2. Acids and Bases Acids and bases Acid-base properties of water (K w ) pH scale Strength of Acids and Bases Weak acid (K a ) Weak base (K b ) Relation between conjugate acid-base ionization constants (K a and K b ) Chemical structure and acidity Acid-base properties of salt solutions Lewis acid-base

3. I. Acid and Base A. Arrehenius acid - a compound that increases [H + ] in water B. Arrehenius Base : a compound that increases [OH - ] in water HCl(aq) +H 2 O(l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq)+H 2 O(l) H 3 O + (aq) + NO 3 - (aq) NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)

4. I. Acid and Base C. Brønsted acid : a proton donor D. Brønsted Base : a proton acceptor NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) acidbaseacidbase E. Conjugate acid-base pair : acid-base pair that are related by a loss or gain of a proton H + acid conjugate base base conjugate acid

5. Example: Conjugate acid-base pair Classify each of the following species a Brønsted acid, base, or both. Give the conjugate acid or base. CO 3 2– base speciesAcid/base/bothConjugate acidConjugate base HCO 3 – H2OH2ObothOH – H3O+H3O+ H 2 SO 4 acid HSO 4 – Amphoteric: a substance that can act either as an acid or a base H 2 PO 4 – bothHPO 4 2– H 3 PO 4

6. II. Acid-base properties of water A. Autoionization of water: water is a very weak electrolyte H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) O H H+ O H H O H HH O H - + [] + H 2 O (l) H + (aq) + OH - (aq) acid conjugate base baseconjugate acid

7. B. Ion-product constant, K w H 2 O (l) H + (aq) + OH - (aq) K C = K w = [H + ][OH - ] (1) K w : the product of the molar concentrations of H + and OH - ions at a particular temperature. K w = [H + ][OH - ] = 1.0 x 10 - 14 at 25 o C [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Neutral solution Acidic solution Basic solution (2) H + and OH – ions are always present in aqueous solutions

8. Q: What is the concentration of H + in a (a) neutral solution and (b) 2.0 x10 -2 M NaOH solution at 25 o C? K w = [H + ][OH - ] = 1.0 x 10 - 14 (b) [OH – ] = 2.0 x10 -2 M [H + ] = KwKw [OH – ] 1.0 x 10 -14 2.0 x10 -2 = = 5.0 x 10 -13 M (basic) (a) Neutral solution [H + ] = [OH – ]= x = 1.0 x 10 -7 M K w = [H + ][OH - ] = 1.0 x 10 - 14 = x 2

9. III. pH scale – a measure of acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = pOH = 7 pH 7 pH > 7 pOH <7 At 25 0 C [H + ]pH pOH = - log [OH – ] As the solution is more acidic when At 25 o C [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00 pOH

10. 16.1

11. Q: If a solution has a pOH value of 10.5, what is the H + ion concentration? pH = 3.5 = - log [H + ] [H + ] = 10 -pH = 10 -3.5 = 3.2 x 10 -4 M pH = 14.0 - pOH = 14.0-10.5 = 3.5 Q: The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40 (acidic) (basic)

12. IV. Strength of Acid and Base A.Strong Electrolyte – 100% dissociation (ionization) - soluble ionic compound, strong acid, strong base NaCl (s) Na + (aq) + Cl - (aq) H2OH2O B.Weak Electrolyte – not completely dissociated - weak acid, weak base CH 3 COOH CH 3 COO - (aq) + H + (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) NaOH (s) Na + (aq) + OH - (aq) H2OH2O NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq)

13. HCl, strong acidHF, weak acid

14. C. Strong acid Strong electrolyte, completely ionize (dissociate) in water HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq) Q: What is the pH of the solution when 8.4 g of HCl is dissolved in 1.2 L of water? HCl is a strong acid – 100% dissociation. H 2 O (l) + HCl (aq) H 3 O + (aq) + Cl - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.19) = 0.72 Initial Equilibrium 0.19 M 0.00 M = 0.19 M Change -0.19 M 0.19 M

15. D. Strong Base Hydroxide of alkali (IA) and alkaline earth (IIA) metal Ionic metal oxide (Na 2 O, CaO), metal hydride, and nitrides Na 2 O (s) + H 2 O (l) 2Na + (aq) + 2OH - (aq) Ex: What is the pH of a 0.048 M Ca(OH) 2 solution? Ca(OH) 2 is a strong base – 100% dissociation. Ca(OH) 2 (aq) Ca 2+ (aq) + 2OH - (aq) pOH = -log [OH - ] = -log(0.096) = 1.02 Initial Equilibrium 0.048 M 0.096 M 0.000 M Change -0.048 M 0.048 M 2 x 0.048 M Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O pH = 14.00 - pOH = 12.98

16. E. Conjugate acid-base pair The conjugate base of a strong acid is extremely weak The conjugate acid of a strong base is extremely weak. The stronger the acid, the weaker the conjugate base The stronger the base, the weaker the conjugate acid Ex: Predict the direction of the following reaction HCN (aq) + F – (aq) HF (aq) + CN – (aq) Which is a stronger acid? HCN or HF Equilibrium will proceed from right to left since HF is a stronger acid and CN – is a stronger base. Which is a stronger base? CN – or F – K C < 1 * Reaction goes to the side that produces gas, solid, or weak/nonelectrolyte

17. 16.2

18. V. Weak Acid A.K a : acid ionization constant HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] KaKa acid strength

20. Ex: What is the pH of a 0.80 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.800.00 -x-x+x+x 0.80 - x 0.00 +x+x xx K a = x2x2 0.80 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.80 = 7.1 x 10 -4 0.80 – x  0.80 K a << 1 x 2 = 5.68 x 10 -4 x = 0.024 M [H + ] = [F - ] = 0.024 M pH = -log [H + ] = 1.62 [HF] = 0.80 – x = 0.78 M

21. When will this approximation work? 0.80 – x  0.80 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.024 0.024 M 0.80 M x 100% = 3.0% Less than 5% Approximation ok. Ex. What is the pH of a 0.010 M HF solution (at 25 0 C)? Ka  Ka  x2x2 0.010 = 7.1 x 10 -4 x = 0.0027 M 0.0027 M 0.010 M x 100% = 27% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or other method. K a = x2x2 0.010 - x = 7.1 x 10 -4 x 2 + 7.1 x 10 -4 x -7.1 x 10 -6 = 0 [H + ] = x = 0.0023 M pH = -log [H + ] = 2.64

22. Steps to solve weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution.

23. V. Weak Acid B. Percent ionization % ionization = ionized acid concentration at equilibrium initial concentration of acid x 100% For a monoprotic acid HA % ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration *Percent ionization as initial concentration increases decreases

24. V. Weak Acid C.Polyprotic acid Acids that have more than one ionizable H H 2 S (aq) H + (aq) + HS - (aq) K a1 = 9.5 x10 -8 HS – (aq) H + (aq) + S 2 - (aq) K a2 = 1 x10 -19 H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) K a1 = 7.5 x10 -3 H 2 PO 4 – (aq) H + (aq) + HPO 4 2– (aq) K a2 = 6.2 x10 -8 HPO 4 2– (aq) H + (aq) + PO 4 3– (aq) K a3 = 4.8 x10 -13 K a3 K a2 K a1 << diprotic triprotic

25. Ex: What is the pH of a 0.024 M H 2 SO 3 solution at 25 o C? (K a1 = 1.3 x 10 -2 and K a2 =6.3 x 10 -8 ) Initial (M) Change (M) Equilibrium (M) 0.0240.00 -x-x+x+x 0.024 - x 0.00 +x+x xx K a1 = x2x2 0.024 - x = 1.3 x 10 -2 x 2 + 1.3 x 10 -2 - 3.12 x 10 -4 = 0 [H + ] = [HSO 3 - ] = 0.0123 M [H 2 SO 3 ] = 0.024 – x = 0.0117 M H 2 SO 3 (aq) H + (aq) + HSO 3 – (aq) x = 0.0123 or -0.025 Not done yet! Next step, let us consider the second ionization of H Treat this type problem with step-by-step ionization

26. Initial (M) Change (M) Equilibrium (M) 0.0123 -y-y+y+y 0.0123 - y 0.00 +y+y 0.0123+yy K a2 = (0.0123+y)y 0.0123 - y = 6.3 x 10 -8 [H + ] = 0.0123 + y = 0.012 M [HSO 3 – ] = 0.0123 – y = 0.012 M HSO 3 – (aq) H + (aq) + SO 3 – (aq) y= 6.3 x10 -8 0.0123 – y  0.0123 K a2 << 1 0.0123 + y  0.0123 [SO 3 2– ] = y = 6.3 x10 -8 M [H 2 SO 3 ] = 0.012 M pH = -log[H + ] =1.92 6.3 x10 -8 M 0.0123 M x 100% << 5 %

27. VI. Weak Base A.K b : base ionization constant B (aq) + H 2 O (l) HB + (aq) + OH - (aq) K b = [HB + ][OH - ] [B] KbKb base strength NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

28. Ex: What is the pH of a 0.50 M C 2 H 5 NH 2 solution at 25 0 C(K b =5.6 x10 -4 )? C 2 H 5 NH 2 ( aq) +H 2 O(l) C 2 H 5 NH 3 + (aq) + OH - (aq) Initial (M) Change(M) Equilibrium(M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K b = x2x2 0.50 - x = 5.6 x 10 -4 Kb  Kb  x2x2 0.50 = 5.6 x 10 -4 0.50 – x  0.50 K b << 1 x 2 = 2.8 x 10 -4 x = 0.017 M [OH – ] = [C 2 H 5 NH 3 + ] = 0.017 M pH = 14.00 - pOH = 14.00 + log [OH – ] = 12.23 [C 2 H 5 NH 2 ] = 0.50 – x = 0.48 M 0.017 M 0.50 M x 100% = 3.4 % assumption okay

29. 16.5

30. VII. Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) C 6 H 5 COO – (aq) + H 2 O (l) OH - (aq) + C 6 H 5 COOH (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak acid (base) and its conjugate base (acid) Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa Ex. What is the reaction and the equilibrium constant of C 6 H 5 COO – in water? C 6 H 5 COO – is a baseand is the conjugate base of C 6 H 5 COOH K b = = KwKw KaKa = 1.5 x 10 -10 1.0 x 10 -14 6.5 x 10 - 5 A – (aq) + H 2 O (l) OH - ( aq) + HA (aq)

31. VIII. Molecular Structure and Acid Strength A.Factors affecting acidity - bond strength : weaker bond, stronger acid - bond polarity: more polar bond, stronger acid B. Binary acid : bond strength is more important - hydrohalic acid HF HCl HBr HI Ex: H 2 O H 2 S H 2 Se<< <<<<

32. C. Oxoacids Z O HZ O-O- + H + -- ++ The O-H bond will be more polar and easier to break if: Z is very electronegative or Z is in a high oxidation state

33. C. Oxoacids Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number (O atom) - Acid strength increases with increasing EN of Z Oxoacids having the same central atom (Z) but different numbers of attached groups. - Acid strength increases as the oxidation no of Z increases H O Cl O O H O Br O O Cl is more electronegative than Br > HBrO HIO > HClO 4 HClO 3 HClO 2 HClO >>>

34. D. Carboxylic acids Acids with functional group -COOH Ex. Predict the strength of the following acids. CF 3 COOH CH 3 COOH H C C O H O H H withdraw electron can be ionized >

35. IX. Salt Solutions Salts are strong electrolytes, completely ionizes in water. The acidity/basicity of salt solutions depends on the ions. Hydrolysis: reaction of ions with H 2 O to produce OH – or H + NO 2 – + H 2 O HNO 2 + OH – NH 4 + + H 2 O NH 3 + H 3 O + A. Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be 2+ ) and the conjugate base of a strong acid (e.g. Cl –, Br –, I –, and NO 3 – ). NaBr (s) Na + (aq) + Br - (aq) H2OH2O KNO 3 (s) K + (aq) + NO 3 – (aq) H2OH2O

36. B. Basic Solutions: Salts derived from a strong base and a weak acid. CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) C. Acidic Solutions: a. Salts derived from a strong acid and a weak base. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) H2OH2O NH 4 + (aq) NH 3 (aq) + H + (aq) b.Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid. For example: AlCl 3 Al(H 2 O) 6 (aq) Al(OH)(H 2 O) 5 (aq) + H + (aq) 3+2+ Al(H 2 O) 6 (aq) +H 2 O (l) Al(OH)(H 2 O) 5 (aq) + H 3 O + (aq) 3+ 2+

37. Acid hydrolysis of Al 3+ Acidity of metal ion increases with smaller size and higher charge.

38. D. Solutions in which both the cation and the anion hydrolyze: K b for the anion > K a for the cation, solution will be basic K b for the anion < K a for the cation, solution will be acidic K b for the anion  K a for the cation, solution will be neutral Example: NH 4 NO 2 NO 2 – : K b = = = 2.2 x 10 -11 KwKw KaKa 1.0 x 10 -14 4.5 x 10 -4 NO 2 – and NH 4 + NH 4 + : K a = = = 5.6 x 10 -10 KwKw KbKb 1.0 x 10 -14 1.8 x 10 -5 K a > K b Weak acid K a,HNO2 =4.5 x10 -4 K b,NH3 =1.8 x10 -5

39. Ex: What is the pH of a 0.10 M solution of a NaF? F – (aq) +H 2 O (l) HF (aq) + OH - (aq) Initial (M) Change (M) Equilibrium (M) 0.100.00 -x-x+x+x 0.10 - x 0.00 +x+x xx Kb  Kb  x2x2 0.10 = 1.4 x 10 -11 0.10 – x  0.10 K b << 1 x 2 = 1.4 x 10 -12 [OH – ] = x = 1.2 x10 -6 M pH = 14.00 -pOH = 14.00 + log [OH – ] = 8.08 K b = [HF][OH - ] [F – ] Acid or basic? H 2 O (l) + F – (aq) HF (aq) + OH - (aq) K b = = = 1.4 x 10 -11 = KwKw KaKa 1.0 x 10 -14 7.1 x 10 -4 x2x2 0.10 - x

40. Basic oxide: group 1A and 2A oxides (except BeO). BaO + H 2 O -> Ba(OH) 2 Amphoteric oxide: BeO and group 3A and 4A oxides. Acid oxide: nonmetal oxide with high oxidation number, such as N 2 O 5, SO 2 etc. N 2 O 5 + H 2 O --> 2 HNO 3 Nonmetal oxide with low oxidation state such as CO or NO are neutral) X. Acidic, basic, and amphoteric oxides

41. XI. Lewis acid and base A.Arrehenius acid/base: substance that increases [H + ]/[OH – ] in water B.B.Brønsted acid/base : proton donor/acceptor C.A Lewis acid is a substance that can accept a pair of electrons (electron pair acceptor) D.A Lewis base is a substance that can donate a pair of electrons (electron pair donor) H+H+ H O H + OH - acidbase N H H H H+H+ + acidbase N H H H H + H H acidbase F B F F + F F N H H H