1. Chapter 18 Acids and Bases 1. Acids and Bases in Water 2. Autoionization of Water and the pH Scale 3. Proton Transfer and the Brønsted-Lowry Acid-Base Definition 4. Solving Problems Involving Weak-Acid Equilibria 5. Weak Bases and Their Relations to Weak Acids 6. Molecular Properties and Acid Strength 7. Acid-Base Properties of Salt Solutions 8. Electron-Pair Donation and the Lewis Acid-Base Definition

2. Table Selected Acids and Bases Acids Strong hydrochloric acid, HCl hydrobromic acid, HBr hydroiodic acid, HI nitric acid, HNO 3 sulfuric acid, H 2 SO 4 perchloric acid, HClO 4 Weak hydrofluoric acid, HF phosphoric acid, H 3 PO 4 acetic acid, CH 3 COOH (or HC 2 H 3 O 2 ) Bases Strong Weak sodium hydroxide, NaOH calcium hydroxide, Ca(OH) 2 potassium hydroxide, KOH strontium hydroxide, Sr(OH) 2 barium hydroxide, Ba(OH) 2 ammonia, NH 3 amine, (CH 3 ) 2 CHNH 2

3. Brønsted-Lowry Acid-Base Definition An acid-base reaction can now be viewed from the standpoint of the reactants AND the products. Acid and base reaction is a proton transfer process. acid-base conjugate pair An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair. An acid is a proton donor, any species which donates a H +. A base is a proton acceptor, any species which accepts a H +. Molecules as Lewis Acids and base An acid is an electron-pair acceptor. A base is an electron-pair donor. acidbase adduct

4. Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction. (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 +

5. 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 - +

6. 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

7. 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 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 acid weaker 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 acidweaker acid Net direction is to the left with K c < 1.

8. Strong acid: HA(g or l) + H 2 O(l) H 3 O + (aq) + A - (aq) The extent of dissociation for strong acids. The extent of dissociation for weak acids. Weak acid: HA(aq) + H 3 O(l) H 2 O + (aq) + A - (aq) Strong acids dissociate completely into ions in water. K c >> 1 Weak acids dissociate slightly (partially) into ions in water. K c << 1 Weak aicd has a K a value ranging from 10 -1 to 10 -12

9. The Acid-Dissociation Constant Weak acids dissociate very slightly into ions in water. Strong acids dissociate completely into ions in water. HA( g or l ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) K c >> 1 K c << 1 K c = [H 3 O + ][A - ] [H 2 O][HA] K c [H 2 O] = K a = [H 3 O + ][A - ] [HA] stronger acid higher [H 3 O + ] larger K a smaller K a lower [H 3 O + ] weaker acid

10. 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.

11. Autoionization of Water and the pH Scale H 2 O(l) H 3 O + (aq) OH - (aq) + + Ball-and-stick model: The two polar O-H bonds. The bent molecule shape give rise to the polar nature of H 2 O.

12. 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 = A change in [H 3 O + ] causes an inverse change in [OH - ]. = 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 - ]

13. The relationship between [H 3 O + ] and [OH - ] and the relative acidity of solutions. [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

14. 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 0 C and obtains a solution with [H 3 O + ] = 3.0  10 -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

15. The pH values of some aqueous solutions pH = - log [H 3 O + ] pOH = - log [OH - ] pH + pOH = 14

16. Summary Acid and base are essential substances in home, industry and environment. In aqueous solution, water combines with the proton released from acid to form hydronium ions, H 3 O +. Acid contains H and yields H 3 O +, while base contain OH and yield OH - in aqueous solution. Strong acid dissociate completely and weak acid dissociate partially. The extend of dissociation is expressed by the acid-dissociation constant, K a. Weak aicd has a K a value ranging from 10 -1 to 10 -12. Water undergoes the self-ionization. Autoionization (Autoprotolysis) process is described by an equilibrium H 2 O (l) + H 2 O (l) = H 3 O + (aq) + OH - (aq) The equilibrium constant is the ion-product for water, K w =1.0  10 -14 We use pH scale to express the values of H 3 O +. pH 7, basic.

17. The relations among [H 3 O + ], pH, [OH - ], and pOH.

18. Take-home message: 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 2.0 M, 0.30 M, and 0.0063M HNO 3. Calculate [H 3 O + ], pH, [OH - ], and pOH of the three solutions at 25 0 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

19. Strengths of conjugate acid- base pairs

20. 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 )

21. 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 %

22. 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

23. 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%

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. 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 )

26. - x-+ x Calculating Equilibrium Concentrations for a Polyprotic Acid 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

27. Summary for last lecture Acid and base are essential substances in home, industry and environment. In aqueous solution, water combines with the proton released from acid to form hydronium ions, H 3 O +. Acid contains H and yields H 3 O +, while base contain OH and yield OH - in aqueous solution. (Arrhenius) Strong acid dissociate completely and weak acid dissociate partially. The extend of dissociation is expressed by the acid-dissociation constant, K a. Weak acid has a K a value ranging from 10 -1 to 10 -12. Water undergoes the self-ionization. Autoionization (Autoprotolysis) process is described by an equilibrium H 2 O (l) + H 2 O (l) = H 3 O + (aq) + OH - (aq) The equilibrium constant is the ion-product for water, K w =1.0  10 -14 We use pH scale to express the values of H 3 O +. pH 7, basic.

28. Summary 1.The Bronsted-Lowry acid and base definition : acid does not require to contain H or base contains OH. 2.Acid base reaction does not require to occur in aqueous solution. 3.An acid is a proton donor, any species which donates a H +. 4.A base is a proton acceptor, any species which accepts a H +. 5.Acid donates its proton and becomes its conjugate base. 6.Base accepts proton and becomes its conjugate acid. 7.A stronger acid gives a weaker conjugate base, or vise versa. 8.The reaction proceeds to the net reaction in which a stronger acid and base to form a weaker acid and base. 9.The common type of weak acid equilibrium problems involve finding K a from known concentration or finding concentration from known K a. 10.We can make assumption to simplify the calculation by neglecting x. 11.Monoprotic acid has one ionizable proton. 12.Polyprotic acid contains more than one ionizable proton.

29. + CH 3 NH 3 + OH - methylammonium ion Abstraction of a proton from water by methylamine. + CH 3 NH 2 H2OH2O methylamine Lone pair binds H + Bronsted-Lowry base: any species accept a proton. B + H 2 O BH + + OH - K b = [HB + ][OH - ] [B][H 2 O ] Base dissociation constant Base hydrolysis constant

30. 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 5.9  10 -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 negligible Initial1.50M0 10 -7 ≈ 0 - Change- x-+ x Equilibrium1.50 - x ≈ 1.5-xx (CH 3 ) 2 NH(aq) + H 2 O(l) (CH 3 ) 2 NH 2 + (aq) + OH - (aq)Concentration

31. Determining pH from K b and Initial [B] continued K b = 5.9  10 -4 = [(CH 3 ) 2 NH 2 + ][OH - ] [(CH 3 ) 2 NH] 5.9x10 -4 = (x) 1.5M x = 3.0  10 -2 M = [OH - ] Check assumption: 3.0  10 -2 M/1.5M  100 % = 2% [H 3 O + ] = K w / [OH - ] = 1.0  10 -14 /3.0  10 -2 = 3.3  10 -13 M pH = -log 3.3  10 -13 = 12.48 pOH = -log 3.0  10 -2 = 1.52 pH = 14- pH = 14 - 1.52 = 12.48 pH calculation: Method 01: pH calculation: Method 02: pH calculation:

32. There exists an important relationship between the K a of HA and the K b of A -. There is the reaction of dissociation of acid and hydrolysis of base. HA + H 2 O  H 3 O + + A -(1) K a A - + H 2 O  HA + OH -(2) K b The sum of the reaction H 2 O + H 2 O  OH - + H + K w =? K w = K a K b =  = [H 3 O + ][OH - ] [A - ][H 3 O + ] [HA][H 2 O] The relation between K a and k b of a conjugate acid-base pair [HA][OH - ] [A - ][H 2 O] K w = K a  K b

33. 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 - ] = K w K a = 5.6  10 -10 M K b = 1.0x10 -14 1.8x10 -5

34. Determining the pH of a Solution of A - continued K b = [HAc][OH - ] [Ac - ] [Ac-] = 0.25M-x ≈ 0.25M 5.6  10 -10 = x 2 /0.25M x = 1.2  10 -5 M = [OH - ] Check assumption: 1.2  10 -5 M/0.25M  100% = 4.8  10 -3 % [H 3 O + ] = K w /[OH - ] = 1.0  10 -14 /1.2  10 -5 = 8.3  10 -10 M pH = -log 8.3  10 -10 = 9.08 pH is unitless.

36. 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.

37. 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. 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)

38. Summary 1.We can construct the reaction table to calculate pH, concentration of a chemical species, or K a /K b. 2.We can make assumption to simplify the calculation by neglecting x. 3.Monoprotic acid has one ionizable proton. 4.Polyprotic acid contains more than one ionizable protons. 5.The first dissociation of polyprotic acid provides virtually all the H 3 O +. 6.The extent to which a weak base abstracts a proton from H 2 O to form OH - is expressed by a base hydrolysis constant K b. 7.By multiplying the expression K a of HA and K b of A -, we obtain K w. 8.The relationship between K a and K b allows us to calculate K a and K b or pH and pOH.