1. 1 The Chemistry of Acids and Bases Chapter 17

2. 2 Acid and Bases

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5. 5 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O(liq) ---> H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.

6. 6 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Strong and Weak Acids/Bases

7. 7 Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Strong and Weak Acids/Bases

8. 8 Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH(aq) ---> Na + (aq) + OH - (aq) NaOH(aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

9. 9 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O(liq) e NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

10. 10 ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theoryThe most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H + IONSACIDS DONATE H + IONS BASES ACCEPT H + IONSBASES ACCEPT H + IONS

11. 11 The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID ACID-BASE THEORIES

12. 12 NH 3 is a BASE in water — and water is itself an ACID NH 3 / NH 4 + is a conjugate pair — related by the gain or loss of H + Every acid has a conjugate base - and vice-versa.

13. 13 Conjugate Pairs

14. 14 More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for autoion = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C

15. 15 More About Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization

16. 16 Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e HO + (aq) + OH - (aq) 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) Le Chatelier predicts equilibrium shifts to the ____________. [HO + ] < 10 -7 at equilibrium. [H 3 O + ] < 10 -7 at equilibrium. Set up a ICE table.

17. 17 Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) initial00.0010 initial00.0010 change+x+x change+x+x equilibx0.0010 + x equilibx0.0010 + x K w = (x) (0.0010 + x) Because x << 0.0010 M, assume [OH - ] = 0.0010 M [H 3 O + ] = K w / 0.0010 = 1.0 x 10 -11 M

18. 18 Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e HO + (aq) + OH - (aq) 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) [HO + ] = K w / 0.0010 = 1.0 x 10 -11 M [H 3 O + ] = K w / 0.0010 = 1.0 x 10 -11 M This solution is _________ because [H 3 O + ] < [OH - ]

19. 19 [H 3 O + ], [OH - ] and pH A common way to express acidity and basicity is with pH pH = log (1/ [H 3 O + ]) = - log [H 3 O + ] In a neutral solution, [HO + ] = [OH - ] = 1.00 x 10 -7 at 25 o C In a neutral solution, [H 3 O + ] = [OH - ] = 1.00 x 10 -7 at 25 o C pH = -log (1.00 x 10 -7 ) = - (-7) = 7

20. 20 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00 General conclusion — Basic solution pH > 7 Basic solution pH > 7 Neutral pH = 7 Neutral pH = 7 Acidic solutionpH < 7 Acidic solutionpH < 7

21. 21 Figure 17.1

22. 22 [H 3 O + ], [OH - ] and pH If the pH of Coke is 3.12, it is ____________. Because pH = - log [HO + ] then Because pH = - log [H 3 O + ] then log [HO + ] = - pH log [H 3 O + ] = - pH Take antilog and get [HO + ] = 10 -pH [H 3 O + ] = 10 -pH [HO + ] = 10 -3.12 = 7.6 x 10 -4 M [H 3 O + ] = 10 -3.12 = 7.6 x 10 -4 M

23. 23 pH of Common Substances

24. 24 Other pX Scales In generalpX = -log X and so pOH = - log [OH - ] K w = [HO + ] [OH - ] = 1.00 x 10 -14 at 25 o C K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C Take the log of both sides -log (10 -14 ) = - log [HO + ] + (-log [OH - ]) -log (10 -14 ) = - log [H 3 O + ] + (-log [OH - ]) pK w = 14 = pH + pOH pK w = 14 = pH + pOH

25. 25 Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K a = 3.2 x 10 -4

26. 26 Equilibria Involving Weak Acids and Bases AcidConjugate Base AcidConjugate Base acetic, CH 3 CO 2 HCH 3 CO 2 -, acetate ammonium, NH 4 + NH 3, ammonia bicarbonate, HCO 3 - CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

27. 27 Equilibria Involving Weak Acids and Bases Consider acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O e H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

28. 28 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

29. 29 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

30. 30 Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

31. 31 Relation of K a, K b, [H 3 O + ] and pH

32. 32 K and Acid-Base Reactions Reactions always go from the stronger A-B pair (larger K) to the weaker A-B pair (smaller K). ACIDSCONJUGATE BASES STRONGSTRONG weakweak weakweak STRONGSTRONG

33. 33 K and Acid-Base Reactions A strong acid is 100% dissociated. Therefore, a STRONG ACID —a good H + donor— must have a WEAK CONJUGATE BASE —a poor H + acceptor. HNO 3 (aq) + H 2 O(liq) e H 3 O + (aq) + NO 3 - (aq) HNO 3 (aq) + H 2 O(liq) e H 3 O + (aq) + NO 3 - (aq) STRONG A base acid weak B STRONG A base acid weak B Every A-B reaction has two acids and two bases.Every A-B reaction has two acids and two bases. Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair. Here K is very large.Here K is very large. Every A-B reaction has two acids and two bases.Every A-B reaction has two acids and two bases. Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair. Here K is very large.Here K is very large.

34. 34 K and Acid-Base Reactions We know from experiment that HNO 3 is a strong acid. 1.It is a stronger acid than H 3 O + 2.H 2 O is a stronger base than NO 3 - 3.K for this reaction is large

35. 35 K and Acid-Base Reactions Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H 2 O e H 3 O + + OAc - WEAK A base acid STRONG B Because [H 3 O + ] is small, this must mean 1.H 3 O + is a stronger acid than HOAc 2.OAc - is a stronger base than H 2 O 3.K for this reaction is small

36. 36 Types of Acid/Base Reactions Strong acid + Strong base H + + Cl - + Na + + OH - e H 2 O + Na + + Cl - Net ionic equation H + (aq) + OH - (aq) e H 2 O(liq) K = 1/K w = 1 x 10 14 Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.

37. 37 Types of Acid/Base Reactions Weak acid + Strong base CH 3 CO 2 H + OH - e H 2 O + CH 3 CO 2 - This is the reverse of the reaction of CH 3 CO 2 - (conjugate base) with H 2 O. OH - stronger base than CH 3 CO 2 - K = 1/K b = 5.6 x 10 4 Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate base. The solution is basic.

38. 38 Types of Acid/Base Reactions Strong acid + Weak base H 3 O + + NH 3 e H 2 O + NH 4 + This is the reverse of the reaction of NH 4 + (conjugate acid of NH 3 ) with H 2 O. H 3 O + stronger acid than NH 4 + K = 1/K a = 5.6 x 10 4 Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate acid. The solution is acid.

39. 39 Types of Acid/Base Reactions Weak acid + Weak base Product cation = conjugate acid of weak base.Product cation = conjugate acid of weak base. Product anion = conjugate base of weak acid.Product anion = conjugate base of weak acid. pH of solution depends on relative strengths of cation and anion.pH of solution depends on relative strengths of cation and anion.

40. 40 Types of Acid/Base Reactions: Summary

41. 41 Calculations with Equilibrium Constants pH of an acetic acid solution. What are your observations? pH of an acetic acid solution. What are your observations? 0.0001 M 0.003 M 0.06 M 2.0 M a pH meter, Screen 17.9

42. 42 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib

43. 43 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] initial1.0000 change-x+x+x equilib1.00-xxx Note that we neglect [H 3 O + ] from H 2 O.

44. 44 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A).

45. 45 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

46. 46 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37

47. 47 Equilibria Involving A Weak Acid Consider the approximate expression For many weak acids [H 3 O + ] = [conj. base] = [K a C o ] 1/2 [H 3 O + ] = [conj. base] = [K a C o ] 1/2 where C 0 = initial conc. of acid Useful Rule of Thumb: If 100K a < C o, then [H 3 O + ] = [K a C o ] 1/2 If 100K a < C o, then [H 3 O + ] = [K a C o ] 1/2

48. 48 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O e HCO 2 - + H 3 O + HCO 2 H + H 2 O e HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = [K a C o ] 1/2 = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = [K a C o ] 1/2 = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47

49. 49 Weak Bases

50. 50 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O e NH 4 + + OH - NH 3 + H 2 O e NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib

51. 51 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O e NH 4 + + OH - NH 3 + H 2 O e NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ] initial0.01000 change-x+x+x equilib0.010 - xx x

52. 52 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O e NH 4 + + OH - NH 3 + H 2 O e NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small (100K b < C o ), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

53. 53 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O e NH 4 + + OH - NH 3 + H 2 O e NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

54. 54 MX + H 2 O ----> acidic or basic solution? Consider NH 4 Cl NH 4 Cl(aq) ----> NH 4 + (aq) + Cl - (aq) (a)Reaction of Cl - with H 2 O Cl - + H 2 O ---->HCl + OH - Cl - + H 2 O ---->HCl + OH - baseacidacidbase baseacidacidbase Cl - ion is a VERY weak base because its conjugate acid is strong. Therefore, Cl - ----> neutral solution Acid-Base Properties of Salts

55. 55 NH 4 Cl(aq) ----> NH 4 + (aq) + Cl - (aq) (b)Reaction of NH 4 + with H 2 O NH 4 + + H 2 O ---->NH 3 + H 3 O + NH 4 + + H 2 O ---->NH 3 + H 3 O + acidbasebaseacid acidbasebaseacid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH 4 + ----> acidic solution See TABLE 17.4 for a summary of acid-base properties of ions. Acid-Base Properties of Salts

56. 56

57. 57 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O e HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 1.Set up concentration table [CO 3 2- ][HCO 3 - ][OH - ] initial [CO 3 2- ][HCO 3 - ][OH - ] initial change equilib equilib Acid-Base Properties of Salts

58. 58 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O e HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 1.Set up ICE table [CO 3 2- ][HCO 3 - ][OH - ] initial0.10 00 [CO 3 2- ][HCO 3 - ][OH - ] initial0.10 00 change-x+x+x equilib0.10 - xxx equilib0.10 - xxx Acid-Base Properties of Salts

59. 59 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O e HCO 3 - +OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Properties of Salts Assume 0.10 - x ≈ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = 0.0046 M Step 2.Solve the equilibrium expression

60. 60 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- + H 2 O e HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Properties of Salts Step 3.Calculate the pH [OH - ] = 0.0046 M pOH = - log [OH - ] = 2.34 pH + pOH = 14, so pH = 11.66, and the solution is ________.