Version 2012 Updated on 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. CaCO 3  Ca 2+ + CO 3 2– CO 3 2– + H +  HCO 3 –

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) mol of acetic acid in water  1L solution CH 3 COOH = CH 3 COO – + H + mass balance : 0.050M = [CH 3 COOH] + [CH 3 COO – ] 2) Aqueous solution of La(IO 3 ) 3 La(IO 3 ) 3 = La IO 3 – initial solid 0 0 final solid x 3x mass balance : [IO 3 – ] = 3 [La 3 + ]

Ex. 3) M HCl + excess solid BaSO 4 Three equilibria : BaSO 4 (s)  Ba 2+ + SO 4 2– SO 4 2– + H 3 O +  HSO 4 – + H 2 O 2H 2 O  H 3 O + + OH – Because the only source for the two sulfate species is the dissolved BaSO 4, the barium ion concentration must equal the total concentration of sulfate containing species. Mass-balance equation: [Ba 2+ ] = [SO 4 2– ] + [HSO 4 – ] The hydronium ions in the solution can exist either as free H 3 O + ions or combined with SO 4 2– to form HSO 4 –. These hydronium ions have two sources: HCl and the dissociation of water. [H 3 O + ] + [HSO 4 – ] = c HCl + [OH – ] = [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: n 1 [C 1 ] + n 2 [C 2 ] + … = m 1 [A 1 ] + m 2 [A 2 ] +... where n i = charge of the ith cation [C i ] = concentration of the ith cation m i = charge of the ith anion [A i ] = concentration of the ith anion

Ex. 1. Charge balance of a solution containing H +, OH –, K +, H 2 PO 4 –, HPO 4 2–, PO 4 3– [H + ] + [K + ] = [OH – ] + [ H 2 PO 4 – ] + 2[ HPO 4 2– ] + 3[ PO 4 3– ] Ex. 2. Charge balance of a solution containing H 2 O, H +, OH –, ClO 4 –, Fe(CN) 6 3–, CN –, Fe 3+, Mg 2+, CH 3 OH, HCN, NH 3, NH 4 + [H + ] + 3[Fe 3+ ] + 2[Mg 2+ ] + [NH 4 + ] = [OH – ] + [ClO 4 – ] + 3[Fe(CN) 6 3– ] + [CN – ] Neutral species (H 2 O, CH 3 OH, HCN, and NH 3 ) do not appear in the charge balance.

Ex M NH 3 solution saturated with AgBr Pertinent equilibria: AgBr (solid)  Ag + (aq) + Br – (aq) Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) + (aq) Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2 + NH 3 + H 2 O  NH OH – 2H 2 O  H 3 O + + OH – Mass balance equations: [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] = [Br – ] c NH3 = [NH 3 ] + [NH 4 + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] = [OH – ] = [NH 4 + ] + [H 3 O + ] Charge balance : [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] + [NH 4 + ] + [H 3 O + ] = [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) 2H 2 O  H 3 O + + OH – 2. Solubility Mg = [Mg +2 ] 3. Ksp = [Mg +2 ][OH – ] 2 = 7.1  10 –12 Kw = [H 3 O + ][OH – ] = 1.0  10 –14 4. Mass balance : [OH – ] = 2[Mg +2 ] + [H 3 O + ] 5. Charge balance: [OH – ] = 2[Mg +2 ] + [H 3 O + ] 6. Three independent algebraic equations and three unknowns 7. Ksp of Mg(OH) 2 is relatively large: [H 3 O + ] << [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 [H 3 O + ] =(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. H 2 O = H + + OH – 2. [H + ] = [OH – ] 3. [H + ] = [OH – ] 4. K w = [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. Hg 2 Cl 2 = Hg Cl –, H 2 O = H + + OH – Step 2. [H + ] + 2 [Hg 2 2+ ] = [Cl – ] + [OH – ] Step 3. [Cl – ] = 2[Hg 2 2+ ] Step 4. Ksp = [Hg 2 2+ ][Cl – ] 2 = (x)(2x) 2 = 4x 3 =1.2 × K w = [H + ][OH – ] = 1.0 × 10 –14 Step 5. Four equations, four unknowns Step 6. [H + ] = [OH – ] = 1.0 × 10 –7 Ksp = [Hg 2 2+ ][Cl – ] 2 = [Hg 2 2+ ](2 [Hg 2 2+ ]) 2 = 1.2 ×  [Hg 2 2+ ] = x =[(1.2× )/4] 1/3 = 6.7 × M [Cl – ] = 2 x = 1.34 × 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 K a F >> K w H 2 O = H + + OH – 2 Both K a F = K w Charge balance : [H + ] = [A – ] + [OH – ] 3 Water K a F << K w Mass balance : F = [HA] + [A – ] Dissociation equilibria constants : K a = [A – ][H + ] / [HA] K w = [H + ][OH – ] Four equarions, four unknowns If the acid dissociation is much greater than the water dissociation, [A – ] >>[OH – ], [A – ]  [H + ] = x K a = [A – ][H + ] / [HA] = (x)(x) / (F –x) (F –x) K a = x 2  x 2 + K a x – K a F = 0 x =( – K a ±  K a 2 –4K a F) / 2 If assume F >>x, F –x  F.  x 2 / (F –x)  x 2 / F = K a  x = [H + ] =  K a F Key points  When F/K a >>10 3 relative error 1.6% Hydronium ion concentrations of weak acids  

Dissociation equilibria : HA = A – + H + initial F 0 0 final F–x x x H 2 O = H + + OH – Charge balance : [H + ] = [A – ] + [OH – ] Mass balance : F = [HA] + [A – ] Dissociation equilibria constants : K a = [A – ][H + ] / [HA] K w = [H + ][OH – ] Four equations, four unknowns K a = [A – ][H + ] / [HA] = ([H + ] –[OH – ])[H + ] / (F –[H + ] + [OH – ] ) K a F – K a [H + ] + K a [OH – ] = [H + ] 2 – [OH – ][H + ] [H + ] 2 + K a [H + ] – (K w + K a F + K a K w / [H + ] )= 0  [H + ] = { – K a +  K a (K w + K a F + K a K w / [H + ] )} / 2

Dissociation equilibria : B + H 2 O = BH + + OH – initial F 0 0 final F–x x x H 2 O = H + + OH – Charge balance : [BH + ] + [H + ] = [OH – ] Mass balance : F = [B] + [BH + ] Dissociation equilibria constants : K b = A BH+ A OH– /A B  [BH + ] [OH – ] / [B] K w = [H + ][OH – ] Four equarions, four unknowns K b = [BH + ] [OH – ] / [B] = (x)(x) / (F –x) (F –x) K b = x 2  x 2 + K b x – K b F = 0 x = [OH – ] = (– K b +  K b 2 +4K b F) / 2 If assume F >>x, F –x  F.  x 2 / (F –x)  x 2 / F = K b  x = [OH – ] =  K b F  [H + ] = K w / [OH – ] = K w /  K b F = K w /  (K w F /K a ) =  (K w K a /F) Weak bases Key points 

The effect of pH on solubility Molar solubility of calcium oxalate at pH = 4.00 Step 1. CaC 2 O 4  Ca 2+ (aq) + C 2 O 4 2 – (aq) H 2 C 2 O 4 + H 2 O  H 3 O + + HC 2 O 4 – HC 2 O 4 – + H 2 O  H 3 O + + C 2 O 4 2 – 2H 2 O = H 3 O + + OH – Step 2. Solubility = [Ca 2+ ] Step 3. K sp = [Ca 2+ ][C 2 O 4 2 – ] = 1.7 × 10 –9 K a1 = [H 3 O + ][HC 2 O 4 – ] / [H 2 C 2 O 4 ] = 5.60 × 10 –2 K a2 = [H 3 O + ][C 2 O 4 2 – ] / [HC 2 O 4 – ] = 5.42 × 10 –5 K w = [H 3 O + ][OH – ] = 1.0 × 10 –14 Step 4. Mass balance : [Ca 2+ ] = [C 2 O 4 2 – ] + [HC 2 O 4 – ] + [H 2 C 2 O 4 ] = solubility if pH = 4.00  [H 3 O + ] = 1.00 × 10 –4 H2C2O4H2C2O4

Step 8. [HC 2 O 4 – ] = [H 3 O + ][C 2 O 4 2 – ] / K a2 =(1.00 ×10 –4 )[C 2 O 4 2 – ] / (5.42 × 10 –5 ) = 1.85 [C 2 O 4 2 – ] [H 2 C 2 O 4 ] = [H 3 O + ][HC 2 O 4 – ] / K a1 = [H 3 O + ](1.85 [C 2 O 4 2 – ]) / K a1 = (10 –4 ) × (1.85 × [C 2 O 4 2 – ]) / 5.60 × 10 –2 = 3.30 × 10 –3 × [C 2 O 4 2 – ]  [Ca 2+ ] = [C 2 O 4 2 – ] + [HC 2 O 4 – ] + [H 2 C 2 O 4 ] = [C 2 O 4 2 – ] [C 2 O 4 2 – ] × 10 –3 [C 2 O 4 2 – ] = 2.85 [C 2 O 4 2 – ] or [C 2 O 4 2 – ] = [Ca 2+ ] / 2.85  K sp = [Ca 2+ ][C 2 O 4 2 – ] = [Ca 2+ ][Ca 2+ ] / 2.85 = 1.7 × 10 –9  [Ca 2+ ] = solubility = (2.85 × 1.7 × 10 –9 ) 1/2 = 7.0 × 10 –5

Dependence of solubility on pH Solubility of CaF 2 Step 1. CaF 2 = Ca F –, F – + H 2 O = HF + OH –, H 2 O = H + + OH – Step 2. [H + ] + 2[Ca 2+ ] = [OH – ] + [F – ] Step 3. [F – ] + [HF] = 2[Ca 2+ ] Step 4. Ksp = [Ca 2+ ][F – ] 2 = (x)(2x) 2 = 4x 3 = 3.9× K b = [HF][OH – ] / [F – ]= 1.5 ×10 -11, K w = [H + ][OH – ] = 1.0 ×10 –14 Step 5. Five equations, five unknowns Step 6. If pH=3.00  [H + ] = 1.0 ×10 –3, [OH – ] = K w / [H + ] = 1.0 ×10 –14 / 1.0 ×10 –3 = 1.0 ×10 –11 K b = [HF][OH – ] / [F – ]  [HF] = [F – ] K b / [OH – ] = 1.5 × [F – ] / 1.0 ×10 –11 = 1.5 [F – ] [F – ] + [HF] = 2[Ca 2+ ]  [F – ] + 1.5[F – ] = 2[Ca 2+ ]  [F – ]=0.80[Ca 2+ ] Ksp = [Ca 2+ ][F – ] 2 = [Ca 2+ ](0.80[Ca 2+ ] ) 2 = 3.9×  [Ca 2+ ] = 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. Ca 10 (PO 4 ) 6 (OH) 2 (s)  10Ca H 2 PO 4 – + 2H 2 O 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 {Ca 10 (PO 4 ) 6 F 2 } 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)  AlF 6 –3 (aq) soluble complex AgCl (solid)  AgCl (aq) AgCl (aq)  Ag + + Cl – AgCl (solid) + Cl –  AgCl 2 – AgCl 2 – + Cl –  AgCl 3 2– The effect of [Cl – ] on the solubility of AgCl.

Solubility of Sulfide (example : HgS ) Step 1. HgS = Hg 2+ +S 2–, S 2– + H 2 O = HS – + OH –, HS – + H 2 O = H 2 S + OH –, H 2 O = H + + OH – Step 2. 2[Hg 2+ ] + [H 1+ ] = 2[S 2– ]+[HS – ]+[OH – ] Step 3. [Hg 2+ ] = [S 2– ]+[HS – ]+[H 2 S] Step 4. Ksp =[ Hg 2+ ][S 2– ] = 5 × K b1 = [HS – ][OH – ] / [S 2 – ] = 0.80, K b2 = [H 2 S][OH – ] / [HS – ]= 1.1 ×10 -7, K w = [H + ][OH – ] = 1.0 ×10 –14 Step 5. six equations, six unknowns Step 6. If pH=8.00  [OH – ] = 1.0 ×10 –6, [H 2 S]= K b2 [H 2 S] / [OH – ] = 0.11 [HS – ] [HS – ]= K b1 [S 2 – ] / [OH – ] =8.0 ×10 5 [S 2 – ] [Hg 2+ ] = [S 2– ]+[HS – ]+[H 2 S] = [S 2– ]+ 8.0 ×10 5 [S 2 – ]+0.11 × 8.0 ×10 5 [S 2 – ] =(8.88 ×10 5 [S 2 – ] Ksp = [Hg 2+ ][S 2 – ] = [Hg 2+ ]{[Hg 2+ ]/ (8.88 ×10 5 )}= 5 ×  [Hg 2+ ] = 2.1×10 –24 M

Sulfide ion concentration as a function of pH in a saturated H 2 S solution. H 2 S 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