acid/base definitions arrhenius model acids produce hydrogen ions in aqueous solutions bases...
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Acid/Base Definitions Arrhenius Model
Acids produce hydrogen ions in aqueous solutions
Bases produce hydroxide ions in aqueous solutions
Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors
Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors
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Acid DissociationAcid Dissociation
HA H+ + A-
Acid Proton Conjugate
base
][
]][[
HA
AHKa
Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)
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Dissociation of Strong AcidsDissociation of Strong AcidsStrong acids are assumed to dissociate completely in solution.
Large Ka or small Ka?Reactant favored or product favored?
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Dissociation Constants: Strong Dissociation Constants: Strong AcidsAcids
Acid Formul
a Conjugate
Base Ka
Perchloric HClO4 ClO4- Very large
Hydriodic HI I- Very large
Hydrobromic HBr Br- Very large
Hydrochloric HCl Cl- Very large
Nitric HNO3 NO3- Very large
Sulfuric H2SO4 HSO4- Very large
Hydronium ion H3O+ H2O 1.0
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Strength of oxyacids
• The more oxygen hooked to the central atom, the more acidic the hydrogen.
• HClO4 > HClO3 > HClO2 > HClO
• Remember that the H is attached to an oxygen atom.
• The oxygens are electronegative
• Pull electrons away from hydrogen
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Strength of oxyacidsStrength of oxyacids
Electron Density
Cl O H
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Strength of oxyacidsStrength of oxyacids
Electron Density
Cl O HO
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Strength of oxyacidsStrength of oxyacids
Cl O H
O
O
Electron Density
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Strength of oxyacids
Cl O H
O
O
O
Electron Density
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Dissociation of Weak AcidsDissociation of Weak AcidsWeak acids are assumed to dissociate only slightly (less than 5%) in solution.
Large Ka or small Ka?Reactant favored or product favored?
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Dissociation Constants: Weak Dissociation Constants: Weak AcidsAcids
Acid Formula Conjugate Base
Ka
Iodic HIO3 IO3- 1.7 x 10-1
Oxalic H2C2O4 HC2O4- 5.9 x 10-2
Sulfurous H2SO3 HSO3- 1.5 x 10-2
Phosphoric H3PO4 H2PO4- 7.5 x 10-3
Citric H3C6H5O7 H2C6H5O7- 7.1 x 10-4
Nitrous HNO2 NO2- 4.6 x 10-4
Hydrofluoric HF F- 3.5 x 10-4
Formic HCOOH HCOO- 1.8 x 10-4
Benzoic C6H5COOH C6H5COO- 6.5 x 10-5
Acetic CH3COOH CH3COO- 1.8 x 10-5
Carbonic H2CO3 HCO3- 4.3 x 10-7
Hypochlorous HClO ClO- 3.0 x 10-8
Hydrocyanic HCN CN- 4.9 x 10-10
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Self-Ionization of WaterSelf-Ionization of Water
H2O + H2O H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14 = [Ka][Kb]
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Calculating pH, Calculating pH, pOHpOHpH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and Relationship between pH and pOHpOH pH + pOH = 14
Finding [HFinding [H33OO++], [OH], [OH--] from pH, pOH] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH
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pH and pOH Calculations
H + O H -
pH pO H
[O H -] = 1 x 10 - 1 4
[H + ]
[H + ] = 1 x 10 - 1 4
[O H -]
p O H = 14 - p H
p H = 14 - p O H
pOH
= -l
og[O
H- ]
pH =
-log
[H+
]
[OH
- ] = 1
0-pO
H
[H+
] = 1
0-pH
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A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?
Step #1: Write the dissociation equation
HC2H3O2 (aq) + H2O(l) C2H3O2-(aq) + H+
(aq)
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A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?
Step #2: ICE it!
HC2H3O2 C2H3O2- + H+
II
CC
EE
0.50 0 0
- x +x +x
0.50 - x xx
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A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?
Step #3: Set up the law of mass action
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
)50.0()50.0(
))((108.1
25 x
x
xxx
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A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?
Step #4: Solve for x, which is also [H+]
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
)50.0(108.1
25 x
x [H+] = 3.0 x 10-3 M
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A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?
Step #5: Convert [H+] to pH
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
pH = - log (3.0 x 10-3) = 2.52
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Dissociation of Strong Dissociation of Strong BasesBases
Strong bases are metallic hydroxides Group I hydroxides (NaOH, KOH) are
very soluble Group II hydroxides (Ca, Ba, Mg, Sr)
are less soluble pH of strong bases is calculated
directly from the concentration of the base in solution
MOH(s) M+(aq) + OH-(aq)
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Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The base reacts with water, producing The base reacts with water, producing its conjugate acid and hydroxide ion:its conjugate acid and hydroxide ion:
CH3NH2 + H2O CH3NH3+ + OH- Kb = 4.38 x 10-
4
4 3 3
3 2
[ ][ ]4.38 10
[ ]b
CH NH OHK x
CH NH
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KKbb for Some Common Weak for Some Common Weak BasesBases
Base Formula Conjugate Acid
Kb
Ammonia NH3 NH4+ 1.8 x 10-5
Methylamine CH3NH2 CH3NH3+ 4.38 x 10-4
Ethylamine C2H5NH2 C2H5NH3+ 5.6 x 10-4
Diethylamine (C2H5)2NH (C2H5)2NH2+ 1.3 x 10-3
Triethylamine (C2H5)3N (C2H5)3NH+ 4.0 x 10-4
Hydroxylamine HONH2 HONH3+
1.1 x 10-8
Hydrazine H2NNH2 H2NNH3+
3.0 x 10-6
Aniline C6H5NH2 C6H5NH3+
3.8 x 10-10
Pyridine C5H5N C5H5NH+ 1.7 x 10-9
Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?
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Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The generic reaction for a base The generic reaction for a base reacting with water, producing its reacting with water, producing its conjugate acid and hydroxide ion:conjugate acid and hydroxide ion:
B + H2O BH+ + OH-
[ ][ ]
[ ]b
BH OHK
B
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A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O NH4+ + OH-
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A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #2: ICE it!
II
CC
EE
0.50 0 0
- x +x +x
0.50 - x xx
NH3 + H2O NH4+ + OH-
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A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
Step #3: Set up the law of mass action
0.50 - x xxEE
)50.0()50.0(
))((108.1
25 x
x
xxx
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
NH3 + H2O NH4+ + OH-
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A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
Step #4: Solve for x, which is also [OH-]
0.50 - x xxEE
)50.0(108.1
25 x
x [OH-] = 3.0 x 10-3 M
NH3 + H2O NH4+ + OH-
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
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A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
Step #5: Convert [OH-] to pH
0.50 - x xxEE
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
NH3 + H2O NH4+ + OH-
pOH = - log (3.0 x 10-3) = 2.52
pH = 14 - pOH = 11.48
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Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is from a strong base, anion from a strong acid
KCl, KNO3
NaCl NaNO3
Both ions are neutral
Neutral
These salts simply dissociate in water:
KCl(s) K+(aq) + Cl-(aq)
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Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is from a strong base, anion from a weak acid
NaC2H3O2
KCN, NaF
Cation is neutral, Anion is basic
Basic
C2H3O2- + H2O HC2H3O2 + OH-
base acid acid base
The basic anion can accept a proton from water:
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Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is the conjugate acid of a weak base, anion is from a strong acid
NH4Cl,
NH4NO3
Cation is acidic, Anion is neutral
Acidic
NH4+(aq) NH3(aq) + H+(aq)
Acid Conjugate Proton base
The acidic cation can act as a proton donor:
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Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid
NH4C2H3O2
NH4CN
Cation is acidic, Anion is basic
See below
IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral
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Acid-Base Properties of SaltsAcid-Base Properties of SaltsType of Salt Exampl
esComment pH of
solutionCation is a highly charged metal ion; Anion is from strong acid
Al(NO3)2
FeCl3
Hydrated cation acts as an acid; Anion is neutral
Acidic
Step #1: AlCl3(s) + 6H2O Al(H2O)6
3+(aq) + Cl-(aq) Salt water Complex ion anion
Step #2: Al(H2O)6
3+(aq) Al(OH)(H2O)52+(aq) + H+(aq)
Acid Conjugate base Proton
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Effect of Structure on Acid-Effect of Structure on Acid-Base Base
PropertiesProperties
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Hydrated metals
• Highly charged metal ions pull the electrons of surrounding water molecules toward them.
• Make it easier for H+ to come off.
Al+3 OH
H
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Acid-Base Properties of Oxides
• Non-metal oxides dissolved in water can make acids.
• SO3 (g) + H2O(l) H2SO4(aq)
• Ionic oxides dissolve in water to produce bases.
• CaO(s) + H2O(l) Ca(OH)2(aq)
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What is the concentration of H3O+ in pure water?
1. 1.00 x 10–7 M
2. 7.00 M3. 1.00 x 10-14 4. 1.00 x 107 M5. 7.00 x 10-14
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[H+] = 8.26 x 10–5 M, what is the pH of the solution?
1.2.162.4.1 3.4.084.4.0835.8.024
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Which of the following is a conjugate acid/base pair?1. HCl/OCl-
2. H2SO4/SO42-
3. NH4+/NH3
4. H3O+/OH-
5. none of these
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Which of the following indicates the most basic solution?1. [H+] = 1 × 10–10 M 2. pOH =6.7 3. [OH–] = 7 × 10–5 M 4. pH = 4.2 5. At least two of the
solutions are equally basic.
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Dissolving Na2SO4 in water will create a ______ solution.
1. Acidic2. Basic3. Neutral4. Cannot be
determined.
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Calculate the pH of a 0.17 M solution of HOCl, Ka = 3.5 10-8.
1.4.112.8.233.9.894.1.005.3.77