1. Acid-Base Concepts -- Chapter 15 1. Arrhenius Acid-Base Concept (last semester) Acid: H + supplier Base: OH – supplier 2. Brønsted-Lowry Acid-Base Concept (more general) (a) Definition (H + transfer) Acid: H + donor Base: H + Acceptor Conjugate Acid-Base Pairs: Base Acid + H + – H +

2. Conjugate Acid-Base Pairs e.g., NH 3(g) +H 2 O (l) NH 4 + (aq) +OH – (aq) baseacid base conjugate pair more examples: NH 2 – NH 3 NH 4 + OH – H2OH2OH3O+H3O+ O 2– OH – H2OH2O HSO 4 – H 2 SO 4 H 3 SO 4 + CH 3 – CH 4 “CH 5 + ” conjugate acids conjugate bases

3. Amphoteric Substances Molecules or ions that can function as both acids and bases (e.g. H 2 O itself!) e.g. the bicarbonate ion, HCO 3 – HCO 3 – + OH – --> H 2 O + CO 3 2– acid base HCO 3 – + HCl --> H 2 CO 3 + Cl – base acid

4. Autoionization of Water Water undergoes auto-ionization to a slight extent: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH – (aq) H 3 O + = hydronium ion OH – = hydroxide ion or, simply, H 2 O (l) H + (aq) + OH – (aq) equilibrium constant: K c = [H + ][OH – ]/[H 2 O] But, [H 2 O] ~ constant ~ 55.6 mol/L at 25 °C so, instead, use the “ion product” for water = K w K w = [H + ][OH – ] = 1.0 x 10 -14 (at 25 °C) In pure water: [H + ] = [OH – ] = 1.0 x 10 -7 M

5. The pH Scale K w = [H + ][OH – ] = 1.0 x 10 -14 The pH scale: pH = – log [H + ] In general: pX = – log X e.g. pOH = – log [OH – ] and, in reverse: [H + ] = 10 –pH mole/L [OH – ] = 10 –pOH mole/L Since K w = [H + ][OH – ] = 1.0 x 10 –14 pK w = pH + pOH = 14.00 Notice the sig figs! two sig figs two sig figs!

6. Relative Acidity of Solutions neutral solution [H + ] = [OH – ] = 1.0 x 10 –7 M pH = pOH = 7.00 acidic solution [H + ] > 10 –7 (i.e. more H + than in pure water) pH 7.00 e.g. if [H + ] = 1.00 x 10 –3 M then pH = 3.000 and pOH = 11.000 Basic solution [H + ] < 10 –7 (i.e. less H + than in pure water) pH > 7.00 [OH – ] > 10 –7 and pOH < 7.00 e.g. if [OH – ] = 1.00 x 10 –3 M then pOH = 3.000 and pH = 11.000

7. Example Problem The water in a soil sample was found to have [OH – ] equal to 1.47 x 10 –9 mole/L. Determine [H + ], pH, and pOH. Answer: [H + ] = K w /[OH – ] = (1.00 x 10 –14 )/(1.47 x 10 –9 ) = 6.80 x 10 –6 pH = – log [H + ] = – log (6.80 x 10 –6 ) = 5.167 (acidic!) pOH = 14.00 – pH = 14.00 – 5.167 = 8.833 {or, pOH = – log [OH – ] = – log (1.47 x 10 –9 ) = 8.833}

8. Strong Acids and Bases Strong Acids (e.g. HCl, HNO 3, etc) ~ 100% ionized HNO 3(aq) + H 2 O H 3 O + (aq) + NO 3 – (aq) or, in simplified form HNO 3(aq) H + (aq) + NO 3 – (aq) [H + ] = initial M of HNO 3 e.g. in a 0.050 M HNO 3 solution: [H + ] = 0.050 and pH = – log (0.050) = 1.30 Strong Bases (metal hydroxides) -- 100% ionized NaOH (aq) Na + (aq) + OH – (aq) [OH – ] = initial M of NaOH 100% Memorize! Table 15.3 p665 Table 15.7 p682

9. Example Problem What mass of Ba(OH) 2 (171.34 g/mole) is required to prepare 250 mL of a solution with a pH of 12.50? First: Write the reaction. Ba(OH) 2(aq) Ba 2+ (aq) + 2 OH – (aq) so, [OH – ] = 2 x M of Ba(OH) 2 solution (2:1 ratio) 100% Second: Determine [OH – ]. pOH = 14.00 – pH = 14.00 – 12.50 = 1.50 [OH – ] = 10 –1.50 = 0.032 M Third: How much Ba(OH) 2 is needed for that much OH – ? 250 mL soln x 0.032 mol OH – 1000 mL soln = 0.0079 mol OH – 0.0079 mol OH – x 1 mole Ba(OH) 2 2 mol OH – x 171.34 g Ba(OH) 2 1 mol Ba(OH) 2 = 0.68 g Ba(OH) 2

10. Sample Problem (a) Determine the pH of a 1.75 M solution of HNO 3. (b) Now suppose that 500.0 mL of a 1.50 M Ba(OH) 2 solution is added to 1.00 L of the above HNO 3 solution. Determine the pH of the resulting solution. (c) What additional volume (in mL) of 1.50 M Ba(OH) 2 must be added to the mixture in part (b) to bring the pH to 7.00?

11. Sample Problem (a)Determine the pH of a 1.75 M solution of HNO 3. a)pH = –0.243 (b) Now suppose that 500.0 mL of a 1.50 M Ba(OH) 2 solution is added to 1.00 L of the above HNO 3 solution. Determine the pH of the resulting solution. b) pH = 0.78 (c) What additional volume (in mL) of 1.50 M Ba(OH) 2 must be added to the mixture in part (b) to bring the pH to 7.00? c) 83 mL Ba(OH) 2 soln

12. Weak Acids and Bases (a) Weak Acids -- Less than 100% ionized (equilibrium !) In general: HA is a weak acid, A – is its conjugate base HA (aq) + H 2 O H 3 O + (aq) + A – (aq) or, simply, HA (aq) H + (aq) + A – (aq) Acid Dissociation Constant: K a K a = [H + ][A – ] [HA] Relative acid strength: Weak acid: K a < ~10 –3 Moderate acid: K a ~ 1 to 10 –3 Strong acid: K a > 1

13. Example Problem Hypochlorous acid, HOCl, has a pK a of 7.52. What is the pH of an 0.25 M solution of HOCl? What is the percent ionization? First: Write equation and ICE table. HOCl (aq) ⇌ H + (aq) + OCl – (aq) 0.25 M00 –x+x 0.25 – xxx I C E Second: Write equilibrium expression. pK a = – log K a, so K a = 10 –pKa = 10 –7.52 = 3.0 x 10 –8 3.0 x 10 –8 = [H + ][Cl – ] [HOCl] = (x)(x) (0.25 –x) = x2x2 Third: Solve for [H + ] and pH. Since K a is very small, assume x << 0.25 x 2 /(0.25) ≈ 3.0 x 10 –8 x ≈ 8.7 x 10 –5 (OK!) pH = –log (8.7 x 10 –5 ) = 4.06 (solution is acidic!)

14. Example Problem, cont. Fourth: Determine percent ionization. % ionization = (amount HA ionized)/(initial) x 100% = (8.7 x 10 –5 )/(0.25) x 100% = 0.035%

15. Weak Bases (b) Weak Bases In general: B is a weak base, HB + is its conjugate acid B (aq) + H 2 O (l) HB + (aq) + OH – (aq) K b = [HB + ][OH – ] [B] e.g. NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH – (aq) K b = [NH 4 + ][OH – ] [NH 3 ] = 1.8 x 10 –5 pK b = -log K b = 4.74 Note: Since OH – rather than H + appears here, first find [OH – ] or pOH, and then convert to pH Example problem: pH of 0.25 M solution of NH 3 ? Set up conc. table as usual, solve for x = [OH – ] [OH – ] = 2.1 x 10 –3 pOH = 2.68 pH = 11.32 (basic!) Base Dissociation Constant:

16. Sample Problem The nitrite ion (NO 2 – ) is a weak base with a pK b value of 10.85. (a) Write a balanced net ionic equation for the major equilibrium reaction that is occurring in an aqueous solution of sodium nitrite (NaNO 2 ). (b) Calculate the pH of a 0.25 M solution of sodium nitrite (NaNO 2 ). Clearly state and justify any assumptions that you make.

17. Sample Problem The nitrite ion (NO 2 – ) is a weak base with a pK b value of 10.85. (a) Write a balanced net ionic equation for the major equilibrium reaction that is occurring in an aqueous solution of sodium nitrite (NaNO 2 ). (b) Calculate the pH of a 0.25 M solution of sodium nitrite (NaNO 2 ). Clearly state and justify any assumptions that you make. (a) NO 2 – (aq) + H 2 O (l) HNO 2(aq) + OH – (aq) (b) 8.28

18. Salts of Weak Acids and Bases A) conjugate acid-base pairs (HA and A – ) –K a : HA H + + A – –K b : A- + H 2 O HA + OH – For any conjugate acid-base pair: K a K b = K w pK a + pK b = 14.00

19. Salt of a Weak Acid (e.g. NaCN) -- basic solution Anion acts as a weak base: K b : CN – (aq) + H 2 O (l) HCN (aq) + OH – (aq) K b = K w /K a = [OH – ][HCN] [CN – ] e.g. K a for HCN is 6.2 x 10 –10 What is pH of a 0.50 M NaCN solution? K b = K w /K a = (1.0 x 10 –14 )/(6.2 x 10 –10 ) = 1.6 x 10 –5 Use a concentration table based on K b reaction above: x = [OH – ] = [HCN] [CN – ] = 0.50 – x ≈ 0.50 (since K b is small) K b = [OH – ][HCN]/[CN – ] ≈ x 2 /0.50 ≈ 1.6 x 10 –5 x = [OH – ] ≈ 2.8 x 10 –3 pOH = 2.55 and pH = 11.45 (basic!)

20. Salt of a Weak Base (e.g. NH 4 Cl) -- Acidic Solution Cation acts as a weak acid: K a : NH 4 + H + + NH 3 K a = K w /K b = [H + ][NH 3 ] [NH 4 + ] e.g. K b for NH 3 is 1.8 x 10 –5 What is pH of a 0.50 M NH 4 Cl solution? K a = K w /K b = (1.0 x 10 –14 )/(1.8 x 10 –5 ) = 5.6 x 10 –10 Use a concentration table based on K a reaction above: x = [H + ] = [NH 3 ] [NH 4 + ] = 0.50 – x ≈ 0.50 (since K a is small) K a = [H + ][NH 3 ]/[NH 4 + ] ≈ x 2 /0.50 ≈ 5.6 x 10 –10 x = [H + ] ≈ 1.7 x 10 –5 pH = 4.77 (acidic!)

21. Sample Problem The pK a value for HCN is 9.21. What molar concentration of NaCN is required to make a solution with a pH of 11.75?

22. Sample Problem The pK a value for HCN is 9.21. What molar concentration of NaCN is required to make a solution with a pH of 11.75? Answer: 2.0 M NaCN

23. Polyprotic Acids e.g. diprotic acids, H 2 A, undergo stepwise dissociation: H 2 A HA – + H + HA – A 2– + H + K a1 = [HA – ][H + ] [H 2 A] K a2 = [A 2– ][H + ] [HA – ] Usually, K a1 >> K a2 so that: The 1st equilibrium produces most of the H + and [HA – ] but the 2nd equilibrium determines [A 2– ]

24. Example Problem Ascorbic acid (vitamin C), H 2 C 6 H 2 O 6, is an example of a diprotic acid with K a1 = 7.9 x 10 –5 and K a2 = 1.6 x 10 –12. For a 0.10 M solution of ascorbic acid, determine the pH and the concentrations of the mono anion, HC 6 H 2 O 6 –, and the dianion, C 6 H 2 O 6 2–. Based on the first equilibrium: x = [H + ] ≅ [HA – ] and [H 2 A] = 0.10 – x ≅ 0.10 K a1 = 7.9 x 10 –5 ≅ x 2 / (0.10) ∴ x ≅ 2.8 x 10 –3 so pH = 2.55 Must use the 2nd equilibrium to find [A 2– ]: K a2 = [A 2– ][H + ] / [HA – ] but, from above [H + ] ≅ [HA – ] ∴ K a2 ≅ [A 2– ] (a general result for H 2 A!) [A 2– ] ≅ 1.6 x 10 –12

25. How to tell if it’s acidic or basic? Anions Anion that is conjugate base of a weak acid is itself a weak base. Exception: H 2 O!! An anion that is a conjugate base of a strong acid is pH-neutral. Anion of a polyprotic acid is amphoteric, e.g. H 2 PO 4 –. Cations Cation that is conjugate acid of a weak base is itself a weak acid. A cation that is a conjugate acid of a strong base is pH-neutral. Small, highly charged metal cations form weakly acidic solutions (not Group I or II), e.g. Al 3+ (aq) + 6 H 2 O (l)  Al(H 2 O) 6 3+ (aq) Neutrals Weak bases: Table 15.8, p683 and Fig. 15.12, p687 Weak acids: Tables 15.4, p666, and 15.12, p687, and Fig. 15.12. Mixed Salts Salts where the cation is acidic and the anion basic form solutions where the pH depends on the relative strengths of the acid and base. weak acid

26. More Sample Problems 1. Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of each of the following. (a) HClO (b) (NH 4 ) 2 SO 4 (c) KCl (e) NaCHO 2 2. Write the appropriate equilibrium constant expressions (K a, etc) for the above reactions. 3. Determine the pH of a 0.100 M solution of each of the above compounds. (Use equilibrium constants from the textbook as needed).

27. Answers 1. (a) HClO (aq) H + (aq) + ClO – (aq) (b) NH 4 + (aq) NH 3(aq) + H + (aq) (c) H 2 O (l) H + (aq) + OH – (aq) (d) CHO 2 – (aq) + H 2 O (aq) HCHO 2(aq) + OH – (aq) 2. (a) K a = [H + ][ClO – ]/[HClO] (b) K a = [NH 3 ][H + ]/[NH 4 + ] (c) K w = [H + ][OH – ] (d) K b = [HCHO 2 ][OH – ]/[CHO 2 – ] 3.(a) 4.26 (b) 4.98 (c) 7.00 (d) 8.38

28. Relative Strengths of Brønsted Acids Binary Acids e.g. HCl, HBr, H 2 S, etc. e.g., relative acidity: HCl > H 2 S (across a period) HI > HBr > HCl > HF (up in a group) Periodic Table Acid Strength Increases

29. Oxo Acids e.g. HNO 3, H 2 SO 4, H 3 PO 4, etc. 1. for same central element, acid strength increases with # of oxygens acid strength increases HClO < HClO 2 < HClO 3 < HClO 4 2. for different central element, but same # of oxygens acid strength increases with electronegativity e.g. H 2 SO 4 > H 2 SeO 4 > H 2 TeO 4 Acid Strength Increases Periodic Table

30. Relative Strengths of Conjugate Acid-Base Pairs For example, HF + H 2 O H 3 O + + F – acid base acid base In this case, the equilibrium lies mainly on the reactant side. Therefore, “HF is a weaker acid than H 3 O + ” In general, weaker Brønsted acids have stronger conjugate bases. (and vice versa)

31. Lewis Acid-Base Concept (most general) Definition (electron pair transfer) Acid: e – pair acceptor Base: e – pair donor Lewis acids -- electron deficient molecules or cations Lewis bases -- electron rich molecules or anions. (have one or more unshared e – pairs)

32. Lewis Acid-Base Reactions (i.e. all non-redox reactions!) OH – (aq) + NH 4 + (aq) --> H 2 O (l) + NH 3(aq) OH – (aq) + CO 2(g) --> HCO 3 – (aq) Review Ch. 9, slides 10-15 from CHEM 10113!

33. Yet More Sample Problems 1)Among the following, which is the strongest acid? HBrO 3 HBrO 4 H 2 SeO 3 H 2 SeO 4 H 2 TeO 3 H 2 TeO 4 2)Among the following, which is the weakest acid? H 2 Se NH 3 PH 3 H 2 S H 2 O AsH 3 3)The conjugate base of CH 3 OH is ______________. 4) The conjugate acid of HF is _________. 5) Write the equation for the reaction of H 2 Se with AsH 3, then answer the questions below. a)Label the Bronsted acid, base, conjugate acid, and conjugate base. b)If K c = 0.078, label which is the strongest acid and strongest base in your equation.

34. Yet More Sample Problems 1)Among the following, which is the strongest acid? HBrO 3 HBrO 4 H 2 SeO 3 H 2 SeO 4 H 2 TeO 3 H 2 TeO 4 HBrO 4 2)Among the following, which is the weakest acid? H 2 Se NH 3 PH 3 H 2 S H 2 O AsH 3 NH 3 3)The conjugate base of CH 3 OH is ______________. CH 3 O – 4) The conjugate acid of HF is _________. H 2 F + 5) Write the equation for the reaction of H 2 Se with AsH 3, then answer the questions below. a)Label the Bronsted acid, base, conjugate acid, and conjugate base. b)If K c = 0.078, label which is the strongest acid and strongest base in your equation. H 2 Se (aq) + AsH 3(aq) AsH 4 + (aq) + HSe – (aq) 1)acid base conj. acid conj. base 2)strongest acid is AsH 4 +, strongest base is HSe –

35. And Finally… Consider the reaction of hydroxide ion with the nitrosyl cation, NO 2 +, to form nitric acid. OH – (aq) + NO 2 + (aq) --> HNO 3(aq) Write Lewis electron dot formulas (including formal charges and/or resonance forms if needed) for all three species in this reaction. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use arrow(s) to illustrate the formation of any new chemical bond(s) during the reaction.

36. And Finally… Consider the reaction of hydroxide ion with the nitrosyl cation, NO 2 +, to form nitric acid. OH – (aq) + NO 2 + (aq) --> HNO 3(aq) Write Lewis electron dot formulas (including formal charges and/or resonance forms if needed) for all three species in this reaction. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use arrow(s) to illustrate the formation of any new chemical bond(s) during the reaction.