations and

unknowns.

By working with n equations and n unknowns, the problem can be solved.

Step 6. By hook, or by crook,

solve all the unknowns.

Ex. Ionization of water

1. H2O = H+ + OH–

2. [H+ ] = [OH–]

3. [H+ ] = [OH–]

4. Kw = [H+ ]f[OH–]f = 1.0 ×10–14

5. Two equations, two unknowns

6. [H+ ]f[OH–]f

= [H+ ](1)[H+](1)= 1.0 × 10–14

[H+ ] = 1.0 × 10–7

pH = – logA H+ = – log [H+ ] f= 7.00

Solubility of mercurous chloride

Step 1. Hg2Cl2 = Hg22+ + 2Cl– , H2O = H+ + OH–

Step 2. [H+ ] + 2 [Hg22+] = [Cl–] + [OH–]

Step 3. [Cl–] = 2[Hg22+]

Step 4. Ksp = [Hg22+][Cl–]2 = (x)(2x)2 = 4x3 =1.2 × 10-18

Kw = [H+ ][OH–] = 1.0 × 10–14

Step 5. Four equations, four unknowns

Step 6. [H+] = [OH–] = 1.0 × 10–7

Ksp = [Hg22+][Cl–]2

= [Hg22+](2 [Hg2

2+])2 = 1.2 × 10-18

[Hg22+] = x =[(1.2×10-18)/4]1/3 = 6.7 × 10-7 M

[Cl–] = 2 x = 1.34 × 10-6 M

Dissociation equilibria : HA = A– + H+ Case Major supplier Necessary

initial F 0 0 of H+ condition

final F–x x x 1 Weak acid KaF >> Kw

H2O = H+ + OH– 2 Both KaF = Kw

Charge balance : [H+] = [A–] + [OH–] 3 Water KaF << Kw

Mass balance : F = [HA] + [A–]

Dissociation equilibria constants : Ka = [A–][H+] / [HA] Kw = [H+][OH–]

Four equarions, four unknowns

If the acid dissociation is much greater than the water dissociation, [A–] >>[OH–],

[A–] [H+] = x

Ka = [A–][H+] / [HA] = (x)(x) / (F –x)

(F –x) Ka = x2 x2 + Kax – KaF = 0

x =( – Ka±Ka2 –4KaF) / 2

If assume F >>x , F –x F .

x2 / (F –x) x2 / F = Ka x = [H+] = KaF

Key points

When F/Ka>>103

relative error 1.6%

Hydronium ion concentrations of weak acids

Dissociation equilibria : HA = A– + H+

initial F 0 0

final F–x x x

H2O = H+ + OH–

Charge balance : [H+] = [A–] + [OH–]

Mass balance : F = [HA] + [A–]

Dissociation equilibria constants : Ka = [A–][H+] / [HA]

Kw = [H+][OH–]

Four equations, four unknowns

Ka = [A–][H+] / [HA] = ([H+] –[OH–])[H+] / (F –[H+] + [OH–] )

KaF – Ka[H+] + Ka[OH–] = [H+]2 – [OH–][H+]

[H+]2 + Ka[H+] – (Kw + KaF + KaKw/ [H+] )= 0

[H+] = { – Ka + Ka2 + 4 (Kw + KaF + KaKw/ [H+] )} / 2

Dissociation equilibria : B + H2O = BH+ + OH–

initial F 0 0

final F–x x x

H2O = H+ + OH–

Charge balance : [BH+ ] + [H+] = [OH–]

Mass balance : F = [B] + [BH+]

Dissociation equilibria constants : Kb = ABH+ AOH– /AB [BH+] [OH–] / [B]

Kw = [H+][OH–]

Four equarions, four unknowns

Kb = [BH+] [OH–] / [B] = (x)(x) / (F –x)

(F –x) Kb = x2 x2 + Kbx – KbF = 0 x = [OH–] = (– Kb+Kb2 +4KbF) / 2

If assume F >>x , F –x F .

x2 / (F –x) x2 / F = Kb x = [OH–] = KbF

[H+] = Kw / [OH–] = Kw / KbF = Kw / (KwF /Ka) = (KwKa/F)

Weak bases

Key points

The effect of pH on solubility

Molar solubility of calcium oxalate at pH = 4.00

Step 1. CaC2O4 Ca2+

(aq) + C2O4

2 – (aq)

H2C2O4 + H2O H3O+ + HC2O4–

HC2O4– + H2O H3O+ + C2O4

2 –

2H2O = H3O+ + OH–

Step 2. Solubility = [Ca2+]

Step 3. Ksp = [Ca2+][C2O42 –] = 1.7 × 10–9

Ka1 = [H3O+][HC2O4–] / [H2C2O4] = 5.60 × 10–2

Ka2 = [H3O+][C2O42 –] / [HC2O4

– ] = 5.42 × 10–5

Kw = [H3O+][OH–] = 1.0 × 10–14

Step 4. Mass balance : [Ca2+] = [C2O42 –] + [HC2O4

– ] + [H2C2O4] = solubility

if pH = 4.00 [H3O+] = 1.00 × 10–4

H2C2O4

Step 8. [HC2O4– ] = [H3O+][C2O4

2 –] / Ka2

=(1.00 ×10–4)[C2O42 –] / (5.42 × 10–5) = 1.85 [C2O4

2 –]

[H2C2O4] = [H3O+][HC2O4–] / Ka1 = [H3O+](1.85 [C2O4

2 –]) / Ka1

= (10–4) × (1.85 × [C2O42 –]) / 5.60 × 10–2

= 3.30 × 10–3 × [C2O42 –]

[Ca2+] = [C2O42 –] + [HC2O4

– ] + [H2C2O4]

= [C2O42 –] + 1.85 [C2O4

2 –] + 3.30 × 10–3 [C2O42 –]

= 2.85 [C2O42 –]

or [C2O42 –] = [Ca2+] / 2.85

Ksp = [Ca2+][C2O42 –] = [Ca2+][Ca2+] / 2.85 = 1.7 × 10–9

[Ca2+] = solubility = (2.85 × 1.7 × 10–9)1/2

= 7.0 × 10–5

Dependence of solubility on pH

Solubility of CaF2

Step 1. CaF2 = Ca2+ + 2F– , F– + H2O = HF + OH– , H2O = H+ + OH–

Step 2. [H+ ] + 2[Ca2+] = [OH–] + [F–]

Step 3. [F–] + [HF] = 2[Ca2+]

Step 4. Ksp = [Ca2+][F–]2 = (x)(2x)2 = 4x3 = 3.9×10-11

Kb = [HF][OH–] / [F–]= 1.5 ×10-11, Kw = [H+ ][OH–] = 1.0 ×10–14

Step 5. Five equations, five unknowns

Step 6. If pH=3.00 [H+] = 1.0 ×10–3 ,

[OH–] = Kw/ [H+ ] = 1.0 ×10–14 / 1.0 ×10–3 = 1.0 ×10–11

Kb = [HF][OH–] / [F–] [HF] = [F–] Kb / [OH–]

= 1.5 ×10-11 [F–] / 1.0 ×10–11 = 1.5 [F–]

[F–] + [HF] = 2[Ca2+] [F–] + 1.5[F–] = 2[Ca2+] [F–]=0.80[Ca2+]

Ksp = [Ca2+][F–]2 = [Ca2+](0.80[Ca2+] )2 = 3.9×10-11

[Ca2+] = 3.9×10-4 M

(left) Transmission electron micrograph of normal human tooth enamel, showing crystals of hydroxyapatite. (right) Decay enamel, showing regions where the mineral dissolved in acid.

Tooth enamel contains the mineral hydroxyapatite, a calcium hydroxyphosphate. This slightly soluble mineral dissolves in acid, because both phosphate and hydroxyl ions react with hydrogen ion.

Ca10(PO4)6(OH)2 (s) 10Ca2+ + 6 H2PO4– + 2H2O

Decay causing bacteria that adhere to tooth produce lactic acid { CH3CH(OH)COOH } by metabolizing sugar. Lactic acid lowers the pH at the surface of the tooth below 5. Fluoride inhibits tooth decay because it forms fluoroapatite {Ca10(PO4)6F2 } which is less soluble and more acid resistant than hydroxyapatite.

Solubility of precipitates in the presence of complexing agents

Al(OH)3 (solid)

Al+3 (aq)

+ 3(OH)3–

(aq)

Al+3 (aq)

+ 6F– (aq)

AlF6–3

(aq)

soluble complex

AgCl (solid) AgCl (aq)

AgCl (aq) Ag+ + Cl–

AgCl (solid) + Cl– AgCl2

–

AgCl2– + Cl– AgCl3

2–

The effect of [Cl– ] on the solubility of AgCl.

Solubility of Sulfide (example : HgS )

Step 1. HgS = Hg2+ +S2– , S2– + H2O = HS– + OH– ,

HS– + H2O = H2S + OH– , H2O = H+ + OH–

Step 2. 2[Hg2+ ] + [H1+] = 2[S2–]+[HS–]+[OH–]

Step 3. [Hg2+ ] = [S2–]+[HS–]+[H2S]

Step 4. Ksp =[ Hg2+][S2–] = 5 ×10-54

Kb1 = [HS–][OH–] / [S2 –] = 0.80, Kb2 = [H2S][OH–] / [HS–]= 1.1 ×10-7,

Kw = [H+ ][OH–] = 1.0 ×10–14

Step 5. six equations, six unknowns

Step 6. If pH=8.00 [OH–] = 1.0 ×10–6 ,

[H2S]= Kb2 [H2S] / [OH–] = 0.11 [HS–]

[HS–]= Kb1 [S2 –] / [OH–] =8.0 ×105 [S2 –]

[Hg2+ ] = [S2–]+[HS–]+[H2S]

= [S2–]+ 8.0 ×105 [S2 –]+0.11 × 8.0 ×105 [S2 –]

=(8.88 ×105 [S2 –]

Ksp = [Hg2+][S2 –] = [Hg2+]{[Hg2+]/ (8.88 ×105 )}= 5 ×10-54

[Hg2+] = 2.1×10–24 M

Sulfide ion concentration as a function of pH in a saturated H2S solution.

H2S is colorless, flammable gas with important chemical toxicological properties. Its noxious odor of rotten eggs permit its detection at extremely low concentration (0.02 ppm). Because the olfactory sense is dulled by its action, however, higher concentrations may be tolerated, and the lethal concentration of 100 ppm may be exceeded.

Summary

Mass balance

Charge balance

Systematic calculation of equilibria

The effect of pH on solubility

Solubility of precipitates in the presence of complexing agents

Solubility of Sulfide