1. The Chemistry of Acids and Bases Chapter 17

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

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

4. 4 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 = HC 2 H 3 O 2 = CH 3 CO 2 H = HOAc = CH 3 COOHOne of the best known is acetic acid = HC 2 H 3 O 2 = CH 3 CO 2 H = HOAc = CH 3 COOH HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + or or HC 2 H 3 O 2 C 2 H 3 O 2 - + H + HC 2 H 3 O 2 C 2 H 3 O 2 - + H + Strong and Weak Acids/Bases

5. 5 Strong Bases 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 CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime)

6. 6 Weak Bases Less than 100% ionized in water One of the best known weak bases is ammonia. ammonia Strong and Weak Acids/Bases NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)

7. 7 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

8. 8 ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID BaseAcid Base NH 4 + + OH - NH 3 + H 2 O

9. 9 ACID-BASE THEORIES NH 3 / NH 4 + is a conjugate pair — related by the gain or loss of H + 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. Every acid has a conjugate base - and vice-versa. NH 3 is a BASE in water — and water is itself an ACID. NH 3 is a BASE in water — and water is itself an ACID.

10. 10 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. Therefore, a STRONG ACID — a good H + donor — must have a WEAK CONJUGATE BASE — a poor H + acceptor. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HNO 3 (aq) + H 2 O (l) 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! Notice that every A-B reaction has two acids and two bases!

11. 11 ACID-BASE THEORIES We know that HNO 3 is a strong acid. We know 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 - WEAK BASE STRONG ACID

12. 12 ACID-BASE THEORIES Acetic acid is only 0.42% ionized when [HC 2 H 3 O 2 ] = 1.0 M. It is a WEAK ACID Acetic acid is only 0.42% ionized when [HC 2 H 3 O 2 ] = 1.0 M. It is a WEAK ACID Because [H 3 O + ] is small, this must mean Because [H 3 O + ] is small, this must mean 1. H 3 O + is a stronger acid than HC 2 H 3 O 2 1. H 3 O + is a stronger acid than HC 2 H 3 O 2 2. C 2 H 3 O 2 - is a stronger base than H 2 O 2. C 2 H 3 O 2 - is a stronger base than H 2 O

13. 13 ACIDS AND BASES Monoprotic acids can donate only one proton, while polyprotic acids can donate two or more protons.Monoprotic acids can donate only one proton, while polyprotic acids can donate two or more protons. Monoprotic bases can accept only one proton, while polyprotic bases can accept two or more protons.Monoprotic bases can accept only one proton, while polyprotic bases can accept two or more protons. Amphiprotic substances can behave as either acids or bases.Amphiprotic substances can behave as either acids or bases.

14. 14 Acid-Base Reactions Now we can describe reactions of acids with bases and the direction of such reaction.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.Consider the acid HF reacting with the base NH 3. HF + NH 3 NH 4 + + F -HF + NH 3 NH 4 + + F - Now we can describe reactions of acids with bases and the direction of such reaction.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.Consider the acid HF reacting with the base NH 3. HF + NH 3 NH 4 + + F -HF + NH 3 NH 4 + + F -

15. 15 Predicting the Direction of Acid-Base Reactions Based on experiment, we can put acids and bases on a chart. Based on experiment, we can put acids and bases on a chart. See Table 17.3 or Appendix H tables 16 ACIDSCONJUGATE BASES STRONG weak weak STRONG ACIDSCONJUGATE BASES STRONG weak weak STRONG This chart can be used to predict the direction of reactions between any A-B pair. 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. Reactions always go from the stronger A-B pair to the weaker A-B pair.

16. 16 ACID-BASE THEORIES Predicting the direction of an acid-base reaction. ACID 1BASE 2ACID 2BASE 1 ++ STRONGweak Reactions always go from the Reactions always go from the stronger A-B pair weaker A-B pair.

17. 17 MORE ABOUT WATER H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION

18. 18 K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M MORE ABOUT WATER Autoionization 2H 2 O H 3 O + OH -

19. 19 Calculating [H + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution HOH H + + OH - HOH H + + OH - Le Chatelier predicts equilibrium shifts to the ____________. [H + ] < 10 -7 at equilibrium. Set up a concentration table. left

20. 20 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 H 2 O (l) H + + OH - H 2 O (l) H + + OH - Calculating [H + ] & [OH - ] initial 0 0.0010 change +x +x change +x +x equilib x 0.0010 + x equilib x 0.0010 + x K w = [H + ][OH - ] = (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 This solution is BASIC because [H 3 O + ] < [OH - ]

21. 21 [H + ], [OH - ] and pH A common way to express acidity and basicity is with pH. pH = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10 -7 at 25 o C pH = -log (1.00 x 10 -7 ) = - (-7.000) = 7.000 pH = -log (1.00 x 10 -7 ) = - (-7.000) = 7.000

22. 22 [H + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? What is the pH of the 0.0010 M NaOH solution? [H + ] = 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

23. 23 [H + ], [OH - ] and pH If the pH of Coke is 3.12, it is ____________. Because pH = - log [H + ] then log [H + ] = - pH log [H + ] = - pH Take antilog and get [H + ] = 10 -pH [H + ] = 10 -3.12 = 7.6 x 10 -4 M Acidic

24. 24 Other pX Scales In generalpX = - log X and so pOH = - log [OH - ] K w = [H 3 O + ] [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 [H 3 O + ] + (- log [OH - ]) - log (10 -14 ) = - log [H 3 O + ] + (- log [OH - ]) pK w = 14.00 = 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 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, HC 2 H 3 O 2 C 2 H 3 O 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%). A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

27. 27 For a weak acid, HA, and its conjugate base, A -, K a. K b = K wFor a weak acid, HA, and its conjugate base, A -, K a. K b = K w Thus, one can calculate K a from K b and K b from K a.Thus, one can calculate K a from K b and K b from K a. Notice as K a increases, K b decreases and visa versa.Notice as K a increases, K b decreases and visa versa. The stronger the acid the weaker the conjugate base.The stronger the acid the weaker the conjugate base. Equilibria Involving Weak Acids and Bases

28. 28 Consider acetic acid HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 - Acid Conj. base Equilibria Involving Weak Acids and Bases (K is designated K a for an ACID) Because [H 3 O + ] and [C 2 H 3 O 2 - ] are SMALL, K a << 1. K a << 1. K a  [H 3 O + C 2 H 3 O 2 - ][C 2 H 3 O 2 - ] C 2 H 3 O 2 [HC 2 H 3 O 2 ]  1.8 x 10 -5

29. 29 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).

30. 30 Increasing Acid Strength Increasing Base Strength Partial table 17.3

31. 31 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concentrations. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ] initial1.00 0 0 change -x +x +x equilib1.00-x x x Note that we neglect [H 3 O + ] from H 2 O.

32. 32 Equilibria Involving A Weak Acid This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A.4). This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A.4). You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Step 2. Write K a expression K a  1.8 x 10 -5 = [H 3 O + ][OAc - ] [HOAc]  x 2 1.00- x

33. 33 And so x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 And so x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 K a  1.8 x 10 -5 = x 2 1.00 First assume x is very small because K a is so small. First assume x is very small because K a is so small. Step 3. Solve K a expression You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Equilibria Involving A Weak Acid K a  1.8 x 10 -5 = [H 3 O + ][OAc - ] [HOAc]  x 2 1.00- x

34. 34 K a  1.8 x 10 -5 = x 2 1.00 Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Equilibria Involving A Weak Acid 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

35. 35 Equilibria Involving A Weak Acid For many weak acids 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 concentration of acid where C 0 = initial concentration of acid Useful Rule of Thumb: Useful Rule of Thumb: If 1000 K a < C o, then [H 3 O + ] = [K a C o ] 1/2 If 1000 K a < C o, then [H 3 O + ] = [K a C o ] 1/2 Consider the approximate expression x  [H 3 O + ]= [K a 1.00] 1/2 K a  1.8 x 10 -5 = x 2 1.00

36. 36 Approximation Rules 10 times K for.1M, so 10  K < [X] 100 times K for.10M, so 100  K < [X] 1000 times K for.100M, so 1000  K < [X] Look at the number of decimal places to determine factor to multiply K by

37. 37 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. 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 NOT Valid pH = 3.47 NOT Valid

38. 38 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calculate the pH. You have 0.010 M NH 3. Calculate 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 K b = 1.8 x 10 -5 Step 1. Define equilibrium concentrations. [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ] initial0.010 0 0 change -x +x +x equilib0.010 - x x x

39. 39 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! K b  1.8 x 10 -5 = [NH 4 + ][OH - ] [NH 3 ] = x 2 0.010- x You have 0.010 M NH 3. Calculate 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 K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Equilibria Involving A Weak Base

40. 40 You have 0.010 M NH 3. Calculate 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 K b = 1.8 x 10 -5 Equilibria Involving A Weak Base Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

41. 41 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K a and the % ionization. 1.4x10 -5 1.2% 2. A solution of propionic acid, CH 3 CH 2 COOH, is 0.20 M. K a = 1.3x10 -5. Calculate the pH and % ionization. 2.79 0.80% 3. A solution of hydrazine, N 2 H 4, is 0.025 M. K b = 8.5x10 -5. Calculate the pH and % ionization. 11.15 5.6% (4.)  Equilibria Involving Weak Acids and Bases

42. 42 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13 2.21, 0.14, 6.1x10 -3, 5.8x10 -8, 2.8x10 -18

43. 43 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 - baseacidacidbase Cl - + H 2 O HCl + OH - baseacidacidbase Cl - ion is a VERY weak base because its conjugate acid is strong. Cl - ion is a VERY weak base because its conjugate acid is strong. Acid-Base Properties of Salts Therefore, Cl - ----> neutral solution X X

44. 44 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 + acidbase base acid NH 4 + + H 2 O NH 3 + H 3 O + acidbase base acid NH 4 + ion is a moderate acid because its conjugate base is weak. 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

45. 45 Determine if the following solutions are acidic, basic, or neutral. KBr,CrCl 3, NaNO 2, KHCO 3, Na 2 CO 3, NH 4 C 2 H 3 O 2

46. 46 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 0.10 0 0 initial 0.10 0 0 change -x +x +x change -x +x +x equilib 0.10 - x x x equilib 0.10 - x x x Acid-Base Properties of Salts

47. 47 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 Assume 0.10 - x ­ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = 0.0046 M K b = 2.1 x 10 -4 = [HCO 3 - ][OH - ] [CO 3 2  ]  x 2 0.10- x

48. 48 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 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 ________. Basic

49. 49 Review Problems Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13 2.21, 0.14, 6.1x10 -3, 5.8x10 -8, 2.8x10 -18 Calculate the pH of a 0.050 M K 2 CO 3 solution.Calculate the pH of a 0.050 M K 2 CO 3 solution.11.50

50. 50 Lewis base = electron pair donor (NH 3 )Lewis base = electron pair donor (NH 3 ) Lewis Acids & Bases Lewis acid = electron pair acceptor (BF 3 )Lewis acid = electron pair acceptor (BF 3 )

51. 51 A Lewis acid and base can interact by sharing an electron pair. Lewis Acids & Bases

52. 52 H H H BASEACID O—HO—H O—H H + + A Lewis acid and base can interact by sharing an electron pair. Formation of hydronium ion is an excellent example. Lewis Acids & Bases

53. 53 Other good examples involve metal ions. Lewis Acids & Bases Colorless M x (NO 3 ) y form colorful hydrated [M x (H 2 O) y ] z+ compounds

54. 54 Such bonds as the H 2 O ---> Co bond are often called COORDINATE COVALENT BONDS because both electrons are supplied by one of the atoms of the bond. Such bonds as the H 2 O ---> Co bond are often called COORDINATE COVALENT BONDS because both electrons are supplied by one of the atoms of the bond. Lewis Acids & Bases Other good examples involve metal ions.

55. 55 The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 ------> COMPLEX IONS The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 ------> COMPLEX IONS All metal ions form complex ions with water — and are of the type [M(H 2 O) x ] n+ where x = 4 and 6. All metal ions form complex ions with water — and are of the type [M(H 2 O) x ] n+ where x = 4 and 6. Lewis Acids & Bases [Cu(NH 3 ) 4 ] 2+

56. 56 Add NH 3 to light blue [Cu(H 2 O) 4 ] 2+ ------> light blue Cu(OH) 2 and then deep blue [Cu(NH 3 ) 4 ] 2+ Add NH 3 to light blue [Cu(H 2 O) 4 ] 2+ ------> light blue Cu(OH) 2 and then deep blue [Cu(NH 3 ) 4 ] 2+ Lewis Acids & Bases

57. 57 [Ni(H 2 O) 6 ] 2+ + 6 NH 3 ---> [Ni(NH 3 ) 6 ] 2+ [Ni(H 2 O) 6 ] 2+ + 6 NH 3 ---> [Ni(NH 3 ) 6 ] 2+ + DMG Lewis Acids & Bases

58. 58 The Fe 2+ in heme can interact with O 2 or CO in a Lewis acid-base reaction. Lewis Acids & Bases

59. 59 Many complex ions containing water undergo HYDROLYSIS to give acidic solutions. [Cu(H 2 O) 4 ] 2+ + H 2 O ---> [Cu(H 2 O) 3 (OH)] + + H 3 O + Lewis Acids & Bases

60. 60 Many complex ions containing water undergo HYDROLYSIS to give acidic solutions. This explains why water solutions of Fe 3+, Al 3+, Cu 2+, Pb 2+, etc. are acidic. Lewis Acids & Bases

61. 61 This explains AMPHOTERIC nature of some metal hydroxides. Al(OH) 3 (s) + 3 H + --> Al 3+ + 3 H 2 O Al(OH) 3 (s) + 3 H + --> Al 3+ + 3 H 2 O Here Al(OH) 3 is a Brønsted base. Al(OH) 3 (s) + OH - --> Al(OH) 4 - Al(OH) 3 (s) + OH - --> Al(OH) 4 - Here Al(OH) 3 is a Lewis acid. Al 3+ O—H - Lewis Acids & Bases

62. 62 Amphoterism of Al(OH) 3 Al(OH) 3 on right Add NaOH Add HCl See Kotz / Treichel 5 th, page 722

63. 63 1. The electronegativity of the O atoms causes the H attached to O to be highly positive. 2. The O—H bond is highly polar. 3. The H atom of O—H is readily attracted to polar H 2 O.

64. 64 Neutral Lewis Acid Carbon dioxide is a neutral (molecular) Lewis acid. +1.5-0.75-0.75

65. 65 Lewis Acids & Bases Many complex ions are very stable. Cu 2+ + 4 NH 3 [Cu(NH 3 ) 4 ] 2+ K for the reaction is called K formation or a “formation constant”. Here K = 6.8 x 10 12. Rxn. is strongly product-favored.

66. 66 Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion. Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion. AgCl(s) + 2 NH 3 Ag(NH 3 ) 2 + + Cl - Lewis Acids & Bases AgCl(s)

67. 67 AgCl(s) Ag + + Cl - K sp = 1.8 x 10 -10 Ag + + 2 NH 3 Ag(NH 3 ) 2 + K form = 1.6 x 10 7 ------------------------------------- AgCl(s) + 2 NH 3 Ag(NH 3 ) 2 + + Cl - K net = K sp K form = 2.9 x 10 -3 K net = K sp K form = 2.9 x 10 -3 Lewis Acids & Bases Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion. Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion.

68. Why?Why? Why are some compounds acids?Why are some compounds acids? Why are some compounds bases?Why are some compounds bases? Why do acids and bases vary in strength?Why do acids and bases vary in strength? Can we predict variations in acidity or basicity?Can we predict variations in acidity or basicity? Why are some compounds acids?Why are some compounds acids? Why are some compounds bases?Why are some compounds bases? Why do acids and bases vary in strength?Why do acids and bases vary in strength? Can we predict variations in acidity or basicity?Can we predict variations in acidity or basicity?

69. 69 Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, which withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive. Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, which withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive. Acetic acid Trichloroacetic acid K a = 1.8 x 10 -5 K a = 0.3

70. 70 Oxyacid Strength For oxyacids, XO n (OH) m, acid strength increases as n increases and is independent of m.For oxyacids, XO n (OH) m, acid strength increases as n increases and is independent of m. Rank in order of increasing acidity: HBrO 3, H 3 BO 3, HIO 4Rank in order of increasing acidity: HBrO 3, H 3 BO 3, HIO 4 BrO 2 (OH), B(OH) 3, IO 3 (OH) B(OH) 3 BrO 2 (OH) IO 3 (OH)

71. 71 These ions are BASES. They become more and more basic as the negative charge increases. As the charge goes up, they interact more strongly with polar water molecules. NO 3 - CO 3 2- Basicity of Oxoanions PO 4 3-

72. 72 Practice Problems 1. Write the equation for the reaction of hypochlorous acid and ammonia. Label the acids and bases. Indicate the conjugate pairs. 2. A 0.15 M weak acid solution is determined to be 2.0% dissociated. Calculate the K a. 3. Calculate the pH of the following solutions: a) 0.50 M nitric acid a) 0.50 M nitric acid b) 0.25 M potassium hydroxide b) 0.25 M potassium hydroxide c) 0.15 M phosphoric acid c) 0.15 M phosphoric acid d) 0.22 M sodium sulfite d) 0.22 M sodium sulfite e) 0.010 potassium cyanide e) 0.010 potassium cyanide

73. 73 Practice Problems 4. Calculate the carbonate ion concentration in a 0.10 M carbonic acid solution. 5. Determine whether a solution of Cs 2 SO 3 is acidic, neutral or basic. 6. Write the Lewis acid/base reaction between PH 3 and BF 3 using dot structures. Indicate the acid and the base.

74. 74 Practice Problems Answers 1. HClO + NH 3 NH 4 + + ClO - acid base acid base acid base acid base 2. 6.0 x 10 -5 3. a).30b) 13.40c) 1.52 d) 10.28e) 10.70 4. 4.8 x 10 -11 5. basic

75. 75 Practice Problems Answers 6.. H H P... H. F F B... F + -->. H H P... H. F F B... F The end.

76. 76 ammonia ammonia NH 3 + H 2 O NH 4 + + OH - or or NH 4 OH NH 4 + + OH - NH 4 OH NH 4 + + OH - Return Return Return

77. 77 Sample Questions Give the conjugate base of HCN. CN - CN - Give the conjugate acid of NO 2 -. HNO 2 Give the conjugate acid of NH 3. NH 4 + The ion HCO 3 - has both a conjugate acid and conjugate base. Give the formula of each. H 2 CO 3 CO 3 2- H 2 CO 3 CO 3 2-

78. 78 Sample Questions Label the acid-base pairs: CN - + HOH HCN + OH - base acid acid base HCO 3 1- + HOH OH - + H 2 CO 3 base acid base acid HCO 3 1- + HOH CO 3 2- + H 3 O + acid base base acid

79. 79 Acid-Base Reactions Consider the acid HF reacting with the base NH 3. Consider the acid HF reacting with the base NH 3. HF + NH 3 NH 4 + + F - acid base acid base Acids: HF > NH 4 + Bases: NH 3 > F - Reaction favors products Consider the acid HF reacting with the base NH 3. Consider the acid HF reacting with the base NH 3. HF + NH 3 NH 4 + + F - acid base acid base Acids: HF > NH 4 + Bases: NH 3 > F - Reaction favors products

80. 80 Increasing Acid Strength Increasing Base Strength

81. 81 Acid-Base Reactions Does the reaction between nitrous acid and ammonia favor reactants or products? Does the reaction between nitrous acid and ammonia favor reactants or products? HNO 2 + NH 3 NH 4 + + NO 2 - acid base acid base acid base acid base Acids: HNO 2 > NH 4 + Bases: NH 3 > NO 2 - Reaction favors products Does the reaction between nitrous acid and ammonia favor reactants or products? Does the reaction between nitrous acid and ammonia favor reactants or products? HNO 2 + NH 3 NH 4 + + NO 2 - acid base acid base acid base acid base Acids: HNO 2 > NH 4 + Bases: NH 3 > NO 2 - Reaction favors products

82. 82 Acid-Base Reactions Does the reaction between hypochlorous acid and fluoride ion favor reactants or products? Does the reaction between hypochlorous acid and fluoride ion favor reactants or products? HClO + F - HF + ClO - acid base acid base acid base acid base Acids: HF > HClO Bases: ClO - > F - Reaction favors reactants Does the reaction between hypochlorous acid and fluoride ion favor reactants or products? Does the reaction between hypochlorous acid and fluoride ion favor reactants or products? HClO + F - HF + ClO - acid base acid base acid base acid base Acids: HF > HClO Bases: ClO - > F - Reaction favors reactants

83. 83 Acid-Base Reactions Does the reaction between ammonium chloride and sodium sulfate favor reactants or products? NH 4 Cl + Na 2 SO 4 ??? Does the reaction between ammonium chloride and sodium sulfate favor reactants or products? NH 4 Cl + Na 2 SO 4 ??? NH 4 + + SO 4 2- NH 3 + HSO 4 1- acid base base acid Acids: HSO 4 1- > NH 4 + Bases: NH 3 > SO 4 2- Reaction favors reactants Does the reaction between ammonium chloride and sodium sulfate favor reactants or products? NH 4 Cl + Na 2 SO 4 ??? Does the reaction between ammonium chloride and sodium sulfate favor reactants or products? NH 4 Cl + Na 2 SO 4 ??? NH 4 + + SO 4 2- NH 3 + HSO 4 1- acid base base acid Acids: HSO 4 1- > NH 4 + Bases: NH 3 > SO 4 2- Reaction favors reactants

84. 84 Sample Questions 1. Determine the pH, pOH, and [OH - ] [H + ] = 2.5 x 10 -6 [H + ] = 2.5 x 10 -6 pH = - log [H + ] = 5.60 pOH = 14 - 5.60 = 8.40 pOH = - log [OH-]or 1.00 x 10 -14 = [H + ][OH - ] 8.40 = - log [OH-] 1.00 x 10 -14 = (2.5 x 10 -6 )[OH - ] [OH - ] = 10 -8.40 [OH - ] = 4.0 x 10 -9 M

85. 85 Sample Questions 2. Determine the pOH, [H + ], and [OH - ] pH = 6.52 pH = 6.52 pOH = 7.48 [H + ] = 3.0 x 10 -7 M [OH - ] = 3.3 x 10 -8 M How can we determine the pH of a solution? Acid/Base indicators is one method.

86. 86 Name of IndicatorpH IntervalColor Change Methyl violet0.2 - 3.0yellow to blue-violet Thymol blue1.2 - 2.8red to yellow Orange IV1.3 - 3.0red to yellow Methyl orange3.1 - 4.4 red to orange to yellow Bromophenol blue3.0 - 4.6 yellow to blue-violet Congo red3.0 - 5.0blue to red Bromocresol green3.8 - 5.4yellow to blue Methyl red4.4 - 6.2red to yellow Bromocresol purple5.2 - 6.8yellow to purple Bromothymol blue6.0 - 7.6yellow to blue Phenol red6.8 - 8.2yellow to red Thymol blue8.0 - 9.6yellow to blue Phenolphthalein8.3 - 10.0colorless to red Thymolphthalein9.3 - 10.5colorless to blue Alizarin yellow10.0 - 12.0yellow to red Indigo carmine11.4 - 13.0blue to yellow

87. 87 Sample Questions 3. Determine the pH, pOH, [H + ], and [OH - ] 0.035 M HCl. Strong acid --> [H + ] = 0.035 pH = 1.46 pOH = 12.54 [OH - ] = 2.9 x 10 -13 M

88. 88 Sample Questions 4. Determine the pH, pOH, [H + ], and [OH - ] 0.15 M NaOH. Strong base --> [OH - ] = 0.15 M pOH = 0.82 pOH = 0.82 pH = 13.18 pH = 13.18 [H + ] = 6.7 x 10 -14 M [H + ] = 6.7 x 10 -14 M

89. 89 Equilibria Involving A Weak Acid You have 0.10 M HC 6 H 4 NO 2. You have 0.10 M HC 6 H 4 NO 2. Its pH is measured to be 2.92 Its pH is measured to be 2.92 Calculate the equilibrium constant. Step 1. Write the equation. Step 1. Write the equation. HA H + + A - HA H + + A -

90. 90 Equilibria Involving A Weak Acid Step 2. Make a chart. HA H + + A - HA H + + A - 0.10 0 0 0.10 0 0 - 0.0012 0.0012 0.10

91. 91 Equilibria Involving A Weak Acid Step 3. Write K a expression, without and with numbers. K a = A - [H + ][A - ] HA [HA] K a =.0012 (0.0012)(0.0012) (0.10 )

92. 92 Equilibria Involving A Weak Acid Step 4. Solve. K a = 1.4 x 10 -5 Step 5. Answer question(s). K a = 1.4 x 10 -5 K a =.0012 (0.0012)(0.0012) (0.10 )

93. 93 Sample Questions 1. You have 0.050 M HCN. Its pH is measured to be 5.4. Calculate the equilibrium constant. HCN H + + CN - 0.050 0 0 - 4 x10 -6 + 4 x10 -6 4 x10 -6 4 x10 -6 4 x10 -60.050 K a = [H + ][CN - ] [HCN] = = 3 x 10 -10 (4 x 10 -6 ) 2 (0.050)

94. 94 Sample Questions 2. You have 0.1000 M HA which is 1.30% dissociated. Calculate the equilibrium constant. Calculate the equilibrium constant. HA H + + A - 0.1000 0 0 0.1000 0 0 - 0.00130 + 0.00130 0.001300.001300.0987 K a = [H + ][A - ] [HA] = = 1.71 x 10 -5 (0.00130) 2 (0.0987)

95. 95 Equilibria Involving A Weak Acid You have 1.00 M HC 2 H 3 O 2. Calculate the equilibrium concentrations of HC 2 H 3 O 2, H +, C 2 H 3 O 2 -, and the pH. Calculate the equilibrium concentrations of HC 2 H 3 O 2, H +, C 2 H 3 O 2 -, and the pH. Step 1. Write the equation. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O 2 -

96. 96 Equilibria Involving A Weak Acid Step 2. Make a chart. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O 2 - 1.00 0 0 1.00 0 0 -x xx +x 1.00-x

97. 97 Equilibria Involving A Weak Acid Step 3. Write K a expression, without and with numbers. K a = C 2 H 3 O 2 - [H + ][C 2 H 3 O 2 - ] HC 2 H 3 O 2 [HC 2 H 3 O 2 ] 1.8 x 10 -5 = x (x)(x) (1.00 - x)

98. 98 Equilibria Involving A Weak Acid Step 4. Solve. 1.8 x 10 -5 = x (x)(x) (1.00 - x) 1.8 x 10 -5 (1.00 - x) = x 2 x 2 + 1.8 x 10 -5 x - 1.8 x 10 -5 = 0 x = 4.2 x 10 -3 M

99. 99 Step 5. Answer question(s). Equilibria Involving A Weak Acid x = [H + ] = [C 2 H 3 O 2 - ] = 0.0042 M x = [H + ] = [C 2 H 3 O 2 - ] = 0.0042 M [HC 2 H 3 O 2 ] = 1.00 - 0.0042 = 1.00 M [HC 2 H 3 O 2 ] = 1.00 - 0.0042 = 1.00 M pH = 2.38 pH = 2.38

100. 100 Equilibria Involving A Weak Acid Note: Assume x is very small because K a is so small. 1.8 x 10 -5 = x (x)(x) (1.00 - x) 1.8 x 10 -5 (1.00) = x 2 x = 4.2 x 10 -3 M 1.8 x 10 -5 = x (x)(x) (1.00) General Rule: Simplfy if [HA] / Ka > 1000

101. 101 Sample Questions 1. Calculate the pH of a 0.100 M HF solution. HF H + + F - 0.100 0 0 - x+ x x x0.100 - x K a = [H + ][F - ] [HF] = = 7.2 x 10 - 4 x 2 (0.100 - x) X = 0.0081 M = {H + ] pH = 2.09

102. 102 Sample Questions 2. Calculate the pH of a 0.20 M HClO solution. HClO H + + ClO - HClO H + + ClO - 0.20 0 0 0.20 0 0 - x+ x x x0.20 K a = [H + ][ClO - ] [HClO] = = 3.5 x 10 - 8 x 2 (0.20) X = 8.4 x 10 - 5 M = {H + ] pH = 4.08

103. 103 Sample Questions 3. Calculate the pH of a 0.20 M HCOOH solution. HA H + + A - HA H + + A - 0.20 0 0 0.20 0 0 - x+ x x x0.20 - x K a = [H + ][A - ] [HA] = = 1.8 x 10 -4 x 2 (0.20 - x) X = 0.0059 M = {H + ] pH = 2.23

104. 104 Sample Questions 4. Calculate the pH of a 0.15 M H 2 CO 3 solution. H 2 CO 3 H + + HCO 3 - 0.15 0 0 0.15 0 0 - x+ x x x0.15 K a = [H + ][HCO 3 - ] [H 2 CO 3 ] = = 4.2 x 10 -7 x 2 0.15 X = 2.5 x 10 -4 M

105. 105 Sample Questions Does the second equilibrium have a significant effect on the pH? Does the second equilibrium have a significant effect on the pH? HCO 3 1- H + + CO 3 2- HCO 3 1- H + + CO 3 2- 2.5 x 10 -4 2.5 x 10 -4 0 2.5 x 10 -4 2.5 x 10 -4 0 - y+ y y2.5 x 10 -4 K a = [H + ][CO 3 2- ] [HCO 3 1- ] = = 4.8 x 10 -11 (2.5 x 10 -4 )y 2.5 x 10 -4 y = 4.8 x 10 -11 M pH = 3.60 NO!

106. 106 Equilibria Involving A Weak Base You have 0.10 M BOH. Its pH is measured to be 10.11 You have 0.10 M BOH. Its pH is measured to be 10.11 Calculate the equilibrium constant. Calculate the equilibrium constant. Step 1. Write the equation. BOH B + + OH - BOH B + + OH - [OH] - = 10 -pOH = 10 -3.89 = 1.3 x 10 -4 M [OH] - = 10 -pOH = 10 -3.89 = 1.3 x 10 -4 M

107. 107 Equilibria Involving A Weak Base Step 2. Make a chart. BOH B + + OH - BOH B + + OH - 0.10 0 0 0.10 0 0 - 0.00013 0.000130.00013 00.10

108. 108 Equilibria Involving A Weak Base Step 3. Write K b expression, without and with numbers. K b = OH - [B + ][OH - ] BOH [BOH] K b =.00013 (0.00013)(0.00013) (0.10)

109. 109 Equilibria Involving A Weak Base Step 4. Solve. K b = 1.7 x 10 -7 K b =.00013 (0.00013)(0.00013) (0.10) Step 5. Answer question(s). K b = 1.7 x 10 -7

110. 110 Sample Question You have 0.0200 M B. You have 0.0200 M B. Its is 3.0% dissociated. Calculate the equilibrium constant. Calculate the equilibrium constant. B + HOH HB + + OH - B + HOH HB + + OH - 0.0200 0 0 - 6.0 x 10 -4 + 6.0 x 10 -4 6.0 x 10 -4 6.0 x 10 -4 0.0194 K b = [HB + ][OH - ] [B] = = 1.9x10 -5 (6.0 x 10 -4 ) 2 (0.0194)

111. 111 Equilibria Involving A Weak Base You have 0.100 M NH 3 Calculate all equilibrium concentrations and the pH. Calculate all equilibrium concentrations and the pH. Step 1. Write the equation. NH 3 + HOH NH 4 + + OH - NH 3 + HOH NH 4 + + OH -

112. 112 Equilibria Involving A Weak Base Step 2. Make a chart. NH 3 + HOH NH 4 + + OH - NH 3 + HOH NH 4 + + OH - 0.100 0 0 0.100 0 0 -x xx +x 0.100

113. 113 Equilibria Involving A Weak Base Step 3. Write K b expression, without and with numbers. K b = OH - [NH 4 + ][OH - ] NH 3 [NH 3 ] 1.8 x 10 -5 = x (x)(x) (0.100)

114. 114 Equilibria Involving A Weak Base Step 4. Solve. 1.8 x 10 -5 = x (x)(x) (0.100) x = 1.3 x 10 -3 M

115. 115 Step 5. Answer question(s). Equilibria Involving A Weak Base [NH 4 + ] = [OH - ] = 0.0013 M [NH 4 + ] = [OH - ] = 0.0013 M [NH 3 ] = 0.099 M [NH 3 ] = 0.099 M pOH = 2.89 pOH = 2.89 pH = 11.11 pH = 11.11

116. 116 Sample Questions Calculate the pH of a 0.015 M B solution. Calculate the pH of a 0.015 M B solution. K b = 2.5 x 10 -7 K b = 2.5 x 10 -7 B + HOH HB + + OH - 0.015 0 0 - x+ x x x0.015 K b = [HB + ][OH - ] [B] = = 2.5x10 -7 x 2 (0.015) X = 6.5 x 10 -5 M = [OH - ] pH = 9.79

117. 117 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K a and the % ionization. Sample Problems: Weak Acids and Bases HA H + + A - 0.10 0 0 - 0.0012+ 0.0012 0.0012 0.10 K a = [H + ][A - ] [HA] = = 1.4 x 10 -5 (0.0012) 2 (0.10) % ionization = 0.0012/0.10 x 100 = 1.2%

118. 118 2. A solution of propionic acid, CH 3 CH 2 COOH, is 0.20 M. K a = 1.3x10 -5. Calculate the pH and % ionization. Sample Problems: Weak Acids and Bases HA H + + A - 0.20 0 0 0.20 0 0 - x+ x x x0.20 K a = [H + ][A - ] [HA] = = 1.3 x 10 -5 x 2 (0.20) % ionization = 0.0016/0.20 x 100 = 0.80% x = 1.6 x 10 -3 = [H + ], pH = 2.80

119. 119 3. A solution of hydrazine, N 2 H 4, is 0.025 M. K b = 8.5x10 -5. Calculate the pH and % ionization. Sample Problems: Weak Acids and Bases N 2 H 4 + HOH N 2 H 5 + + OH - 0.0250 0 - x+ x x x0.025 K b = [N 2 H 5 + ][OH - ] [N 2 H 4 ] = = 8.5 x 10 -5 x 2 (0.025) % ionization = 0.0015/0.025 x 100 = 6.0% x = 1.5 x 10 -3 = [OH - ], pOH = 2.82 pH = 11.18

120. 120 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13 H 3 AsO 4 H + + H 2 AsO 4 - 0.15 0 0 0.15 0 0 - x+ x x x0.15 - x K a = [H + ][H 2 AsO 4 - ] [H 3 AsO 4 ] = = 2.5 x 10 -4 x 2 (0.15 - x) [H 3 AsO 4 ] = 0.14 M [H 2 AsO 4 - ] = 6.0 x 10 -3 M x = 6.0 x 10 -3 M = [H + ], pH = 2.22

121. 121 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13 H 2 AsO 4 - H + + HAsO 4 2- 6.0 x 10 -3 6.0 x 10 -3 0 6.0 x 10 -3 6.0 x 10 -3 0 - y+ y y 6.0 x 10 -3 K a = [H + ][HAsO 4 2- ] [H 2 AsO 4 - ] = = 5.6 x 10 -8 6.0 x 10 -3 y 6.0 x 10 -3 y = 5.6 x 10 -8 M = [HAsO 4 2- ]

122. 122 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of H 3 AsO 4. K 1 = 2.5x10 -4, K 2 = 5.6x10 -8, K 3 = 3.0x10 -13 HAsO 4 2- H + + AsO 4 3- 5.6 x 10 -8 6.0 x 10 -3 0 5.6 x 10 -8 6.0 x 10 -3 0 - z+ z z 6.0 x 10 -3 5.6 x 10 -8 K a = [H + ][AsO 4 3- ] [HAsO 4 2- ] = = 3.0 x 10 -13 6.0 x 10 -3 z 5.6 x 10 -8 z = 2.8 x 10 -18 M = [AsO 4 3- ]

123. 123 Determine if the following solutions are acidic, basic, or neutral. KBr K + + HOH KOH + H + Br - + HOH HBr + OH - Acid-Base Properties of Salts CrCl 3 Cr 3+ + HOH CrOH 2+ + H + Cl - + HOH HCl + OH - X X X X neutral X X acidic

124. 124 Determine if the following solutions are acidic, basic, or neutral. NaNO 2 Na + + HOH NaOH + H + NO 2 - + HOH HNO 2 + OH - Acid-Base Properties of Salts KHCO 3 K + + HOH KOH + H + HCO 3 - + HOH H 2 CO 3 + OH - K b = 2.4x10 -8 HCO 3 - + HOH CO 3 2- + H 3 O + K a = 4.8x10 -11 X X basic X X

125. 125 Determine if the following solutions are acidic, basic, or neutral. Na 2 CO 3 Na + + HOH NaOH + H + CO 3 2- + HOH HCO 3 - + OH - Acid-Base Properties of Salts NH 4 C 2 H 3 O 2 NH 4 + + HOH NH 3 + H 3 O + K a = 5.6x10 -10 C 2 H 3 O 2 - + HOH HC 2 H 3 O 2 +OH - K b =5.6x10 -10 X X basic neutral

126. 126 Acid-Base Properties of Salts You have 0.10 M Na 2 CO 3 Calculate the pH. Calculate the pH. Step 1. Write the equation. CO 3 2- + HOH HCO 3 - + OH - CO 3 2- + HOH HCO 3 - + OH -

127. 127 Acid-Base Properties of Salts Step 2. Make a chart. CO 3 2- + HOH HCO 3 - + OH - CO 3 2- + HOH HCO 3 - + OH - 0.10 0 0 0.10 0 0 -x xx +x 0.10 - x

128. 128 Acid-Base Properties of Salts Step 3. Write K b expression, without and with numbers. K b = OH - [HCO 3 - ][OH - ] CO 3 2- [CO 3 2- ] 2.1 x 10 -4 = x (x)(x) (0.10 - x)

129. 129 Acid-Base Properties of Salts Step 4. Solve. 2.1 x 10 -4 = x (x)(x) (0.10 - x) x = 4.5 x 10 -3 M

130. 130 Step 5. Answer question(s). Acid-Base Properties of Salts [OH - ] = 0.0045 M [OH - ] = 0.0045 M pOH = 2.35 pOH = 2.35 pH = 11.65 pH = 11.65

131. 131 Sample Questions 1. Calculate the pH of a 0.10 M K 3 PO 4 solution. PO 4 3- + HOH HPO 4 2- + OH - 0.10 0 0 - x+ x x x0.10 - x K b = [HPO 4 2- ][OH - ] [PO 4 3- ] = = 0.028 x 2 (0.10 - x) X = 0.041 pH = 12.61

132. 132 Sample Questions 2. Calculate the pH of a 0.10 M Na 2 C 2 O 4 solution. C 2 O 4 2- + HOH HC 2 O 4 - + OH - 0.100 0 - x+ x x x 0.10 K b = C 2 O 4 [HC 2 O 4 - ][OH - ] C 2 O 4 [C 2 O 4 2- ] = =1.6x10 -10 x 2 (0.10) X = 4.0 x 10 -6 pH = 8.60

133. 133 Sample Questions 3. Calculate the pH of a 0.045 M KCN solution. CN - + HOH HCN + OH - CN - + HOH HCN + OH - 0.045 0 0 - x+ x x x 0.045 K b = CN [HCN][OH - ] CN [CN - ] = = 2.5x10 -5 x 2 (0.045) x = 1.1 x 10 -3 = [OH - ] pOH = 2.96, pH = 11.04