General and Inorganic Chemistry I.

General and Inorganic Chemistry I. Lecture 7 István Szalai Eötvös University István Szalai (Eötvös University) Lecture 1 1 / 80 Chemical Equilibrium A + B C + D C + D A + B A + B C + D István Szalai (Eötvös University) Lecture 1 2 / 80

Chemical Equilibrium A + B C + D rate forward = k f [A][B] rate backward = k b [C][D] k f [A][B] = k b [C][D] k f = [C][D] k b [A][B] K c = [C][D] [A][B] István Szalai (Eötvös University) Lecture 1 3 / 80 Mass Action Law Guldberg and Waage, 1867 aa + bb cc + dd K c = [C]c [D] d [A] a [B] b The value of K c is constant at a given temperature; changes if the temperature changes; does not depend on the initial concentrations. István Szalai (Eötvös University) Lecture 1 4 / 80

Partial Pressure and the Equilibrium Constant Solving the ideal gas equation, pv = nrt, for pressure gives p = n V (RT ) = crt N 2 (g) + 3H 2 (g) 2NH 3 (g) K p = p2 NH 3 p N2 p 3 H 2 Relationship between K p and K c : [NH 3 ] 2 = ( pnh3 RT ) 2 [N2 ] = ( pn2 RT ) [H 2 ] 3 = ( ph2 RT ) 3 István Szalai (Eötvös University) Lecture 1 5 / 80 Relationship between K p and K c ( pnh3 ) 2 K c = ( pn2 RT RT ) ( ph2 RT ) 3 In general the relationship is: K c = p2 NH 3 p N2 p 3 H 2 (RT ) 2 1 3 K c = p2 NH 3 p N2 p 3 H 2 (RT ) 2 K p = K c (RT ) n n = (n gas prod ) (n gas react ) István Szalai (Eötvös University) Lecture 1 6 / 80

Finding Equilibrium Concentrations For the following reaction, the equilibrium constant is 49.0, at a certain temperature. If 0.400 mol each of A and B are placed in a 2.00-liter container, what concentration of all species are present at equilibrium? The initial concentrations are A + B C + D [A] 0 = 0.400mol 2.00L = 0.200 M [B] 0 = 0.400mol 2.00L = 0.200 M [C] 0 = 0 M [D] 0 = 0 M Let x=moles per liter of C and D formed. At equilibrium [A] eq =(0.200 x) M [B] eq =(0.200 x) M [C] eq =x M [D] eq =x M István Szalai (Eötvös University) Lecture 1 7 / 80 Finding Equilibrium Concentrations K c = K c = [C][D] [A][B] = 49.0 (x)(x) (0.200 x)(0.200 x) = 49.0 K c = (x) 2 (0.200 x) 2 = 49.0 x = 0.175 [A] eq =(0.200 x) = 0.025 M [B] eq =(0.200 x) = 0.025 M [C] eq =x = 0.175 M [D] eq =x = 0.175 M István Szalai (Eötvös University) Lecture 1 8 / 80

Disturbing a System at Equilibrium Le Chatelier s Principle: If a system at equilibrium is disturbed by changing its conditions (applying a stress), the system shifts in the direction that reduces the stress. If given sufficient time, a new state of equilibrium is established. Changes in concentration Stress Direction of shift of A + B C + D Increase conc. of A or B right Increase conc. of C or D left Decrease conc. of A or B left Decrease conc. of C or D right István Szalai (Eötvös University) Lecture 1 9 / 80 Disturbing a System at Equilibrium Changes in volume and pressure Stress Direction of shift of A(g) 2D(g) Volume decrease, Toward smaller number of moles of gas pressure increase (left for this reaction) Volume increase, Toward larger number of moles of gas pressure decrease (right for this reaction) If there is no change in the total number of moles of gases in the balanced chemical equation, volume (pressure) change does not affect the position of equilibrium. István Szalai (Eötvös University) Lecture 1 10 / 80

Disturbing a System at Equilibrium Changes in temperature A + B C + D + heat Heat is produced by the forward (exothermic) reaction. Increasing the temperature at constant pressure favors the reaction to the left, removing some of the extra heat. Lowering the temperature favors the to the right. W + X + heat Y + Z By contrast, for an endothermic reaction at equilibrium an increase in temperature at constant pressure favors the reaction to the right. A decrease in temperature favors the reaction to the left. István Szalai (Eötvös University) Lecture 1 11 / 80 Temperature dependence of equilibrium constant van t Hoff equation: or d ln K dt = H0 RT 2 ln K T2 ln K T1 = H0 R ( 1 T 2 1 T 1 exothermic reaction: H 0 < 0, T 2 > T 1, K 2 < K 1 endothermic reaction: H 0 > 0, T 2 > T 1, K 2 > K 1 ) István Szalai (Eötvös University) Lecture 1 12 / 80

Acids, Bases and Salts Arrhenius theory (1884) An acid is a substance that contains hydrogen and produces H + in aqueous solution. A base is a substance that contains the OH (hydroxyl) group and produces hydroxide, OH, in aqueous solution. Neutralization: H + (aq) + OH (aq) H 2 O(l) The Arrhenius theory of acid-base behavior satisfactorily explained reactions of protic acids with metal hydroxides. The hydronium ion H + (H 2 O) n, H 3 O + (if n = 1), (H + (H 2 O) 4 or H 9 O + ) The hydrated hydrogen ion is the species that gives aqueous solutions of acids their characteristic acidic properties. István Szalai (Eötvös University) Lecture 1 13 / 80 Acids, Bases and Salts Brønstedt-Lowry theory (1923) An acid is defined as a proton donor, and a base is defined as a proton acceptor. An acid-base reaction is the transfer of a proton from an acid to a base. H 2 O(l)+ HCl(aq) H 3 O + + Cl (aq) base 1 + acid 2 acid 1 + base 2 H 2 O(l)+ HF(aq) H 3 O + + F (aq) base 1 + acid 2 acid 1 + base 2 We can describe Brønstedt-Lowry theory acid-base reactions in terms conjugate acid-base pairs. These are species that differ by a proton. The stronger the acid, the weaker is its conjugate base, the weaker the acid, the stronger is its conjugate base. István Szalai (Eötvös University) Lecture 1 14 / 80

Acid-Base Equilibrium Acids: HA + H 2 O A + H 3 O + K = [A ][H 3 O + ] [HA][H 2 O] K a = [A ][H 3 O + ] [HA] The larger the value of K a, the stronger is the acid. Strong acids ionize completely in dilute aqueous solution, whereas weak acids ionize only slightly. István Szalai (Eötvös University) Lecture 1 15 / 80 Acid-Base Equilibrium Bases: B + H 2 O BH + + OH K = [BH+ ][OH ] [B][H 2 O] K b = [BH+ ][OH ] [B] István Szalai (Eötvös University) Lecture 1 16 / 80

The Autoinization of Water Careful measurements show that pure water ionizes ever so slightly to produce equal number of hydrated hydrogen ions and hydroxide ions. H 2 O(l) + H 2 O(l) H 3 O + + H (aq) base 1 + acid 2 acid 1 + base 2 K w = [H 3 O + ][OH ] = 1 10 14 (at 25 C) Water is said to be amphiprotic; that is H 2 O molecules can both donate and accept protons. Whether water acts as an acid or a base depends on the other species present. István Szalai (Eötvös University) Lecture 1 17 / 80 The ph Scale ph = -log[h 3 O + ] and poh = -log[oh ] H 2 O(l) + H 2 O(l) H 3 O + + H (aq) K w = [H 3 O + ][OH ] = 1 10 14 (at 25 C) (-log[h 3 O + ]) + (-log[oh ]) = (-log (1 10 14 )) ph + poh =14.0 Solution At 25 C acidic [H 3 O + ] > 1.0 10 7 [OH ] neutral [H 3 O + ] = 1.0 10 7 = [OH ] basic [H 3 O + ] < 1.0 10 7 [OH ] István Szalai (Eötvös University) Lecture 1 18 / 80

The ph Scale ph range for a few common substances Substance ph range gastric contents (human) 1.6-3.0 soft drinks 2.0-4.0 lemons 2.2-2.4 tomatoes 4.0-4.4 beer 4.0-5.0 milk (cow s) 6.3-6.6 blood plasma (human) 7.3-7.5 egg white 7.6-8.0 István Szalai (Eötvös University) Lecture 1 19 / 80 Acid-Base Equilibrium Relationship between the acid and base dissociation constants of conjugated acid-base pairs: HA + H 2 O A + H 3 O + K a = [A ][H 3 O + ] [HA] A + H 2 O HA + OH K b = [HA][OH ] [A ] K a K b = [A ][H 3 O + ] [HA] [HA][OH ] [A ] K a K b = K w István Szalai (Eötvös University) Lecture 1 20 / 80

Amphoterism Amphoterism describes the ability of a substance to react either as an acid or as a base. H 2 O(l) + H 2 O(l) H 3 O + + H (aq) base 1 + acid 2 acid 1 + base 2 Several insoluble metal hydroxide are amphoteric: Al(OH) 3 (s) + 3H + (aq) Al 3+ (aq) + H 2 O(l) Al(OH) 3 (s) + OH (aq) Al(OH) 4 (aq) István Szalai (Eötvös University) Lecture 1 21 / 80 Strength of Acids Binary acids The ease of ionization of binary protic acids depends on both the ease of breaking H X bonds and the stability of the resulting ions in solutions. Let us consider the strength of the hydrohalic acids. HF ionizes only slightly in dilute aqueous solution. H 2 O(l)+ HF(aq) H 3 O + + F (aq) HCl, HBr and HI ionize completely or nearly completely. H 2 O(l)+ HX(aq) H 3 O + + X (aq) X=Cl, Br, I The order of bond strengths: (strongest) HF HCl > HBr > HI (weakest) István Szalai (Eötvös University) Lecture 1 22 / 80

Strength of Acids Ternary acids Acid strengths of most ternary acids containing the same central element increase with increasing number of oxygen atoms. The following orders of increasing acid strength are typical. H 2 SO 3 < H 2 SO 4 HNO 2 < HNO 3 HClO < HClO 2 < HClO 3 < HClO 4 István Szalai (Eötvös University) Lecture 1 23 / 80 Strength of Acids Ternary acids For most ternary acids containing different elements in the same oxidation state from the same group in the periodic table, acid strengths increase with increasing electronegativity of the central element. H 2 SeO 4 < H 2 SO 4 H 2 SeO 3 < H 2 SO 3 H 3 PO 4 < HNO 3 HBrO 3 < HClO 3 István Szalai (Eötvös University) Lecture 1 24 / 80

Strength of Acids Polyprotic acids The first step in the ionization of a polyprotic acid always occurs to a greater extent than the second step because it is easier to remove a proton from a neutral acid molecule the from a negatively charged anion. H 2 SO 4 (aq) H + + HSO 4 (aq) HSO 4 (aq) H+ + SO 2 4 (aq) K a1 K a2 István Szalai (Eötvös University) Lecture 1 25 / 80 Strength of Acids Comparison of extents of ionization of some acids Acid K a ph % ionization 0.10 M HCl very large 1.00 100 0.10 M CH 3 COOH 1.8 10 5 2.89 1.3 0.10 M HOCl 3.5 10 8 4.23 0.059 István Szalai (Eötvös University) Lecture 1 26 / 80

Strength of Acids and Bases The hydronium ion is the strongest acid that can exist in aqueous solution. All acids stronger than H 3 O + react completely with water to produce H 3 O + and their conjugate bases. Strong acids: HCl, HBr, HI, HClO 4, HClO 3, HNO 3, H 2 SO 4, H 2 O(l)+ HClO 4 (aq) H 3 O + + ClO 4 (aq) The hydroxide ion is the strongest base that can exist in aqueous solution. Bases stronger than OH react completely with water to produce OH and their conjugate acids. Strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2, NaOH(s) H 2O Na + (aq) + OH (aq) István Szalai (Eötvös University) Lecture 1 27 / 80 Strength of Acids and Bases Relative strengths of conjugate acid-base pairs Acid Base HClO 4 100% ionized in Negligible base ClO 4 HI dilute aq. soln. strength in water I HBr Br HCl Cl HNO 3 NO 3 H 3 O + H 2 O HF Equilibrium mixture F CH 3 COOH of noionized molecules CH 3 COO HCN of acid, conjugate CN NH + 4 base and H + NH 3 H 2 O OH NH 3 Reacts completely with water NH 2 István Szalai (Eötvös University) Lecture 1 28 / 80

Hydrolysis Hydrolysis is the reaction of a substance with water. Acid-base properties of dilute aqueous solutions of salts: Salt of strong bases and strong acids They give neutral solutions because neither the cation nor the anion reacts appreciably with water. NaCl(s) H 2O Na + (aq) + Cl (aq) (100%) István Szalai (Eötvös University) Lecture 1 29 / 80 Hydrolysis Salt of strong bases and weak acids They give basic solutions because anions of weak acids react with water to from hydroxide ions. NaCH 3 COO(s) H 2O Na + (aq) + CH 3 COO (aq) (100%) hydrolysis of acetate ions: CH 3 COO (aq) + H 2 O CH 3 COOH(aq) + OH K b = [CH 3COOH][OH ] [CH 3 COO ] K w = K a K b István Szalai (Eötvös University) Lecture 1 30 / 80

Hydrolysis Salt of weak bases and strong acids They give acidic solutions because cations of weak bases react with water to from H 3 O + ions. NH 4 Cl(s) H 2O NH + 4 (aq) + Cl (aq) (100%) hydrolysis of ammonium ions: NH + 4 (aq) + H 2O NH 3 (aq) + H 3 O + K a = [NH 3][H 3 O + ] [NH + 4 ] K w = K a K b István Szalai (Eötvös University) Lecture 1 31 / 80 Hydrolysis Salt of weak bases and strong acids Salts that contain small, highly charged cations. They give acidic solutions because cations react with water to from H 3 O + ions. hydrolysis of hydrated Al 3+ : [Al(H 2 O) 6 ] 3+ + H 2 O [Al(OH)(H 2 O) 5 ] 3+ + H 3 O + K a = [Al(OH)(H 2 O)2+ 5 ][H 3O + ] [ Al(H 2 O) 3+ 6 ] István Szalai (Eötvös University) Lecture 1 32 / 80

Hydrolysis Salt of weak bases and weak acids The common example of a salt of this type is ammonium acetate, NH 4 CH 3 COO. CH 3 COO (aq) + H 2 O CH 3 COOH(aq) + OH NH + 4 (aq) + H 2O NH 3 (aq) + H 3 O + Neutral solution if parent K b = K a. Basic solution if parent K b > K a. Acidic solution if parent K b < K a. István Szalai (Eötvös University) Lecture 1 33 / 80 Hydrolysis Acid-base properties of some common ions in water solution Neutral Basic Acidic Anion Cl, NO 3 CH 3 COO, CN HSO 4 Br, ClO 4 F, NO 2 H 2 PO 4 I CO 2 3, HCO 3 S 2, HS PO 3 4, HPO2 4 Cation Li +, Ca 2+ H 2 N-CH 2 -CH 2 -NH + 3 NH + 4, Mg2+ Na +, Ba 2+ Al 3+ K +, Rb + transition metal ions István Szalai (Eötvös University) Lecture 1 34 / 80

Buffers A buffer solution contains a conjugate acid-base pair with both the acid and the base in reasonable concentrations. Changes in ph are in minimized in buffer solutions because the acidic components reacts with added (strong) bases, while the basic component reacts with added (strong) acids. CH 3 COOH and NaCH 3 COO CH 3 COOH(aq) + H 2 O CH 3 COO (aq) + H 3 O + CH 3 COO (aq) + H 2 O CH 3 COOH(aq) + OH [H + ] = K a [CH 3 COOH] [CH 3 COO ] ph = pk a + log [CH 3COO ] [CH 3 COOH] István Szalai (Eötvös University) Lecture 1 35 / 80 Buffers Acid/salt buffer ph = K a [HA] [A ] Base/salt buffer ph = pk a + log [A ] [HA] poh = K b [B] [BH + ] poh = pk b + log [BH+ ] [B] István Szalai (Eötvös University) Lecture 1 36 / 80

Buffers Approximate ph Weak acid Weak base pk a 5 CH 3 COOH CH 3 COO 4.74 6 H 2 CO 3 HCO 3 6.36 7 H 2 PO 4 HPO 2 4 7.21 9 NH + 4 NH 3 9.25 10 HCO 3 CO 2 3 10.32 István Szalai (Eötvös University) Lecture 1 37 / 80 ph calculations Strong acids and bases: [strong acid] > 10 5 M [H + ] = [strong acid] [OH ] = [strong base] István Szalai (Eötvös University) Lecture 1 38 / 80

ph calculations Weak acids: HA + H 2 O A + H 3 O + K a = [A ][H 3 O + ] [HA] [A ] = [H 3 O + ] [HA] = [HA] T [H 3 O + ] K a = [H 3 O + ] 2 [HA] T [H 3 O + ] [HA] T [H 3 O + ] [H 3 O + ] = K a [HA] T István Szalai (Eötvös University) Lecture 1 39 / 80 ph calculations Weak base: [B] T [OH ] [OH ] = K b [B] T István Szalai (Eötvös University) Lecture 1 40 / 80

ph calculations Hydrolysis hydrolysis of ammonium ions: NH + 4 (aq) + H 2O NH 3 (aq) + H 3 O + K a = [NH 3][H 3 O + ] [NH + 4 ] [NH + 4 ] T [H 3 O + ] [H 3 O + ] = K a [NH + 4 ] T [H 3 O + K w ] = [NH + 4 K ] T b(ammonia) István Szalai (Eötvös University) Lecture 1 41 / 80 Exercises What is the ph of a 10 7 M HCl solution? HCl is a strong acid: HCl + H 2 O H 3 O + + Cl [H + ] = [HCl] [H + ] = 10 7 M ph = 7 neutral?? wrong!!! H 2 O + H 2 O H 3 O + + OH K w = [H 3 O + ][OH ] = 1 10 14 [H 3 O + ] = [OH ] + [Cl ] [H 3 O + ] = K w [H 3 O + ] + [HCl] István Szalai (Eötvös University) Lecture 1 42 / 80

Exercises [H 3 O + ] 2 [HCl] [H 3 O + ] K w = 0 [H 3 O + ] 2 10 7 [H 3 O + ] 10 14 = 0 [H 3 O + ] = 10 7 ± (10 7 ) 2 + 4 1 10 14 2 [H 3 O + ] = 1.628 10 7 M ph = 6.79 István Szalai (Eötvös University) Lecture 1 43 / 80 Exercises Calculate the ph of 0.1 M (and 0.001 M) solutions of acetic acid. (K a = 2 10 5 ) CH 3 COOH + H 2 O CH 3 COO + H 3 O + K a = [CH 3COO ][H + ] [CH 3 COOH] [CH 3 COOH] T = [CH 3 COOH] + [CH 3 COO ] [H + ] = [CH 3 COO ] K a = [H + ] 2 [CH 3 COOH] T [H + ] István Szalai (Eötvös University) Lecture 1 44 / 80

Exercises K a = [H+ ] 2 0.1 [H + ] [H + ] 2 + 2 10 5 [H + ] 2 10 5 0.1 = 0 [H + ] = 2 10 5 ± (2 10 5 ) 2 + 4 1 (2 10 5 0.1) 2 [H + ] = 1.41 10 3 M ph = 2.85 István Szalai (Eötvös University) Lecture 1 45 / 80 Exercises K a = [H+ ] 2 0.1 [H + ] 0.1 [H + ] [H + ] = K a 0.1 [H + ] = 2 10 5 0.1 [H + ] = 1.41 10 3 M ph = 2.85 István Szalai (Eötvös University) Lecture 1 46 / 80

Exercises Calculate the ph of 0.1 M (and 0.001 M) sodium acetate solutions. CH 3 COO (aq) + H 2 O CH 3 COOH(aq) + OH K b = [CH 3COOH][OH ] [CH 3 COO ] K w = K a K b [CH 3 COO ] T = [CH 3 COO ] + [CH 3 COOH] [OH ] = [CH 3 COOH] K b = [OH ] 2 [CH 3 COO ] T [OH ] István Szalai (Eötvös University) Lecture 1 47 / 80 Exercises K b = [OH ] 2 0.1 [OH ] K b = K w = 10 14 = 5 10 10 K a 2 10 5 [OH ] 2 + 5 10 10 [OH ] 5 10 10 0.1 = 0 [OH ] = 5 10 10 ± (5 10 10 ) 2 + 4 1 (5 10 10 0.1) 2 [OH ] = 7.07 10 6 M poh = 5.15 ph = 14 poh = 8.85 István Szalai (Eötvös University) Lecture 1 48 / 80

Exercises K b = [OH ] 2 0.1 [OH ] 0.1 [OH ] [OH ] = Kw K a 0.1 [OH ] = 5 10 10 0.1 [OH ] = 7.07 10 6 M ph = 8.85 István Szalai (Eötvös University) Lecture 1 49 / 80 Exercises What is the ph of the solution which contains 0.1 M of acetic acid and 0.05 M sodium acetate? CH 3 COOH + H 2 O CH 3 COO + H 3 O + K a = [CH 3COO ][H + ] [CH 3 COOH] [CH 3 COOH] T [CH 3 COOH] [CH 3 COO ] T [CH 3 COO ] [H + ] [CH 3 COO ] K a = [CH 3COO ] T [H + ] [CH 3 COOH] T István Szalai (Eötvös University) Lecture 1 50 / 80

Exercises [H + ] = K a [CH 3COOH] T [CH 3 COO ] T [H + ] = 2 10 5 0.1 0.05 [H + ] = 4 10 5 M ph = 4.40 István Szalai (Eötvös University) Lecture 1 51 / 80 Exercises What will be the ph if 8 ml of 0.1 M NaOH is added to 250 ml of a solution, which is 0.1 M in acetic acid and 0.05 M in sodium acetate? CH 3 COOH + H 2 O CH 3 COO + H 3 O + CH 3 COOH + NaOH CH 3 COO + Na + + H 2 O K a = [CH 3COO ][H + ] [CH 3 COOH] n(ch 3 COOH) = n(ch 3 COOH) 0 n(naoh) n(ch 3 COO ) = n(ch 3 COO ) 0 + n(naoh) n(ch 3 COOH) = 0.1mol/dm 3 250 10 3 dm 3 0.1mol/dm 3 8 10 3 dm 3 n(ch 3 COO ) = 0.05mol/dm 3 250 10 3 dm 3 +0.1mol/dm 3 8 10 3 dm 3 István Szalai (Eötvös University) Lecture 1 52 / 80

Exercises n(ch 3 COOH) = 0.025 0.0008 = 0.0242mol n(ch 3 COO ) = 0.0125 + 0.0008 = 0.0117mol [H + ] = K a n(ch 3COOH)/V n(ch 3 COO )/V [H + ] = K a 0.025 0.0008 0.0125 + 0.0008 [H + ] = 3.63 10 5 M ph = 4.44 István Szalai (Eötvös University) Lecture 1 53 / 80 Lewis Theory An acid is any species that can accept a share in an electron pair. A base is any species that can make available, or donate, a share in an electron pair. Neutralization is defined as a coordinate covalent bond formation. BCl 3 (g) + NH 3 (g) Cl 3 B:NH 3 István Szalai (Eötvös University) Lecture 1 54 / 80

Complex Formation Equilibrium The most common type of complexes is: metal-ligand complexes. in which the ligand (a Lewis base) donates an electron pair to the metal ion (a Lewis acid). Classification of ligands: Me(H 2 O) 2+ 6 + L Me(H 2 O) 5 L 2+ + H 2 O Me(H 2 O) 5 L 2+ + L Me(H 2 O) 4 L 2+ 2 + H 2 O 1 by charge: anion (F, Cl, CN ), neutral molecule (NH 3, CO, H 2 O) 2 by the number of electron pair donating groups: monodentate (Cl, CN, NH 3 ), bidentate ( OOC-COO ), polydentate (EDTA) István Szalai (Eötvös University) Lecture 1 55 / 80 Complex Formation Equilibrium Stepwise equilibria Stepwise equilibrium Commulative equilibrium constant constant Me 2+ + L MeL 2+ K 1 = [MeL2+ ] [Me 2+ ][L] β 1 = K 1 = [MeL2+ ] [Me 2+ ][L] MeL 2+ + L MeL 2+ 2 ] β 2 = K 1 K 2 = 2 K 2 = [MeL2+ [MeL 2+ ][L] MeL 2+ 2 + L MeL 2+ 3 K 3 = [MeL2+ [MeL 2+ MeL 2+ n 1 + L MeL2+ n K n = [MeL2+ 3 ] [MeL2+ 2 ] [Me 2+ ][L] 2 2 ][L] β 3 = K 1 K 2 K 3 = n ] [MeL 2+ n 1 ][L] In the majority of cases K 1 > K 2 > K 3 > K 4. If K n+1 > K n the system is cooperative. [MeL 2+ 3 ] [Me 2+ ][L] 3 β n = n i=1 K i = [MeL2+ n ] [Me 2+ ][L] n István Szalai (Eötvös University) Lecture 1 56 / 80

Complex Formation Equilibrium Classification of ligands by the character of the donor group (in relation with the acceptor atom/ion) Lewis, Pearson Acid Base Hard Soft Hard Soft little great Polarizability little great small large Size small large ionic covalent Bonding tendency ionic covalent great little EN great little great little Charge great little István Szalai (Eötvös University) Lecture 1 57 / 80 Complex Formation Equilibrium Classification of ligands by the character of the donor group (in relation with the acceptor atom/ion) Lewis, Pearson István Szalai (Eötvös University) Lecture 1 58 / 80

Complex Formation Equilibrium Classification of ligands by the character of the donor group (in relation with the acceptor atom/ion) Lewis, Pearson Ag + (soft acid) prefers to form complexes with soft bases. Ag + + 2CN [AgCN 2 ] β 2 = 5.5 10 18 Ag + + 2I [AgI 2 ] β 2 = 1.0 10 11 Ag + + 2Br [AgBr 2 ] β 2 = 1.3 10 7 Ag + + 2Cl [AgCl 2 ] β 2 = 2.5 10 5 István Szalai (Eötvös University) Lecture 1 59 / 80 Complex Formation Equilibrium Problems related to the complex equilibria What is the predominant species in a solution [Al 3+ ] = 10 6 mol/dm 3 and [F ] = 10 3 mol/dm 3? Calculate the c Al 3+ c F values! log K 1 = 6.1, log K 2 = 5.0, log K 3 = 3.85, log K 4 = 2.7 Al 3+ + F AlF 2+ K 1 = [AlF2+ ] [Al 3+ ][F ] AlF 2+ + F AlF + 2 K 2 = [AlF+ 2 ] [AlF 2+ ][F ] AlF + 2 + F AlF 3 K 1 = [AlF 3] [AlF + 2 ][F ] AlF 3 + F AlF 4 K 1 = [AlF 4 ] [AlF ] 3 [F ] István Szalai (Eötvös University) Lecture 1 60 / 80

Complex Formation Equilibrium [AlF 2+ ] = β 1 [Al 3+ ][F ] = 1.258 10 3 [AlF + 2 ] = β 2[Al 3+ ][F ] 2 = 1.258 10 1 [AlF 3 ] = β 3 [Al 3+ ][F ] 3 = 8.913 10 1 [AlF 4 ] = β 4[Al 3+ ][F ] 4 = 4.467 10 1 c Al 3+ = [Al 3+ ] + [AlF 2+ ] + [AlF + 2 ] + [AlF 3] + [AlF 4 ] c Al 3+ = 1.465 mol/dm 3 c F = [F ] + [AlF 2+ ] + 2[AlF + 2 ] + 3[AlF 3] + 4[AlF 4 ] c F = 4.175 mol/dm 3 István Szalai (Eötvös University) Lecture 1 61 / 80 Complex Formation Equilibrium What is the concentration of silver ions in a solution c Ag + = 0.2 mol/dm 3 and [NH 3 ] = 0.15 mol/dm 3? log K 1 = 3.3, log K 2 = 3.9 Ag + + NH 3 AgNH + 3 K 1 = [AgNH+ 3 ] [Ag + ][NH 3 ] AgNH + 3 + NH 3 Ag(NH 3 ) + 2 K 1 = [Ag(NH 3 )+ 2 ] [AgNH + 3 ][NH 3] c Ag + = [Ag + ] + [AgNH + 3 ] + [Ag(NH 3 )+ 2 ] c Ag + = [Ag + ] + β 1 [Ag + ][NH 3 ] + β 2 [Ag + ][NH 3 ] 2 [Ag + ] = c Ag + 1 + β 1 [NH 3 ] + β 2 [NH 3 ] 2 István Szalai (Eötvös University) Lecture 1 62 / 80

Complex Formation Equilibrium [Ag + ] = 0.2 1 + 10 3.3 0.15 + 10 3.3+3.9 (0.15) 2 [Ag + ] = 5.603 10 7 mol/dm 3 István Szalai (Eötvös University) Lecture 1 63 / 80 Micro and Macro Equilibria István Szalai (Eötvös University) Lecture 1 64 / 80

Micro and Macro Equilibria Micro constants: k A = [ o +] [ o o][[h + ], ka C = [ + + ] [ + o ][[H + ] Macro constants: k C = [ + o ] [ o o][[h + ], kc A = [ + + ] [ o + ][[H+ ] K 1 = [HG] [G ][H + ], K 2 = [H 2G + ] [HG][H + ] István Szalai (Eötvös University) Lecture 1 65 / 80 Micro and Macro Equilibria Relationships between macro and micro constants: K 1 = [ o +] + [ + o ] [ o o][[h + ] = k A + k C K 1 K 2 = [ + + ] [ o o][[h + ] 2 = ka k C A = kc k A C number of number of number of groups micro species micro constants 2 4 4 3 8 12 n 2 n n2 n István Szalai (Eötvös University) Lecture 1 66 / 80

Heterogeneous Equilibria By definition at least two phases are present. Solid-gas equilibria Decomposition of carbonates CaCO 3 (s) CaO(s) + CO 2 (g) Decomposition of oxides 2 Ag 2 O(s) 4Ag(s) + O 2 (g) Decomposition of acid salts 2Na 2 HPO 4 (s) Na 4 P 2 O 7 (s) + H 2 O(g) Decomposition of crystal-water-containing compounds CuSO 4 5H 2 O(s) CuSO 4 (s) + 5H 2 O(g) István Szalai (Eötvös University) Lecture 1 67 / 80 Heterogeneous Equilibria Solid-gas equilibria CaCO 3 (s) CaO(s) + CO 2 (g) The equilibrium constant contains pressures in the gas phase: K p = p CO2 The pressure of CO 2, above CaCO 3 and CaO if the system is at equilibrium, is constant at a given temperature. István Szalai (Eötvös University) Lecture 1 68 / 80

Heterogeneous Equilibria Liquid-gas equilibria Example: O 2 uptake of hemoglobin Hem(H 2 O) 4 (l) + O 2 (g) Hem(H 2 O) 3 O 2 (l) + H 2 O(l) Hem(H 2 O) 3 O 2 (l) + O 2 (g) Hem(H 2 O) 2 (O 2 ) 2 (l) + H 2 O(l) Hem(H 2 O) 2 (O 2 ) 2 (l) + O 2 (g) Hem(H 2 O)(O 2 ) 3 (l) + H 2 O(l) Hem(H 2 O)(O 3 ) 2 (l) + O 2 (g) Hem(O 2 ) 4 (l) + H 2 O(l) The equilibrium constant: K p = p O2 K 1 < K 2 < K 3 < K 4 very unusual, cooperative system, to provide the cells with O 2. István Szalai (Eötvös University) Lecture 1 69 / 80 Heterogeneous Equilibria Liquid-solid equilibria Biological significance: tooth and bone formation and decay, kidney and bile stones... Tooth and bone formation: apatite: Ca(OH) 2 3Ca 3 (PO 4 ) 2 flouroapatite: CaF 2 3Ca 3 (PO 4 ) 2 (more resistant to acids) István Szalai (Eötvös University) Lecture 1 70 / 80

Heterogeneous Equilibria Liquid-solid equilibria - slightly soluble salts AgCl(s) Ag + (aq) + Cl (aq) In equilibria that involve slightly soluble compounds in water, the equilibrium constant is called the solubility product constant, K sp K sp = [Ag + ][Cl ] István Szalai (Eötvös University) Lecture 1 71 / 80 Heterogeneous Equilibria General example: Me x A y (s) xme y+ (aq) + ya x (aq) Solubility in pure water: K sp = [Me y+ ] x [A x ] y S = 1 x [Mey+ ] = 1 y [Ax ] István Szalai (Eötvös University) Lecture 1 72 / 80

Heterogeneous Equilibria K sp = (xs) x (ys) y K sp x x y y = S x S y K sp x x y y = S x+y S = x+y K sp x x y y István Szalai (Eötvös University) Lecture 1 73 / 80 Heterogeneous Equilibria Calculate the molar solubility of Ag 2 SO 4 (a) in pure water (b) in 0.10 mol/dm 3 K 2 SO 4 (c) in 0.10 mol/dm 3 AgNO 3 solution! (K sp = 1.7 10 5 ) Ag 2 SO 4 (s) 2Ag + (aq) + SO 2 4 (aq) K sp = [Ag + ] 2 [SO 2 4 ] István Szalai (Eötvös University) Lecture 1 74 / 80

Heterogeneous Equilibria (a) Solubility in pure water: S = 1 2 [Ag+ ] = [SO 2 4 ] S = 1 2 [Ag+ ] = [SO 2 4 ] K sp = (2S) 2 S K sp 2 2 = S 2 S K sp 4 = S 3 S = 3 Ksp 4 S = 3 1.7 10 5 4 S = 0.0161 mol/dm 3 István Szalai (Eötvös University) Lecture 1 75 / 80 Heterogeneous Equilibria (b) Solubility in 0.10 mol/dm 3 K 2 SO 4 S = 1 2 [Ag+ ] [SO 2 4 ] K sp = (2S) 2 [SO 2 4 ] K sp 2 2 [SO 2 4 ] = S 2 K sp S = 4[SO 2 4 ] 1.7 10 5 S = 4 0.10 S = 0.00651 mol/dm 3 István Szalai (Eötvös University) Lecture 1 76 / 80

Heterogeneous Equilibria (c) Solubility in 0.10 mol/dm 3 AgNO 3 S = [SO 2 4 ] 1 2 [Ag+ ] K sp = [Ag + ] 2 S S = K sp [Ag + ] 2 = S 1.7 10 5 0.10 2 S = 0.0017 mol/dm 3 István Szalai (Eötvös University) Lecture 1 77 / 80 Heterogeneous Equilibria Conclusion: The common ions in excess decrease the solubility. The ion on the higher power in the K sp decreases the solubility more. István Szalai (Eötvös University) Lecture 1 78 / 80

Heterogeneous Equilibria Limestone Cave Formation Groundwater entering the cave is saturated with CaCO 3 under fairly acidic conditions with a high partial pressure of CO 2. When cave air mixes with the air from outside, the partial pressure of CO 2 decreases. As a droplet hangs on the wall excess CO 2 comes out from the solution. CO 2 (aq) CO 2 (g) Carbonic acid concentration decreases as the equilibrium shifts to the right. H 2 CO 3 (aq) CO 2 (aq) + H 2 O(l) István Szalai (Eötvös University) Lecture 1 79 / 80 Heterogeneous Equilibria Limestone Cave Formation The decrease in carbonic acid causes a decrease in hydrogen carbonate concentration and the equilibrium shifts to the right. 2HCO 3 (aq) H 2CO 3 (aq) + CO 2 3 (aq) Carbonate ion is formed. The shift decrease the solubility of CaCO 3. This leads to the precipitation of limestone. Ca 2+ (aq) + CO 2 3 (aq) CaCO 3(s) István Szalai (Eötvös University) Lecture 1 80 / 80