10 Effect of Electrolytes on Chemical Equilibria 10A The Effect of Electrolytes on Chemical Equilibria H3AsO4 + 3I- + 2H+ ↔ H3AsO3 + I3- + H2O Added electrolyte: Ba(NO3)2, K2SO4, NaClO4; → [I3-]↓ , color ↓ ← Fig. 10-1 Effect of electrolyte conc. on conc.-based equilibrium constants. K a' × 105 K w' × 1014 CKNO 3 concentration Concentration NaCl, M Fig. 10-2 Effect of electrolyte conc. on the solubility of some salts. 10A-1 The Effect of Ionic Charges on Equilibria Fig. 10-2 10A-2 The Effect of Ionic Strength 1 2 ionic strength = μ = ([A ]Z A + [B]Z B2 + [C]Z C2 + ⋅ ⋅ ⋅) 2 [A], [B], [C], … molar species conc.; ZA, ZB, ZC, … ionic charge Ex.10-1 Calculate the ionic strength of (a) a 0.1 M soln of KNO3 and (b) a 0.1 M soln of Na2SO4. 1 (a) For KNO3 soln, μ = (0.1 × 12 + 0.1 × 12 ) = 0.1 2 (b) For Na2SO4 soln, μ = 1 (0.2 × 12 + 0.1 × 2 2 ) = 0.3 2 Ex. 10-2 What is the ionic strength of a solution that is 0.05 M in KNO3 and 0.1 M in Na2SO4. 1 μ = (0.05 × 12 + 0.05 × 12 + 0.2 × 12 + 0.1 × 2 2 ) = 0.35 M 2 48 Effect of Charge on Ionic Strength (C = molarity of the salt) Type Example 1:1 NaCl 1:2 Ba(NO3)2, Na2SO4 1:3 Al((NO3)3, Na3PO4 2:2 MgSO4 μ C 3C 6C 4C 10B Activity Coefficients aX = γ X [ X ] aX : activity of species X [X] : molar conc. γX : activity coefficient m n for ppt Xm Yn, thermodynamic equilibrium constant : K sp = aX ⋅ aY K sp = [X] m [Y] n ⋅ γ Xm ⋅ γ Yn = K sp' ⋅ γ Xm ⋅ γ Yn ' Concentration solubility product constant: K sp 10B-1 Properties of Activity Coefficients Activity: a measure of its effective conc. as determined by the lowering of the freezing point of water, by electrical conductivity and by the mass action effect. 1. μ↑→ γx↓, at moderate μ, γx < 1, As μ → 0, γx → 1, ' → K . ax →[X] and K sp sp Fig.10-3 Effect of ionic strength on activity coefficient. 2. γx independent of the electrolyte, dependent only μ 3. For a giving μ, species charge↑→ γx↓(departs farther from unity) 4. For a giving μ, γx of ions of the same charge are approxi. equal. 5. γx → effective behavior in all equilibria in which it participates HCN + H2O ↔ H3O+ + CNAg+ + CN- ↔ AgCN(s) Ni2+ + 4CN- ↔ Ni(CN)4249 10B-2 The Debye-Hückel Equation 0.51Z x2 μ − log γ x = 1 + 3.3α x μ when μ << 0.01 Eq.10-5 1+ μ ≈ 1 − log γ x = 0.51Z x2 μ 1923, P Debye & E. Hückel γx = activity coefficient of the species X Zx = charge on the species X μ = ionic strength of the solution αx = effective diameter of the hydrated ion X in nanometers (10-9 m) Debye-Hückel limiting law (DHLL) Tab 10-1 Activity Coefficients for Ions at 25℃ Activity Coefficient at Indicated Ionic Strength Ion αx, nm 0.001 0.005 0.01 0.05 0.1 + H3O 0.9 0.967 0.934 0.913 0.85 0.83 + Li , C6H5COO 0.6 0.966 0.930 0.907 0.83 0.80 + Na , IO3 , HSO3 , HCO3 , H2PO4 , 0.4-0.45 0.965 0.927 0.902 0.82 0.77 H2AsO4-, OAcOH-, F-, SCN-, HS-, ClO3-, ClO4-, BrO3-, 0.35 0.965 0.926 0.900 0.81 0.76 IO4-, MnO4K+, Cl-, Br-, I-, CN-, NO2-, NO3-, HCOO0.3 0.965 0.925 0.899 0.81 0.75 + + + + + Rb , Cs , Tl , Ag , NH4 0.25 0.965 0.925 0.897 0.80 0.75 2+ 2+ Mg , Be 0.8 0.872 0.756 0.690 0.52 0.44 2+ 2+ 2+ 2+ 2+ 2+ 2+ Ca , Cu , Zn , Sn , Mn , Fe , Ni , 0.6 0.870 0.748 0.676 0.48 0.40 Co2+, Phthalate2Sr2+, Ba2+, Cd2+, Hg2+, S20.5 0.869 0.743 0.668 0.46 0.38 2+ 22Pb , CO3, SO3 , C2O4 0.45 0.868 0.741 0.665 0.45 0.36 2+ 2222Hg2 , SO4 , S2O3 , CrO4 , HPO4 0.40 0.867 0.738 0.661 0.44 0.35 3+ 3+ 3+ 3+ 3+ Al , Fe , Cr , La , Ce 0.9 0.737 0.540 0.443 0.24 0.18 33PO4 , Fe(CN)6 0.4 0.726 0.505 0.394 0.16 0.095 4+ 4+ 4+ 4+ Th , Zr , Ce , Sn 1.1 0.587 0.348 0.252 0.10 0.063 4Fe(CN)6 0.5 0.569 0.305 0.200 0.047 0.020 Ex.10-3 Use Eq. 10-5 to calculate the activity coefficient for Hg2+ in a solution that has an ionic strength of 0.085. Use 0.5 nm for the effective diameter of the ion. Compare the calculated value with γHg2+ obtained by interpolation of data from Table 10-1. 0.51(2) 2 0.085 (1). − log γ Hg = = 0.4016, γ Hg = antilog(-0.4016) = 0.397 ≈ 0.40 1 + 3.3(0.5) 0.085 2+ 2+ (2). Table 10-1: μ = 0.100 → γHg2+ = 0.38; μ = 0.050 → γHg2+ = 0.46 γ Hg 2 + = 0.38 + 0.015 (0.46 − 0.38) = 0.404 ≈ 0.40 0.050 50 10B-3 Equilibrium Calculations using Activity Coefficients Ex. 10-4 Find the relative error introduced by neglecting activities in calculating the solubility of Ba(IO3)2 in a 0.033 M solution of Mg(IO3)2. Ksp of Ba(IO3)2 is 1.57 × 10-9. aBa 2 + ⋅ a 2 IO3- K sp' = = K sp = 1.57 × 10−9 = [Ba 2 + ]γ Ba 2 + ⋅ [IO3- ]2 γ 2 IO3- K sp (γ Ba 2 + )(γ 2 IO 3- ) = K sp = [ Ba 2 + ][ IO 3- ] 2 1 1 μ = ([Mg 2+ ] × 22 + [IO3- ] ×12 ) = (0.033 × 4 + 0.066) = 0.099 ≈ 0.1 M 2 2 Table 10-1: μ = 0.100 → γBa2+ = 0.38; γ - = 0.77 IO3 K sp' = 1.57 × 10 −9 = 6.97 × 10 − 9 = [Ba 2 + ][IO3- ]2 2 0.38 × 0.77 [ IO 3- ] = 2 × 0.033 + 2 [ Ba 2 + ] ≈ 0.066 [Ba2+]×(0.066)2 = 6.97 × 10-9; [Ba2+] = solubility = 1.60 × 10-6 If neglect activities [Ba2+] × (0.066)2 = 1.57 × 10-9; [Ba2+] = solubility = 3.60 × 10-7 3.60 × 10 −7 − 1.60 × 10 −6 relative error = × 100% = −77% 1.60 × 10 − 6 Ex. 10-5 Use activities to calculate the [H3O+] in a 0.120 M HNO2 solution that is also 0.050 M in NaCl. What is the relative percent error incurred by neglecting activity corrections? 1 μ = (0.0500 × 12 + 0.0500 × 12 ) = 0.0500 M 2 μ = 0.050 → γ γ HNO 2 = 1.0 = 0.85 ; γ NO - = 0.81; H 3O + 2 K a ⋅ γ HNO2 [H 3 O + ][ NO -2 ] 7.1×10 −4 ×1.0 ' Ka = = = = 1.03 ×10 −3 [HNO 2 ] 0.85 × 0.81 γ H O+ γ NO- ( 3 )( ) 2 [ H 3 O + ] = 0.120 × 1.03 × 10 −3 = 1.11 × 10 −2 M If assuming all γ = 1: [H 3O + ] = 0.120 × 7.1× 10 − 4 = 9.2 × 10-3 relative error = 9.2 × 10 −3 − 1.11 × 10 −2 × 100% = −17% 1.11 × 10 − 2 10B-4 Omitting Activity Coefficients in Equilibrium Calculations 51
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