1. Chemistry 102(060) Summer 2015 Instructor: Dr. Upali Siriwardane
Office: CTH 311
Phone
Office Hours: M,Tu, W,Th,F 9:00-11:00 am or by appointment..;
Test Dates:
July 20, 2015 (Test 1): Chapter 13
July 27, 2015 (Test 2): Chapter 14
August 4, 2015 (Test 3): Chapter 15
August 12, 2015 (Test 4): Chapter 17
August 13, 2015 (Make-up test) comprehensive:
Chapters 13-17

2. Chapter 15. Acids and Bases
15.1 Heartburn 659
15.2 The Nature of Acids and Bases 660
15.3 Definitions of Acids and Bases 662
15.4 Acid Strength and the Acid Dissociation Constant (Ka) 665
15.5 Autoionization of Water and pH 668
15.6 Finding the and pH of Strong and Weak Acid Solutions 673
15.7 Base Solutions 682
15.8 The Acid–Base Properties of Ions and Salts 685
15.9 Polyprotic Acids 693
Acid Strength and Molecular Structure 698
Lewis Acids and Bases 700
Acid Rain 701

3. Types of Reactions a) Precipitation Reactions.
Reactions of ionic compounds or salts
b) Acid/base Reactions.
Reactions of acids and bases
c) Redox Reactions.
reactions of oxidizing & reducing agents

4. What are Acids &Bases? Definition? a) Arrhenius b) Bronsted-Lowry
c) Lewis

5. Arrhenius Definitions
Arrhenius, Svante August ( ), Swedish chemist, 1903 Nobel Prize in chemistry
Acid Anything that produces hydrogen ions in a water solution.
HCl (aq) H+ ( aq) + Cl- ( aq)
Base Anything that producs hydroxide ions in a water solution.
NaOH (aq) Na+ ( aq) + OH- ( aq)
Arrhenius definitions are limited proton acids and hydroxide bases to aqueous solutions.

6. Brønsted-Lowry definitions
Expands the Arrhenius definitions to include many bases other than hydroxides and gas phase reactions
Acid Proton donor
Base Proton acceptor
This definition explains how substances like ammonia can act as bases.
Eg. HCl(g) + NH3(g) > NH4Cl(s)
HCl (acid), NH3 (base).
NH3(g) + H2O(l) NH OH-

7. Lewis Definition
G.N. Lewis was successful in including acid and bases without proton or hydroxyl ions.
Lewis Acid: A substance that accepts an electron pair.
Lewis base: A substance that donates an electron pair.
E.g. BF3(g) + :NH3(g) F3B:NH3(s)
the base donates a pair of electrons to the acid forming a coordinate covalent bond common to coordination compounds. Lewis acids/bases will be discussed later in detail

8. 1) Acids and bases can be defined in several ways
1) Acids and bases can be defined in several ways. Which definitions of the bases that fits the description below? a) a compound that produces more OH- ions in water: b) a proton acceptor: c) an electron pair donor:

9. Types of Acids and Bases
Binary acids
Oxyacid
Organic acids
Acidic oxides
Basic oxides
Amine
Polyprotic acids

10. Types of Acids and Bases
Binary acids: HCl, HBr, HI, H2S
More than two elements: HCN
Oxyacid: HNO3, H2SO4, H3PO4
Polyprotic acids: H2SO4, H3PO4
Organic acids: R-COOH, R= CH3-, CH3CH2-
Acidic oxides: SO3, NO2, CO2,
Basic oxides: Na2O, CaO
Amine: NH3. R-NH2, R= CH3-, CH3CH2- : primary
R2-NH : secondary, R3-N: tertiary
Lewis acids & bases: BF3 and NH3

11. These are usually oxides of non-metallic elements such as P, S and N.
Acidic Oxides
These are usually oxides of non-metallic elements such as P, S and N.
E.g. NO2, SO2, SO3, CO2
They produce oxyacids when dissolved in water
SO3 + H2O ---> H2SO4
CO2 + H2O ---> H2CO3
NO2 + H2O ---> HNO3

12. Basic Oxides
Oxides oxides of metallic elements such as Na, K, Ca. They produce hydroxyl bases when dissolved in water.
e.g. CaO + H2O ---> Ca(OH)2
Na2O + H2O ---> 2 NaOH

13. Protic Acids
Monoprotic Acids: The form protic refers to acidity due to protons. Monoprotic acids have only one acidic proton. e.g. HCl.
Polyprotic Acids: They have more than one acidic proton.
e.g. H2SO diprotic acid
H3PO triprotic acid.

14. Polyprotic Acids
acids where more than one hydrogen per molecule is released

15. Amines
Class of organic bases derived from ammonia NH3 by replacing hydrogen by organic groups. They are defined as bases similar to NH3 by Bronsted-Lowery or Lewis acid/base definitions.

16. Amines

17. 2) Identify types of acids/bases as: binary acids, oxy acids, organic acids, acidic oxides, basic oxides, amine and polyprotic acids.
a) HF b) HBr c) H3PO4
d) H2SO e) HNO f) R-COOH
g) NO h) SO j)CaO
k) R-NH2

18. Influence of Molecular Structure on Acid Strength
Binary Hydrides
hydrogen & one other element
Bond Strengths
weaker the bond, the stronger the acid
Stability of Anion
higher the electronegativity, stronger the acid

19. Binary Acids
Compounds containing acidic protons bonded to a more electronegative atom.
e.g. HF, HCl, HBr, HI, H2S
The acidity of the haloacid
(HX; X = Cl, Br, I, F)
Series increase in the following order:
HF < HCl < HBr < HI

20. Oxyacids Compounds containing acidic - OH groups in the molecule.
Acidity of H2SO4 is greater than H2SO3 because of the extra O (oxygens)
The order of acidity of oxyacids from the a halogen (Cl, Br, or I) shows a similar trend.
HClO4 > HClO3 > HClO2 > HClO
perchloric chloric chlorus hyphochlorus

21. Influence of Molecular Structure on Acid Strength
Oxyacids
hydrogen, oxygen, & one other element
H-O-E
higher the electronegativity on E, stronger the acid as this weakens the bond between the O and H

22. Oxo Acid
<
<
<
<

23. 3) Which of the following is stronger acid or base:
H2SO4 or H2SO3:
HCl or HI:
HClO or HClO3:
H2S or HF:
CF3COOH or CH3COOH:
CH3COOH or CH3CH2COOH

24. 3) Which of the following is stronger acid or base:
H2SO4 or H2SO3: Ka H2SO4>> 1 ; H2SO3= 1.4 x 10-2
HCl or HI:
HClO or HClO3:
H2S or HF: H2S: Ka:H2S=6.3 x 10-8 ; HF= 6.3 x 10-4
CF3COOH or CH3COOH: 2.5 x 10-3 and 1.8 x 10-5
CH3COOH or CH3CH2COOH: 1.8 x 10-5 and 1.2 x 10-5

25. Dissociation Strong Acids: HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq)
H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq)
Dissociation Equilibrium Weak Acid/base:
H2O(l) + H2O(l) H3+O(aq) + OH-(aq)
This dissociation is called autoionization of water.
HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq)
NH3 (aq) + H2O(l) NH4+ + OH-(aq)
Equilibrium constants: Ka, Kb and Kw

26. 4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations? a) HCl: b) H2SO4 :

27. 4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations? c) H2O (autoionization): What is auto ionization? d) HC2H3O(acetic acid): e) NH3:

28. Bronsted acid/conjugate base and base/conjugate acid pairs in acid/base equilibria
HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq)
HCl(aq): acid
H2O(l): base
H3+O(aq): conjugate acid
Cl-(aq): conjugate base
H2O/ H3+O: base/conjugate acid pair
HCl/Cl-: acid/conjugate base pair

29. Brønsted-Lowry Definitions
Conjugate acid-base pairs.
Acids and bases that are related by loss or gain of H+ as H3O+ and H2O.
Examples. Acid Base
H3O + H2O
HC2H3O2 C2H3O2-
NH4 + NH3
H2SO4 HSO4-
HSO4- SO42-

30. Select acid, base, acid/conjugate base pair, base/conjugate acid pair
H2SO4(aq) + H2O(l) H 3+O(aq) + HSO4-(aq)
acid
base
conjugate acid
conjugate base
base/conjugate acid pair
acid/conjugate base pair

31. 5) For HOCl write: a) Dissociation equilibrium reaction for the HOCl: b) Identify the acid/conjugate base pair: c) Identify the base/conjugate acid pair: d) The equilibrium constant expression:

32. Strong Acid vs. Weak Acids
completely ionized
Hydrioidic HI Ka ~ pKa = -11
Hydrobromic HBr Ka ~ 109 pKa = -9
Perchloric HClO4 Ka ~ 107 pKa = -7
Hyrdrochloric HCl Ka ~ 107 pKa = -7
Chloric HClO3 Ka ~ 103 pKa = -3
Sulfuric H2SO4 Ka ~ 102 pKa = -2
Nitric HNO3 Ka ~ 20 pKa = -1.3
Weak acid
partially ionized
Hydrofluoric acid HF Ka = 6.6x pKa = 3.18
Formic acid HCOOH Ka = 1.77x pKa = 3.75
Acetic acid CH3COOH Ka = 1.76x pKa = 4.75
Nitrous acid HNO2 Ka = 4.6x pKa = 3.34
Acetyl Salicylic acid C9H8O4 Ka = 3x pKa = 3.52
Hydrocyanic acid HCN Ka = 6.17x pKa = 9.21

33. Strong Base vs. Weak Base
completely ionized
Lithium hydroxide LiOH
Sodium hydroxide NaOH
Potassium hydroxide KOH Kb~
Rubidium hydroxide RbOH
Cesium hydroxide CsOH
Boarder-line Bases
Magnesium hydroxide Mg(OH)2
Calcium hydroxide Ca(OH)2
Strotium hydroxide Sr(OH) Kb~ 0.01 to0.1
Barium hydroxide Ba(OH)2
Weak Base
partially ionized
Ammonia NH3 Kb=1.79x10-5 pKb = 4.74
Ethyl amine CH3CH2NH2 Kb=5.6x10-4 pKb = 3.25

34. Acid and Base Strength
Strong acids Ionize completely in water HCl, HBr, HI, HClO3,
HNO3, HClO4, H2SO4.
Weak acids Partially ionize in water.
Most acids are weak.
Strong bases Ionize completely in water Strong bases are metal hydroxides - NaOH, KOH
Weak bases Partially ionize in water.

35. Common Acids and Bases Acids Formula Molarity* nitric HNO3 16
hydrochloric HCl
sulfuric H2SO4 18
acetic HC2H3O2 18
Bases
ammonia NH3(aq) 15
sodium hydroxide NaOH solid
*undiluted.

36. 6. Identify stronger and weaker acids:

37. Autoionization of Water
Autoionization When water molecules react with one another to form ions.
Acids and bases alter the dissociation equilibrium of water based on Le Chaterlier’s principle
Kw = [ H3O+ ] [ OH- ]
= 1.0 x at 25oC
Note: [H2O] is constant and is included in Kw.
H2O(l) + H2O(l) H3O+(aq) + OH-(aq) (10-7M) (10-7M)
ion product
of water

38. What is pH? Kw = [H3+O][OH-] = 1 x 10-14 [H3+O][OH-] = 10-7 x 10-7
Extreme cases:
Basic medium
[H3+O][OH-] = x 100
Acidic medium
[H3+O][OH-] = 100 x 10-14
pH value is -log[H+]
spans only 0-14 in water.

39. pH and other “p” scales
Substance pH
1 M HCl
Gastric juices
Lemon juice
Classic Coke
Coffee
Pure Water
Blood
Milk of Magnesia
Household ammonia
1M NaOH
We need to measure and use acids and bases over a very large concentration range.
pH and pOH are systems to keep track of these very large ranges.
pH = -log[H3O+]
pOH = -log[OH-]
pH + pOH = 14

40. pH scale
A logarithmic scale used to keep track of the large changes in [H+].
10 0 M M M
Very Neutral Very
acidic Basic
When you add an acid to, the pH gets smaller.
When you add a base to, the pH gets larger.

41. pH of Aqueous Solutions

42. pH, pKw and pOH The relation of pH, Kw and pOH Kw = [H+][OH-]
log Kw = log [H+] + log [OH-]
-log Kw= -log [H+] -log [OH-] ;
previous equation multiplied by -1
pKw = pH + pOH; pKw = 14
since Kw =1 x 10-14
14 = pH + pOH
pH = 14 - pOH
pOH = 14 - pH

43. Measuring pH
Arnold Beckman
inventor of the pH meter
father of electronic instrumentation

44. 7. Identify the following as acidic/basic/neutral and estimate/calculate pH.
Solution Acidic/basic/neutral pH of the solution
a) [ H+] > [OH-] and [H+] > 1.0 x 10-7 M : :
b) [H+] < [OH-] and [H+] < 1.0 x 10-7 M : :
c) [H+] = [OH-] = 1.0 x 10-7 M : :
d) [H+] > [OH-] = 1.0 × M : :
e) [H+]< [OH-] = 1.0 x 10+7 M : :

45. pH and pOH calculations of acid and base solutions
a) Strong acids/bases
dissociation is complete for strong acid such as HNO3 or base NaOH
[H+] is calculated from molarity (M) of the solution
b) weak acids/bases
needs Ka , Kb or percent(%)dissociation

46. pH of 0.5 M H2SO4 Solution HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq)
H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq)
HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq)
[H3+O][HSO4-]
H2SO4 ; Ka1 =
[H2SO4]
[H3+O][SO42-]
H2SO4 ; Ka2 = ; Ka2 ignored
[HSO4-]

47. pH of 0.5 M H2SO4 Solution pH = -log(0.5) pH = 0.30
H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq)
the moles of H+ ions in the solution is equal to moles of H2SO4 at the beginning.
[H2SO4] = [H+] = 0.5 mole/L
pH = -log [H+]
pH = -log(0.5)
pH = 0.30

48. 1.5 x 10-2 M NaOH.
1.5 x 10-2 M NaOH.
NaOH is also a strong base dissociates completely in water.
[NaOH] = [HO- ] = 1.5 x 10-2 mole/L
pOH = -log[HO-]= -log(1.5 x 10-2)
pOH = 1.82
As defined and derived previously:
pKw= pH + pOH; pKw= 14
pH = pKw + pOH
pH = 14 - pOH
pH = ; pH = 12.18

49. How many OH- are in the compound?
8) For a 0.10 M solution of Ba(OH)2.Is it a strong base?
How many OH- are in the compound?
b) Calculate the [OH-] and [H+]:
c) pH of the solution:

50. b) Is it polyprotic acid? c) Dissociation equilibria:
9) Calculate the pH of the strong acid 0.2 M H2SO4.
a) Is it a strong acid?
b) Is it polyprotic acid?
c) Dissociation equilibria:
d) Why second dissociation equilibria is not considered for [H+] concentration?
e) Calculate the [H+]
f) pH of the solution:

51. pH of Mixtures of Strong and Weak Acids
the presence of the strong acid retards the dissociation of the weak acid
The pH of the solution is mainly based on the strong acid
Eg. 1.0 M HCl and 1.0 HC2H3O2
HCl(aq) + H2O(l)  H3+O(aq) + Cl-(aq)
HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq)

52. pH of Mixtures of Acids and Bases
The pH of the solution is mainly based on the excess acid or base present
Eg. 10 mL of 1.0 M HCl and 20 mL 1.0 NaOH
Moles of excess NaOH ( M x L) = 1.0 x 0.001= 0.001
Mixed together volume = 30 mL = L
Molarity of excess NaOH = 0.001/0.030= 0.030
Calculate pOH and then pH

53. Equilibrium, Constant, Ka & Kb
Ka: Acid dissociation constant for a equilibrium reaction.
Kb: Base dissociation constant for a equilibrium reaction.
Acid: HA + H2O H3+O + A-
Base: BOH + H2O B+ + OH-
[H3+O][ A-] [B+ ][OH-]
Ka = ; Kb =
[HA] [BOH]

54. Acid Dissociation Constant
HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq)
[H3+O][Cl-]
Ka=
[HCl]
[H+][Cl-]

55. Base Dissociation Constant
NH3 + H2O NH OH-
[NH4+][OH-]
K =
[NH3]

56. Comparing Kw and Ka & Kb
Any compound with a Ka value greater than Kw of water will be a an acid in water.
Any compound with a Kb value greater than Kw of water will be a base in water.

57. Ionization Constants for Acids

58. WEAKER/STRONGER Acids and Bases & Ka and Kb values
A larger value of Ka or Kb indicates an equilibrium favoring product side.
Acidity and basicity increase with increasing Ka or Kb.
pKa = - log Ka and pKb = - log Kb
Acidity and basicity decrease with increasing pKa or pKb.

59. Which is weaker? a. HNO2 ; Ka= 4.0 x 10-4. b. HOCl2 ; Ka= 1.2 x 10-2.
c. HOCl     ;  Ka= 3.5 x 10-8.
d. HCN      ;  Ka= 4.9 x

60. What is Ka1 and Ka2? H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq)
HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq)

61. Ka Examples H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq)
HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq)
[H3+O][HSO4-]
H2SO4 ; Ka1 =
[H2SO4]
[H3+O][SO42-]
H2SO4 ; Ka2 =
[HSO4-]

62. Ka Examples HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq) [H+][C2H3O2-]
H C2H3O2; Ka=
[H C2H3O2]
NH3 (aq) + H2O(l) NH4+ + OH-(aq)
[NH4+][OH-]
NH3; Kb=
[ NH3]

63. % Dissociation gives x (amount dissociated) need for pH calculation
% Dissoc. = x 100
Initial amount/con. x
% Dissoc. = x 100
concentration

64. How do you calculate pH of weak acids/bases?
From % dissociation
From Ka or Kb
What is % dissociation
Amount dissociated
% Dissoc. = x 100
Initial amount

65. How do you calculate % dissociation from Ka or Kb
1.00 M solution of HCN; Ka = 4.9 x 10-10
What is the % dissociation for the acid?

66. 1.00 M solution of HCN; Ka = 4.9 x 10-10
First write the dissociation equilibrium equation:
HCN(aq) + H 2O(l) <===> H 3+O(aq) + CN-(aq)
[HCN] [H+ ] [CN- ]
Ini. Con M M M
Cha. Con x x x
Eq. Con x x x
[H 3+O ][CN-] x2
Ka = =
[HCN] x

67. 1.00 M solution of HCN; Ka = 4.9 x 10-10
x ~ since x is small x2
Ka = ; Ka = 4.9 x = x2
1.0
x = x = x
Amount disso x
x 100 = x 100
Ini. amount
% Diss. =2.21 x x 100 = %

68. a) Dissociation equilibria: b) ICE setup: c) Amount dissociated:
10) Calculate the % dissociation of 2.00 M solutions of HCN (Ka= 4.9 x 10-10)
a) Dissociation equilibria:
b) ICE setup:
c) Amount dissociated:
d) % dissociation::

69. Calculate the pH of a weak acid from % dissociation
1 M HF, 2.7% dissociated
Notice the conversion of % dissociation to a fraction (x): 2.7/100=0.027) x=0.027

70. Calculate the pH of a weak acid from % dissociation
HF(aq) + H 2O(l) <===> H 3+O(aq) + F-(aq)
[H+][F-]
Ka =
[HF]
[HF] [H+ ] [F- ]
Ini. Con M 0.0 M 0.00 M
Chg. Con -x x x
Eq.Con
pH = -log [H+]
pH = -log(0.027)
pH = 1.57

71. 11) Calculate the Ka of if 5. 0 M HF, 2
11) Calculate the Ka of if 5.0 M HF, 2.7% dissociated: a) Dissociation equilibrium: b) ICE setup: c) Amount dissociated: d) Ka :

72. HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)
Weak acid Equilibria
Example
Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = x 10-5
HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)
The first step is to write the equilibrium expression
Ka =
[H3O+][Bz-]
[HBz]

73. Weak acid Equilibria HBz H3O+ Bz- Initial conc., M 0.10 0.00 0.00
Change, DM -x x x
Eq. Conc., M x x x
[H3O+] = [Bz-] = x
We’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.

74. Weak Acid Equilibria Solve the equilibrium equation in terms of x
Ka = x =
x = (6.28 x )(0.10)
H3O+ = M
pH = 2.60 x2
0.10

75. pH from Ka or Kb 1.00 M solution of HCN; Ka = 4.9 x 10-10
First write the dissociation equilibrium equation:
HCN(aq) + H 2O(l) H 3+O(aq) + CN-(aq)
[HCN] [H+ ] [CN- ]
Ini. Con M M M
Chg. Con x x x
Eq. Con x x x

76. Weak Acid Equilibria [H 3+O ][CN-] x2
[HCN] x
x ~ since x is small x2
Ka = ; Ka = 4.9 x = x2
1.0
x = x = x
pH = -log [H+]
pH = -log(2.21 x 10-5)
pH = 4.65

77. The Conjugate Partners of Strong Acids and Bases
The conjugate acid/base of a strong base/acid has no net effect on the pH of a solution
The conjugate base of a weak acid hydrolyze in water and basic or
pH of a solution > E.g. Na+C2H3O2- sodium acetate
The conjugate acid of a weak base hydrolyze in water and acidic or
pH of a solution < 7.00 E.g NH4Cl

78. 12) Calculate the [H+], [OH-] and pH of 0. 90 M HC2H3O2; Ka= 1
12) Calculate the [H+], [OH-] and pH of 0.90 M HC2H3O2; Ka= 1.8 x a) Dissociation equilibria: b) ICE setup: c) [H+] and [OH-]: d) pH:

79. 13) Calculate the [H+], [OH-] pOH and pH 5. 0 M NH3; Kb = 1
13) Calculate the [H+], [OH-] pOH and pH 5.0 M NH3; Kb = 1.8 x 10-5 a) Dissociation equilibria: b) ICE setup: c) [H+] and [OH-]: d) pOH and pH:

80. 14) Calculate the pH of a 0. 015 M solution of lactic acid
14) Calculate the pH of a M solution of lactic acid. The Ka for lactic acid is 1.4 x 10-4.

81. 15) Calculate the pH of a 0. 14 M solution of an acid with Ka = 6
15) Calculate the pH of a 0.14 M solution of an acid with Ka = 6.2 x 10-8 (pH = 4.03)

82. Acid-Base Properties of Typical Ions

83. Hydrolysis
Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-base reaction.
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
NH4+(aq) + H2O(l) NH3 (aq) + H3O+(aq)
This type of reaction is given a special name.
Hydrolysis
The reaction of an anion with water to produce the conjugate acid and OH-.
The reaction of a cation with water to produce the conjugate base and H3O+.

84. What salt solutions would be acidic, basic and neutral?
1) strong acid + strong base = neutral
2) weak acid + strong base = basic
3) strong acid + weak base = acidic
weak acid + weak base = neutral,
basic or an acidic solution depending on the relative strengths of the acid and the base.

85. What pH? Neutral, basic or acidic?
a)NaCl
neutral
b) NaC2H3O2
basic
c) NaHSO4
acidic
d) NH4Cl

86. 1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic? a) Solid sodium carbonate-(Na2CO3): b) Sodium chloride- (NaCl): c) Sodium acetate- (NaC2H3O2): d) Ammonium sulfate-((NH4)2SO4):

87. How do you calculate pH of a salt solution?
Find out the pH, acidic or basic?
If acidic it should be a salt of weak base
If basic it should be a salt of weak acid
if acidic calculate Ka from Ka= Kw/Kb
if basic calculate Kb from Kb= Kw/Ka
Do a calculation similar to pH of a weak acid or base

88. What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)
Find out the pH, acidic
if acidic calculate Ka from Ka= Kw/Kb
Ka= Kw/Kb = 1 x /1.8 x 10-5)
Ka= X 10-10
Do a calculation similar to pH of a weak acid

89. Continued NH4+ + H2O H 3+O + NH3 [NH4+] [H3+O ] [NH3 ]
Ini. Con M M 0.00 M
Change x x x
Eq. Con x x x
[H 3+O ] [NH3 ]
Ka(NH4+) = =
[NH 4+] x2
; appro.: x
( x)

90. Continued pH of 0.5 M NH4Cl solution is 4.77 (acidic) x2
Ka(NH4+) = = 5.56 x
0. 5
x2 = x x 0.5 = x
x= x = x 10-5
[H+ ] = x = 1.66 x 10-5 M
pH = -log [H+ ] = - log x 10-5
pH = 4.77
pH of 0.5 M NH4Cl solution is 4.77 (acidic)