1. The Chemistry of Acids and Bases Chapter 17 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2. 2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

3. 3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

4. 4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Strong and Weak Acids/Bases

5. 5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 = HOAc HOAc(aq) + H 2 O(liq) OAc - (aq) + H 3 O + (aq) OAc - = CH 3 CO 2 - = acetate ion Strong and Weak Acids/Bases

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

8. 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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) NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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) NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

10. 10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID ACID-BASE THEORIES

12. 12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES NH 3 is a BASE in water — and water is itself an ACID

13. 13 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES 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 +

14. 14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES 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.

15. 15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES 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) H 3 O + (aq) + NO 3 - (aq) HNO 3 (aq) + H 2 O(liq) H 3 O + (aq) + NO 3 - (aq) STRONG A base acid weak B STRONG A base acid weak B Notice that every A-B reaction has two acids and two bases!

16. 16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES 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 -

17. 17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H 2 O 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

18. 18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Now we can describe reactions of acids with bases and the direction of such reaction. Consider the acid HF reacting with the base NH 3. HF + NH 3 --> NH 4 + + F - Now we can describe reactions of acids with bases and the direction of such reaction. Consider the acid HF reacting with the base NH 3. HF + NH 3 --> NH 4 + + F -

19. 19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Acid-Base Reactions Now we can describe reactions of acids with bases and the direction of such reaction. Consider the acid HF reacting with the base NH 3. HF + NH 3 --> NH 4 + + F - Now we can describe reactions of acids with bases and the direction of such reaction. Consider the acid HF reacting with the base NH 3. HF + NH 3 --> NH 4 + + F -

20. 20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Predicting the Direction of Acid-Base Reactions Based on experiment, we can put acids and bases on a chart. See Table 17.3 (page 794) ACIDSCONJUGATE BASES ACIDSCONJUGATE BASES STRONGweak STRONGweak weakSTRONG weakSTRONG This chart can be used to predict the direction of reactions between any A-B pair. Reactions always go from the stronger A-B pair to the weaker A-B pair.

21. 21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ACID-BASE THEORIES Predicting the direction of an acid-base reaction. Use Table 17.3 Reactions always go from the stronger A-B pair to the weaker A-B pair.

22. 22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved MORE ABOUT WATER H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION

23. 23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

24. 24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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) HO + (aq) + OH - (aq) 2 H 2 O(liq) 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 concentration table.

25. 25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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) HO + (aq) + OH - (aq) 2 H 2 O(liq) 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 [HO + ] = K w / 0.0010 = 1.0 x 10 -11 M [H 3 O + ] = K w / 0.0010 = 1.0 x 10 -11 M

26. 26 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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) HO + (aq) + OH - (aq) 2 H 2 O(liq) 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 BASIC because [HO + ] < [OH - ] This solution is BASIC because [H 3 O + ] < [OH - ]

27. 27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [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

28. 28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [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

29. 29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved [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

30. 30 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 - ]) 14 = pH + pOH

31. 31 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K a = 3.2 x 10 -4

32. 32 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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%).

33. 33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving Weak Acids and Bases Consider acetic acid HOAc + H 2 O H 3 O + + OAc - Acid Conj. base

34. 34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving Weak Acids and Bases Consider acetic acid HOAc + H 2 O 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.

35. 35 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving Weak Acids and Bases Values of K a for acid and K b for bases are found in TABLE 17.4 — page 799 Notice the relation of TABLE 17.4 to the table of relative acid/base strengths (Table 17.3).

36. 36 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Determining the pH of an acetic acid solution. See Screen 17.8. Determining the pH of an acetic acid solution. See Screen 17.8. 0.0001 M 0.003 M 0.06 M 2.0 M a pH meter

37. 37 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib

38. 38 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. [HOAc][H 3 O + ][OAc - ] [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.

39. 39 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

40. 40 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

41. 41 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 2. Write K a expression This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A). You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

42. 42 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

43. 43 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a expression First assume x is very small because K a is so small. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

44. 44 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a expression First assume x is very small because K a is so small. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

45. 45 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a expression First assume x is very small because K a is so small. And so x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

46. 46 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

47. 47 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

48. 48 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression 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 You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

49. 49 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Acid Consider the approximate expression

50. 50 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

51. 51 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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:

52. 52 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 100 K a < C o, then [H 3 O + ] = [K a C o ] 1/2 If 100 K a < C o, then [H 3 O + ] = [K a C o ] 1/2

53. 53 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 2 - + H 3 O + HCO 2 H + H 2 O 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

54. 54 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib

55. 55 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ] initial0.01000 change-x+x+x equilib0.010 - xx x

56. 56 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression

57. 57 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O 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

58. 58 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O 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 and [NH 3 ] = 0.010 - 4.2 x 10 -4 ­ 0.010 M

59. 59 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O 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 and [NH 3 ] = 0.010 - 4.2 x 10 -4 ­ 0.010 M The approximation is valid!

60. 60 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O 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

61. 61 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

62. 62 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 17.5 for a summary of acid-base properties of ions. Acid-Base Properties of Salts

63. 63 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase 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 - ] [CO 3 2- ][HCO 3 - ][OH - ] initial initial change equilib equilib Acid-Base Properties of Salts

64. 64 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase 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 - ] [CO 3 2- ][HCO 3 - ][OH - ] initial0.1000 initial0.1000 change-x+x+x equilib0.10 - xxx equilib0.10 - xxx Acid-Base Properties of Salts

65. 65 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 2.Solve the equilibrium expression Acid-Base Properties of Salts

66. 66 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 2.Solve the equilibrium expression Acid-Base Properties of Salts

67. 67 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 2.Solve the equilibrium expression Assume 0.10 - x ­ 0.10, because 100K b < C o Acid-Base Properties of Salts

68. 68 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 2.Solve the equilibrium expression Assume 0.10 - x ­ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = 0.0046 M Acid-Base Properties of Salts

69. 69 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 3.Calculate the pH [OH - ] = 0.0046 M Acid-Base Properties of Salts

70. 70 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 3.Calculate the pH [OH - ] = 0.0046 M pOH = - log [OH - ] = 2.34 Acid-Base Properties of Salts

71. 71 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 3.Calculate the pH [OH - ] = 0.0046 M pOH = - log [OH - ] = 2.34 pH + pOH = 14, Acid-Base Properties of Salts

72. 72 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 HCO 3 - +OH - baseacidacidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 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 ________. Acid-Base Properties of Salts