1 2 IV. GENERAL PRINCIPLES OF EQUILIBRIUM Combining eq.const. ? Topics : Equilibrium _________________________________________________________ + - + - -14 Equilibrium Constants ex) H2O ⇔ H + OH Kw=[H ][OH ]=1.0x10 + - -5 Le Chatelier Principles NH3(aq) + H2O ⇔ NH4 + OH KNH3 = 1.8x10 Activity find the Eq.Const. for + + Chemical Equilibrium ⇐ Basis in Chemical Analysis NH4 ⇔ NH3 (aq) + H (aq) _________________________________________________________ solubility, complex formation, acid-base reactions, etc.. 1 The Equilibrium Constant a A + b B ⇔ c C + d D a b ratef = kf [A] [B] c d rateb = kb [C] [D] c d k f [c] [D] 2. Le Chatelier's Principle Eq. Constant K = = a b kb [A] [B] : When a stress or change is given in rxn, [A]: molar concentration of A the system moves its eqn. in the way the change is partially offset. if K > 1, forward rxn. is favored. For example, - +3 ⇔ - -2 + * [A] : for gas, --- atm (PA) BrO3 + 2 Cr + 4H2O Br + Cr2O7 + 8 H for pure solid & liq, ---- 1 1) if the conc. of one specie changes, Types of Equilibrium Constants rxn. moves where it is consumed. Acid-Base Solubility [Br − ][Cr O−2 ][H+ ]8 K = 2 7 = 1.0×1011 Complex formation − +3 2 [BrO3 ][Cr ] Redox Phase distribution + -2 +3 [H ]=5.0M, [Cr2O7 ]=0.01M, [Cr ]=0.0030M - - [Br ]=1.0M, [BrO3 ]=0.043M -2 increase [Cr2O7 ] to 0.020M, rxn goes left. 3 4 Use Q (rxn quotiont) 4. Separation by Precipitation 8 (1.0)(0.020)(5.0) 11 Q 2.0 10 K +2 +2 = 2 = × > Consider 0.010 M Pb , Hg ions, (0.043)(0.0030) 2 each forms insoluble iodide. thus, rxn should go back until Q=K. 2+ - -9 PbI2(s) ⇔ Pb + 2 I Ksp = 7.9 x 10 2) if T changes, +2 - -28 Hg2I2 (s) ⇔ Hg2 + 2 I Ksp = 1.1 x 10 o o o ∆G ∆H ∆S +2 +2 K = exp(− ) = exp(− )exp( ) Is it possible to separate Pb and Hg2 ions ? RT RT R +2 (instead of complete separation, 99.99 % of Hg2 ) in case of endothermic, +2 0.01% Hg2 ion exists --- + ∆ Ho : rxn goes forward if T raised. +2 - 2 -6 - 2 -28 [Hg2 ][I ] = (1.0×10 ) [I ] = 1.1×10 3) if pressure changes, x = [I-] = 1.0×10-11 M shift in the direction that results in a reduction of the volume - -11 +2 of the system. when [I ] larger than 1.0×10 M, Hg2 further precipitates. 3. Solubility Product Eq. Const. for rxn in which solid salt dissolves. +2 - Hg2Cl2 ⇔ Hg2 + 2 Cl initial: 0 0.030 final: x 2x+0.030 +2 - 2 2 Ksp = [Hg2 ][Cl ] = x (2x+0.030) formerly x = 6.7 × 10-7 M 2 -18 ⇒ x(0.030) = Ksp = 1.2 × 10 x = 1.3 × 10-15 M solubility is REDUCED by adding COMMON ION. 5 6 5. Complex Formation − + - X + M → MX → MX2 high X- (complex ions) 1) Lewis acids & bases + - -2 - PbI , PbI3 , PbI4 I : ligand Pb ← :I:- → [Pb-I]+ ⇑ electron donation electron acceptor donor = coordinate covalent bond [Lewis Acid] [Lewis base] 6. Protic Acids and Bases protic : transfer of H+ from one molecule to another + + H : proton, H3O : hydronium ion 1) Bronsted-Lowry Acids & Bases Acid : proton donor Base : proton acceptor + - HCl + H2O = H3O + Cl acid base + - HCl + NH3 = NH4 Cl Applicable to all solvents Arrhenius B-L Lewis + + - Acid H3O donor H donor e pair acceptor Base OH- donor H+ acceptor e- pair donor All solvents All substances 7 8 2) Salts : ionic solid iii) Aprotic : trace level of water determines product of acid-base rxn. (neutralization) their effective acid-base prop benzene, hexane, chloroform, CCl . 3) Conjugate acids & bases 4 CH COOH + CH NH ⇔ CH COO- + CH NH + 3 3 2 3 3 3 2) Intrinsic Acidity & Basicity acid base base acid conjugate pair : tendency to split a proton off the acid or add a proton to the conjugate base 4) Nature of H+ and OH- (H + )(S − ) Autoprotolysis (self-ionization of water) HS ⇔ H+ + S- K SH = a (HS) + - + H2O + H2O ⇔ H3O + OH + + SH (H 2S ) HS + H ⇔ H2S K = b (HS)(H + ) -14 K = 1.0x10 = Kw + − + - SH (H 2S )(S ) 2 HS ⇔ H2S + S K = 7. Acid-Base Equilibria in Nonaqueous Systems s (HS)2 autoprotolysis (HS) ~ 1 1) solvent classifications activity of solvent i) Amphiprotic solvents : both Bronsted acid and base K SH = (H S + )(S − ) = K SH K SH - protogenic : much stronger acid than water s 2 a b H O + − H2SO4, HF, HCOOH, CH3COOH 2 if HS = H2O Ks = Kw = (H3O )(OH ) EtOH + − - protophilic : much stronger base than water = EtOH Ks = (EtO2 )(EtO ) NH 3 + − NH3(l), butylamine, ethylenediamine = NH3(l) Ks = (NH 4 )(NH 2 ) - hydroxylic solvent : between protogenic & protophilic The solvated proton - lyonium ion water, methanol, ethanol Anion - lyate ion ii) weak basic but negligibly acidic : ketones, ethers, nitriles, esters, DMSO 9 10 1) weak acids & bases + - HA + H2O ⇔ H3O + A ka [H+ ][A− ] k = small ka - weak acid a [HA] -5 ex) CH3COOH ----- ka = 1.75x10 + - B + H2O ⇔ BH + OH kb [BH+ ][OH− ] k = small k = weak base b [B] b -5 ex) NH3 --- kb = 1.75x10 8. Strengths of Acids & Bases (strong or weak) → dissociate completely ? or partly ? 2) ka & kb in a conjugate pair [H+ ][A− ] HA ⇔ H+ + A- k = Strong acids : HCl hydrochloric acid a [HA] H SO sulfuric acid 2 4 [HA][OH− ] HI, HNO , HClO (perchloric acid) - - 3 4 A + H O ⇔ HA + OH kb = 2 [A− ] Strong bases : LiOH lithium hydroxide ----------------------------------------------------------------------------- + - NaOH Sodium hydroxide H2O ⇔ H + OH kw = ka kb KOH, RbOH, CsOH, -5 R4NOH (quaternary ammonium -) ex) ka (CH3COOH) = 1.75x10 - kb of CH3COO = 11 12 9. Activity decrease in product conc. by being Activity ⇒ Effective concentration under ionic surroundings attracted to the previously dissolved ions 1). The effect of Ionic Strength on Solubility of salts. → Le Chatelier's principle 2+ * Consider a saturated sol. Hg2(IO3)2 , concentration of Hg2 ? → MX dissolves more 2+ - -18 Hg2(IO3)2 ⇔ Hg2 + 2 IO3 Ksp = 1.3 x 10 What is Ionic Strength ? 2+ - 2 2 Ksp = [Hg2 ][IO3 ] = x(2x) : measure the total conc. of ions in solutions. 2+ - -7 x = [Hg2 ]=[IO3 ] = 6.9 x 10 M 1 2 2 1 2 Ionic Strength I = (c1z1 + c2z2 + ... ) = ∑cizi 2 2 2+ this is the max. [Hg2 ] that can be dissolved. ci : concentration of ith species ⇒ only when they are in distilled water without any other ions. zi : charge (+1, +2) + - * If 0.050 M KNO3 is added to the sol, (K does not react with IO3 ) 2) Activity Coefficient 2+ -6 ⇒ more solid (Hg2(IO3)2) dissolves until = [Hg2 ]= 1.0 x 10 M (50% increase) For expressing the concentration influenced by other ionic presence, Activity must be used. Addition of "inert" salt (KNO3) to insoluble salt (Hg2(IO3)2) ⇒ Increase in solubility Activity of C : Ac = [C]γc γc : activity coeff. Why ? aA + bB → cC + dD Ac Ad [C]c γc [D]d γd K 0 = C D = C D With more ions, ionic atmosphere builds up eq Aa Ab [A]a γa [B]b γb Thermodynamic eq. const. : A B A B → net attraction between cation - anion becomes smaller than c d γC γ D pure cation and anion. = Keq a b (atmospheric counter charge decreases the ionic charge) γ Aγ B c d ⇒ increasing the solubility [C] [D] Thermodynamic eq. const. : K = eq [A]a[B]b ∴ Increasing ionic strength promotes the dissociation of compounds into ions Activity Coefficient ∝ ionic strength ≅ 1 at low ionic strength + - MX ⇔ M + X < 1 at high ionic strength 13 14 3) Activity Coefficient of Ions b) with 0.0125 M MgSO4 2 2+ − 2 2 Ksp = ACa 2+ AF − = [Ca ]γCa 2+ [F ] γF − Debye Huckel's definition 2 2 = x(2x) γCa 2+ γF − − 0.51z 2 I 0 logγ = ( I ≤ 0.1M ) at 25 C 2 2 1+ (0.33α I / 305) µ = 1/2(0.0125x2 + 0.0124x(-2) = 0.0500 M γCa 2+ = 0.485, γF − = 0.81 α : ion size parameter : effective hydrated radius small, highly charged ions binds solvent molecule x = 3.1 x 10-4 M = [Ca+2] more tightly, and have larger hydrated radii. 45% increase in solubility αF − > αI − 0 ii. The Common Ion Effect For singly charged ions α ∼ 3 A multiply 11 In the case of common ion existed +2 [Ca ] of CaF2 with 0.050 M NaF (I=0.05), [F-] : common ion 4). Using Activity Coefficient +2 - CaF2 ⇔ Ca + 2 F initial 0 0.05 i. A Simple Solubility problem final x 2x+0.05 +2 _______________________________ * Concentration of Ca saturated with CaF2 in a 0.0125 M soln. of If x is very small, 2x << 0.05. MgSO 4 Since I=0.05, γ is the same as above. 2+ - -11 CaF2 (s) ⇔ Ca + 2 F Ksp = 3.9 x 10 2+ − 2 2 K = [Ca ]γ 2+ [F ] γ − initial: 0 0 sp Ca F final: x 2x = x(0.485)(0.050)2 (0.81)2 = 3.9x10−11 ______________________________ x = [Ca+2] = 4.9 x 10-8 M a) without MgSO4 2 -11 Ksp = x(2x) = 3.9 x 10 ⇒ Decrease in solubility due to common ion -4 +2 x = 2.136 x 10 M = [Ca ] which affects the total product Ksp 15 5). pH Revisited (from the point of Activity) pH = - log [H+] + pH = −log AH + = −log[H ]γH + if we use pH meter, it sees actual hydrogen ions only. ex) pH of water containing 0.10 M KCl at 250C. µKCl = 0.10M γH + = 0.83, γOH − = 0.76 -14 Kw = x(0.83)x(0.76) = 1.0 x 10 x = 1.26 x 10-7 M pH = 6.98.