1. Chapter 18 – Acid-Base Equilibria 18.1 – Acids and Bases in Water 18.2 – Autoionization of Water and the pH Scale 18.3 – Proton Transfer and the Br ø nsted-Lowry Acid-Base Definition 18.4 – Solving Problems Involving Weak-Acid Equilibria 18.5 – Weak Bases and Their Relation to Weak Acids 18.6 – Molecular Properties and Acid Strength 18.7 – Acid-Base Properties of Salt Solutions 18.8 – Generalizing the Br ø nsted-Lowry Concept: The Leveling Effect 18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition 1

2. 18.1 – Acids and Bases in Water 2

3. 3 Strong acid: HA( g or l ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) Weak acid: HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) Zn with 1M HCl( aq ) vs. 1M CH 3 COOH( aq )

4. 18.1 – Acids and Bases in Water 4 ACID STRENGTH

5. 18.1 – Acids and Bases in Water 5 SAMPLE PROBLEM 18.1: SOLUTION: Classifying Acid and Base Strength from the Chemical Formula PROBLEM:Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. (a) H 2 SeO 4 (b) (CH 3 ) 2 CHCOOH(c) KOH(d) (CH 3 ) 2 CHNH 2 PLAN:Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. (a) Strong acid - H 2 SeO 4 - the number of O atoms exceeds the number of ionizable protons by 2. (b) Weak acid - (CH 3 ) 2 CHCOOH is an organic acid having a -COOH group. (c) Strong base - KOH is a Group 1A(1) hydroxide. (d) Weak base - (CH 3 ) 2 CHNH 2 has a lone pair of electrons on the N and is an amine.

6. 18.2 – Autoionization of Water and the pH Scale 6 H 2 O( l ) H 3 O + ( aq )OH - ( aq ) + +

7. 18.2 – Autoionization of Water and the pH Scale 7 K c = [H 3 O + ][OH - ] [H 2 O] 2 K c [H 2 O] 2 =[H 3 O + ][OH - ] The Ion-Product Constant for Water, K w : K w = A change in [H 3 O + ] causes an inverse change in [OH - ], and vice versa. = 1.0 x 10 -14 at 25  C H 2 O( l ) + H 2 O( l ) H 3 O + ( aq ) + OH - ( aq ) in an acidic solution, [H 3 O + ] > [OH - ] in a basic solution, [H 3 O + ] < [OH - ] in a neutral solution, [H 3 O + ] = [OH - ]

8. 18.2 – Autoionization of Water and the pH Scale 8 [H 3 O + ][OH - ] Divide into K w ACIDIC SOLUTION BASIC SOLUTION [H 3 O + ] > [OH - ] [H 3 O + ] = [OH - ][H 3 O + ] < [OH - ] NEUTRAL SOLUTION

9. 18.2 – Autoionization of Water and the pH Scale 9 SAMPLE PROBLEM 18.2:Calculating [H 3 O + ] and [OH - ] in an Aqueous Solution PROBLEM:A research chemist adds a measured amount of HCl gas to pure water at 25 ° C and obtains a solution with [H 3 O + ] = 3.0x10 -4 M. Calculate [OH - ]. Is the solution neutral, acidic, or basic? SOLUTION: PLAN:Use the K w at 25 0 C and the [H 3 O + ] to find the corresponding [OH - ]. K w = 1.0x10 -14 = [H 3 O + ] [OH - ] so [OH - ] = K w / [H 3 O + ] = 1.0x10 -14 /3.0x10 -4 = [H 3 O + ] is > [OH - ] and the solution is acidic. 3.3x10 -11 M

10. 18.2 – Autoionization of Water and the pH Scale pH = -log [H 3 O + ] # of sig figs pH of a neutral soln = 7.00 pH of an acidic soln < 7.00 pH of a basic soln > 7.00 1 pH unit = 10x change [H 3 O + ] = 10 -pH p-scales – pOH = -log [OH - ] – pK = -log K 10

11. 18.2 – Autoionization of Water and the pH Scale 11 Table 18.3 The Relationship Between K a and pK a Acid Name (Formula)K a at 25  CpK a Hydrogen sulfate ion (HSO 4 - ) 1.02x10 -2 Nitrous acid (HNO 2 ) Acetic acid (CH 3 COOH) Hypobromous acid (HBrO) Phenol (C 6 H 5 OH) 7.1x10 -4 1.8x10 -5 2.3x10 -9 1.0x10 -10 1.991 3.15 4.74 8.64 10.00

12. 18.2 – Autoionization of Water and the pH Scale 12

13. 18.2 – Autoionization of Water and the pH Scale 13 SAMPLE PROBLEM 18.3:Calculating [H 3 O + ], pH, [OH - ], and pOH PROBLEM:In an art restoration project, a conservator prepares copper- plate etching solutions by diluting concentrated HNO 3 to 2.0M, 0.30M, and 0.0063M HNO 3. Calculate [H 3 O + ], pH, [OH - ], and pOH of the three solutions at 25 ° C. SOLUTION: PLAN:HNO 3 is a strong acid so [H 3 O + ] = [HNO 3 ]. Use K w to find the [OH - ] and then convert to pH and pOH. For 2.0M HNO 3, [H 3 O + ] = 2.0M and -log [H 3 O + ] = -0.30 = pH [OH - ] = K w / [H 3 O + ] = 1.0x10 -14 /2.0 = 5.0x10 -15 M; pOH = 14.30 [OH - ] = K w / [H 3 O + ] = 1.0x10 -14 /0.30 = 3.3x10 -14 M; pOH = 13.48 For 0.3M HNO 3, [H 3 O + ] = 0.30M and -log [H 3 O + ] = 0.52 = pH [OH - ] = K w / [H 3 O + ] = 1.0x10 -14 /6.3x10 -3 = 1.6x10 -12 M; pOH = 11.80 For 0.0063M HNO 3, [H 3 O + ] = 0.0063M and -log [H 3 O + ] = 2.20 = pH

14. 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition 14 (acid, H + donor)(base, H + acceptor) HClH2OH2O + Cl - H3O+H3O+ + Lone pair binds H + (base, H + acceptor)(acid, H + donor) NH 3 H2OH2O + NH 4 + OH - + Lone pair binds H +

15. 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition 15 Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions BaseAcid+ Base+ Conjugate Pair Reaction 4H 2 PO 4 - OH - + Reaction 5H 2 SO 4 N2H5+N2H5+ + Reaction 6HPO 4 2- SO 3 2- + Reaction 1HFH2OH2O+F-F- H3O+H3O+ + Reaction 3NH 4 + CO 3 2- + Reaction 2HCOOHCN - +HCOO - HCN+ NH 3 HCO 3 - + HPO 4 2- H2OH2O+ HSO 4 - N 2 H 6 2+ + PO 4 3- HSO 3 - +

16. 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition 16 SAMPLE PROBLEM 18.4:Identifying Conjugate Acid-Base Pairs PROBLEM:The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H 2 PO 4 - ( aq ) + CO 3 2- ( aq ) HPO 4 2- ( aq ) + HCO 3 - ( aq ) (b) H 2 O( l ) + SO 3 2- ( aq ) OH - ( aq ) + HSO 3 - ( aq ) SOLUTION: PLAN:Identify proton donors (acids) and proton acceptors (bases). (a) H 2 PO 4 - ( aq ) + CO 3 2- ( aq ) HPO 4 2- ( aq ) + HCO 3 - ( aq ) proton donor proton acceptor proton donor conjugate pair 1 conjugate pair 2 (b) H 2 O( l ) + SO 3 2- ( aq ) OH - ( aq ) + HSO 3 - ( aq ) conjugate pair 2 conjugate pair 1 proton donor proton acceptor proton donor

17. 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition 17

18. 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition 18 SAMPLE PROBLEM 18.5:Predicting the Net Direction of an Acid-Base Reaction PROBLEM:Predict the net direction and whether K a is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2- ( aq ) + NH 4 + ( aq ) SOLUTION: PLAN:Identify the conjugate acid-base pairs and then consult Figure 18.10 to determine the relative strength of each. The stronger the species, the more preponderant its conjugate. (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2- ( aq ) + NH 4 + ( aq ) stronger acidweaker acidstronger baseweaker base Net direction is to the right with K c > 1. (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) stronger baseweaker basestronger acid weaker acid Net direction is to the left with K c < 1.

19. 18.4 – Solving Problems Involving Weak-Acid Equilibria 19 SAMPLE PROBLEM 18.6:Finding the K a of a Weak Acid from the pH of Its Solution PROBLEM:Phenylacetic acid (C 6 H 5 CH 2 COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.13M HPAc is 2.62. What is the K a of phenylacetic acid? PLAN:Write out the dissociation equation. Use pH and solution concentration to find the K a. K a = [H 3 O + ][PAc - ] [HPAc] Assumptions: With a pH of 2.62, the [H 3 O + ] HPAc >> [H 3 O + ] water. [PAc - ] ≈ [H 3 O + ]; since HPAc is weak, [HPAc] initial ≈ [HPAc] initial - [HPAc] dissociation SOLUTION:HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq )

20. 18.4 – Solving Problems Involving Weak-Acid Equilibria 20 SAMPLE PROBLEM 18.6:Finding the K a of a Weak Acid from the pH of Its Solution continued Concentration(M)HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq ) Initial0.12-1x10 -7 0 Change--x-x+x+x+x+x Equilibrium-0.12-xxx +(<1x10 -7 ) [H 3 O + ] = 10 -pH = 2.4x10 -3 M which is >> 10 -7 (the [H 3 O + ] from water) x ≈ 2.4x10 -3 M ≈ [H 3 O + ] ≈ [PAc - ][HPAc] equilibrium = 0.12-x ≈ 0.12 M So K a = (2.4x10 -3 ) 0.12 = 4.8 x 10 -5 Be sure to check for % error.= 4x10 -3 % x100 [HPAc] dissn ; 2.4x10 -3 M 0.12M [H 3 O + ] from water ; 1x10 -7 M 2.4x10 -3 M x100 = 2.0 %

21. 18.4 – Solving Problems Involving Weak-Acid Equilibria 21 SAMPLE PROBLEM 18.7:Determining Concentrations from K a and Initial [HA] PROBLEM:Propanoic acid (CH 3 CH 2 COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H 3 O + ] of 0.10M HPr (K a = 1.3x10 -5 )? SOLUTION: PLAN:Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute. x = [HPr] diss = [H 3 O + ] from HPr = [Pr - ] Assumptions:For HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq ) K a =[H 3 O + ][Pr - ] [HPr] HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq )Concentration(M) Initial0.10-00 Change--x-x+x+x+x+x Equilibrium-0.10-xxx Since K a is small, we will assume that x << 0.10

22. 18.4 – Solving Problems Involving Weak-Acid Equilibria 22 SAMPLE PROBLEM 18.7:Determining Concentrations from K a and Initial [HA] continued (x)(x) 0.10 1.3x10 -5 = [H 3 O + ][Pr - ] [HPr] = = 1.1x10 -3 M = [H 3 O + ] Check: [HPr] diss = 1.1x10 -3 M/0.10 M x 100 = 1.1%

23. 18.4 – Solving Problems Involving Weak-Acid Equilibria 23 Percent HA dissociation = [HA] dissociated [HA] initial x 100

24. 18.4 – Solving Problems Involving Weak-Acid Equilibria 24 Polyprotic acids acids with more than more ionizable proton H 3 PO 4 ( aq ) + H 2 O( l ) H 2 PO 4 - ( aq ) + H 3 O + ( aq ) H 2 PO 4 - ( aq ) + H 2 O( l ) HPO 4 2- ( aq ) + H 3 O + ( aq ) HPO 4 2- ( aq ) + H 2 O( l ) PO 4 3- ( aq ) + H 3 O + ( aq ) K a1 = [H 3 O + ][H 2 PO 4 - ] [H 3 PO 4 ] K a2 = [H 3 O + ][HPO 4 2- ] [H 2 PO 4 - ] K a3 = [H 3 O + ][PO 4 3- ] [HPO 4 2- ] K a1 > K a2 > K a3 = 7.2x10 -3 = 6.3x10 -8 = 4.2x10 -13

25. 18.4 – Solving Problems Involving Weak-Acid Equilibria 25 ACID STRENGTH

26. 18.4 – Solving Problems Involving Weak-Acid Equilibria 26 SAMPLE PROBLEM 18.8:Calculating Equilibrium Concentrations for a Polyprotic Acid PROBLEM:Ascorbic acid (H 2 C 6 H 6 O 6 ; H 2 Asc for this problem), known as vitamin C, is a diprotic acid (K a1 = 1.0x10 -5 and K a2 = 5x10 -12 ) found in citrus fruit. Calculate [H 2 Asc], [HAsc - ], [Asc 2- ], and the pH of 0.050M H 2 Asc. SOLUTION: PLAN:Write out expressions for both dissociations and make assumptions. K a1 >> K a2 so the first dissociation produces virtually all of the H 3 O +. K a1 is small so [H 2 Asc] initial ≈ [H 2 Asc] diss After finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation. K a1 = [HAsc - ][H 3 O + ] [H 2 Asc] = 1.0x10 -5 K a2 = [Asc 2- ][H 3 O + ] [HAsc - ] = 5x10 -12 H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq ) HAsc - ( aq ) + H 2 O (l ) Asc 2- ( aq ) + H 3 O + ( aq )

27. 18.4 – Solving Problems Involving Weak-Acid Equilibria 27 - x-+ x continued H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq )Concentration(M) Initial0.050-00 Equilibrium0.050 - x-xx K a1 = [HAsc - ][H 3 O + ]/[H 2 Asc] = 1.0x10 -5 = (x)(x)/0.050 M pH = -log(7.1x10 -4 ) = 3.15 7.1x10 -4 M-00 Change- x-+ x 7.1x10 -4 - x-xx Equilibrium Change Initial x x = 7.1x10 -4 M HAsc - ( aq ) + H 2 O (l ) Asc 2- ( aq ) + H 3 O + ( aq )Concentration(M) x = 6x10 -8 M

28. 18.5 – Weak Bases and Their Relation to Weak Acids 28 BASE STRENGTH K b = [BH + ][OH - ] [B]

29. 18.5 – Weak Bases and Their Relation to Weak Acids 29 + CH 3 NH 3 + OH - methylammonium ion + CH 3 NH 2 H2OH2O methylamine Lone pair binds H +

30. 18.5 – Weak Bases and Their Relation to Weak Acids 30 SAMPLE PROBLEM 18.9:Determining pH from K b and Initial [B] PROBLEM:Dimethylamine, (CH 3 ) 2 NH, a key intermediate in detergent manufacture, has a K b of 5.9x10 -4. What is the pH of 1.5M (CH 3 ) 2 NH? SOLUTION: PLAN:Perform this calculation as you did those for acids. Keep in mind that you are working with K b and a base. (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq ) Assumptions: [(CH 3 ) 2 NH 2 + ] = [OH - ] = x ; [(CH 3 ) 2 NH 2 + ] - x ≈ [(CH 3 ) 2 NH] initial K b >> K w so [OH - ] from water is neglible Initial1.50M00- Change- x-+ x Equilibrium1.50 - x-xx (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq )Concentration

31. 18.5 – Weak Bases and Their Relation to Weak Acids 31 K b = 5.9x10 -4 = [(CH 3 ) 2 NH 2 + ][OH - ] [(CH 3 ) 2 NH] 5.9x10 -4 = (x) 1.5M x = 3.0x10 -2 M = [OH - ] Check assumption:3.0x10 -2 M/1.5M x 100 = 2% [H 3 O + ] = K w /[OH - ] = 1.0x10 -14 /3.0x10 -2 = 3.3x10 -13 M pH = -log 3.3x10 -13 = 12.48

32. 18.5 – Weak Bases and Their Relation to Weak Acids 32 SAMPLE PROBLEM 18.10:Determining the pH of a Solution of A - PROBLEM:Sodium acetate (CH 3 COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? K a of acetic acid (HAc) is 1.8x10 -5. SOLUTION: PLAN:Sodium salts are soluble in water so [Ac - ] = 0.25M. Write the association equation for acetic acid; use the K a to find the K b. Initial0.25M-00 Change-x-x+x+x+x- Equilibrium-0.25M-xxx Ac - ( aq ) + H 2 O( l ) HAc( aq ) + OH - ( aq )Concentration K b = [HAc][OH - ] [Ac - ] = KwKw KaKa = 5.6x10 -10 M K b = 1.0x10 -14 1.8x10 -5

33. 18.5 – Weak Bases and Their Relation to Weak Acids 33 K b = [HAc][OH - ] [Ac - ] [Ac-] = 0.25M-x ≈ 0.25M 5.6x10 -10 = x 2 /0.25M x = 1.2x10 -5 M = [OH - ] Check assumption:1.2x10 -5 M/0.25M x 100 = 4.8x10 -3 % [H 3 O + ] = K w /[OH - ]= 1.0x10 -14 /1.2x10 -5 = 8.3x10 -10 M pH = -log 8.3x10 -10 M = 9.08

34. 18.6 – Molecular Properties and Acid Strength 34 6A(16) H2OH2O H2SH2S H 2 Se H 2 Te 7A(17) HF HCl HBr HIHI Electronegativity increases, acidity increases Bond strength decreases, acidity increases Figure 18.12 The effect of atomic and molecular properties on nonmetal hydride acidity.

35. 18.6 – Molecular Properties and Acid Strength 35 HO I HOBrHOCl>> HO O O O << Figure 18.13 The relative strengths of oxoacids.  HOCl  

36. 18.6 – Molecular Properties and Acid Strength 36 Table 18.7 K a Values of Some Hydrated Metal Ions at 25 0 C Free IonHydrated IonKaKa Fe 3+ Fe(H 2 O) 6 3+ ( aq ) 6 x 10 -3 Sn 2+ Sn(H 2 O) 6 2+ ( aq ) 4 x 10 -4 Cr 3+ Cr(H 2 O) 6 3+ ( aq ) 1 x 10 -4 Al 3+ Al(H 2 O) 6 3+ ( aq ) 1 x 10 -5 Cu 2+ Cu(H 2 O) 6 2+ ( aq ) 3 x 10 -8 Pb 2+ Pb(H 2 O) 6 2+ ( aq ) 3 x 10 -8 Zn 2+ Zn(H 2 O) 6 2+ ( aq ) 1 x 10 -9 Co 2+ Co(H 2 O) 6 2+ ( aq ) 2 x 10 -10 Ni 2+ Ni(H 2 O) 6 2+ ( aq ) 1 x 10 -10 ACID STRENGTH

37. 18.6 – Molecular Properties and Acid Strength 37 Al(H 2 O) 5 OH 2+ Al(H 2 O) 6 3+ Figure 18.13 The acidic behavior of the hydrated Al 3+ ion.H2OH2O H3O+H3O+ Electron density drawn toward Al 3+ Nearby H 2 O acts as base

38. 18.7 – Acid-Base Properties of Salt Solutions 38

39. 18.7 – Acid-Base Properties of Salt Solutions 39 SAMPLE PROBLEM 18.11:Predicting Relative Acidity of Salt Solutions PROBLEM:Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water: (a) Potassium perchlorate, KClO 4 (b) Sodium benzoate, C 6 H 5 COONa (c) Chromium trichloride, CrCl 3 (d) Sodium hydrogen sulfate, NaHSO 4 SOLUTION: PLAN:Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. (a) The ions are K + and ClO 4 -, both of which come from a strong base(KOH) and a strong acid(HClO 4 ). Therefore the solution will be neutral. (b) Na + comes from the strong base NaOH while C 6 H 5 COO - is the anion of a weak organic acid. The salt solution will be basic. (c) Cr 3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H 3 O +. Cl - comes from the strong acid HCl. Acidic solution. (d) Na + comes from a strong base. HSO 4 - can react with water to form H 3 O +. So the salt solution will be acidic.

40. 18.7 – Acid-Base Properties of Salt Solutions 40 SAMPLE PROBLEM 18.12:Predicting the Relative Acidity of Salt Solutions from K a and K b of the Ions PROBLEM:Determine whether an aqueous solution of zinc formate, Zn(HCOO) 2, is acidic, basic, or neutral. SOLUTION: PLAN:Both Zn 2+ and HCOO - come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables 18.2 and 18.7. K a Zn(H 2 O) 6 2+ = 1x10 -9 K a HCOO - = 1.8x10 -4 ; K b = K w /K a = 1.0x10 -14 /1.8x10 -4 = 5.6x10 -11 K a for Zn(H 2 O) 6 2+ >>> K b HCOO -, therefore the solution is acidic. Zn(H 2 O) 6 2+ ( aq ) + H 2 O( l ) Zn(H 2 O) 5 OH + ( aq ) + H 3 O + ( aq ) HCOO - ( aq ) + H 2 O( l ) HCOOH( aq ) + OH - ( aq )

41. 18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition 41 acidbaseadduct An acid is an electron-pair acceptor. A base is an electron-pair donor. M 2+ H 2 O( l ) M(H 2 O) 4 2+ ( aq ) adduct

42. 18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition 42 Figure 18.15 The Mg 2+ ion as a Lewis acid in the chlorophyll molecule.

43. 18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition 43 SAMPLE PROBLEM 18.13:Identifying Lewis Acids and Bases PROBLEM:Identify the Lewis acids and Lewis bases in the following reactions: (a) H + + OH - H 2 O (b) Cl - + BCl 3 BCl 4 - (c) K + + 6H 2 O K(H 2 O) 6 + SOLUTION: PLAN:Look for electron pair acceptors (acids) and donors (bases). (a) H + + OH - H 2 O acceptor donor (b) Cl - + BCl 3 BCl 4 - donor acceptor (c) K + + 6H 2 O K(H 2 O) 6 + acceptor donor