Chapter 16 - Acids and...
Transcript of Chapter 16 - Acids and...
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16.1 Acids and Bases: The Brønsted–Lowry Model 16.2 pH and the Autoionization of Water 16.3 Calculations Involving pH, Ka and Kb 16.4 Polyprotic Acids
Chapter 16 - Acids and Bases
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16.1
Acids and Bases: Basic Definitions
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Sour Taste React with “active” metals (Al, Zn, Fe) to yield H2 gas:
Corrosive React with carbonates to produce CO2:
Change color of vegetable dyes Turn blue litmus red
React with bases to form salts
Properties of Acids
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Properties of Bases
Bitter Taste
Also Known as Alkalies
Solutions feel slippery
Change color of vegetable dyes
Turn red litmus blue
React with acids to form salts
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Acids and Bases in Solution
Acids ionize in water to form H+ ions. (More precisely, the H+ from the acid molecule is donated to a
water molecule to form hydronium ion, H3O+)
Bases dissociate in water to form OH- ions. (Bases, such as NH3, that do not contain OH- ions, produce OH-
by pulling H+ off water molecules.)
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AcidsAcids are molecular compounds that ionize when they dissolve in
water. The molecules are pulled apart by water.
The percentage of molecules that ionize varies.
Acids that ionize virtually 100% are called strong acids.
HCl(aq) ➜ H+(aq) + Cl−(aq)
Acids that only ionize a small percentage are called weak acids.
HF(aq) ⇔ H+(aq) + F−(aq)
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Strong Acid
Weak Acid
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Molecular Models of Selected Acids
Binary Acid
“Oxy" Acids
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Common AcidsName Formula Uses Strength
Perchloric HClO4 explosives, catalysts Strong
Nitric HNO3 explosives, fertilizers, dyes, glues Strong
Sulfuric H2SO4 explosives, fertilizers, dyes, glue, batteries Strong
Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid Strong
Phosphoric H3PO4 fertilizers, plastics, food preservation Moderate
Chloric HClO3 explosives Moderate
Acetic HC2H3O2 plastics, food preservation, vinegar Weak
Hydrofluoric HF metal cleaning, glass etching Weak
Carbonic H2CO3 soda water, blood buffer Weak
Hypochlorous HClO sanitizer Weak
Boric H3BO3 eye wash Weak
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Strong Acids and Their Ionization in Water
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Common Weak Acids and Their Ionization in Water
4.74
3.75
3.17
7.54
3.15
pKa = -log KaWhy do we not define Ka for a strong acid ??
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Structure of Bases
Most contain OH- ions
Some contain CO32- ions
“Molecular” bases contain structures which react with H+
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Name Formula Common Name Uses Strength
Sodium Hydroxide NaOH Lye, Caustic Soda soap, plastic production, petroleum refining
Strong
Potassium Hydroxide KOH Caustic Potash soap, cotton processing,
electroplatingStrong
Calcium Hydroxide
Ca(OH)2 Slaked Lime cement Strong
Sodium Bicarbonate
NaHCO3 Baking Soda food preparation, antacids Weak
Magnesium Hydroxide Mg(OH)2 Milk of Magnesia antacids Weak
Ammonium Hydroxide
NH4OH Ammonia Water fertilizers, detergents, explosives Weak
Common Inorganic Bases
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Strong Bases and Their Ionization in Water
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Weak Bases and Their Ionization in Water
pKb = -log Kb
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acetic acid
Arrhenius Theory
Bases dissociate in water to produce OH- and cations:
Acids ionize in water to produce H+ and anions:
acetate anion
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Arrhenius Theory
An Arrhenius Acid-Base Reaction
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Problems with Arrhenius Theory
Does not explain why some molecular substance dissolve in water to give basic solutions, but do
not contain OH-. (NH3)
Does not explain why some ionic substances dissolve in water to give basic solutions, but do
not contain OH-. (NaCO3)
Does not explain why some molecular substance dissolve in water to give acidic solutions, but do
not contain H+. (CO2)
Only applicable to aqueous solutions.
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Brønsted-Lowry Theory
Acid are proton donors
Bases are proton acceptors
Acid-Base Reactions involve proton transfer
Base structure includes a pair of unshared electrons
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Brønsted-Lowry Theory
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In a Brønsted-Lowry reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a
base in the reverse process.
Each reactant and product are a conjugate pair.
The original base becomes a conjugate acid, and the original acid becomes a conjugate base.
Conjugate Pairs
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H-A + :B ⇔ :A- + H-B+
acid base conjugate conjugate base acid
HCHO2 + H2O ⇔ CHO2- + H3O+
acid base conjugate conjugate base acid
H2O + NH3: ⇔ HO- + NH4+
acid base conjugate conjugate base acid
Conjugate Pairs
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Conjugate Pairs
H2O and OH- are an acid/base conjugate
pair.
NH3 and NH4+ are an base/acid conjugate
pair.
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Lewis Acid-Base Theory
A Lewis base is an electron donor or nucleophile.
A Lewis acid is an electron acceptor or electrophile.
When a nucleophile donates a pair of electrons to an electrophile, a covalent bond forms.
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16.2
pH and the Autoionization of Water
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Autoionization of WaterWater is an extremely weak electrolyte.
One out of every 10 million water molecules forms ions.
All aqueous solutions, therefore, contain some H+ and OH-.
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Autoionization of Water
The Ion Product of Water is Defined:
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at 25ºC in the absence of any other acids or
bases.
A change in [H3O+] causes and inverse change in [OH-].
Autoionization of Water
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Basic SolutionNeutral SolutionAcidic Solution
Relationship between [H3O+] and [OH-]
[H3O+] = [OH-][H3O+] > [OH-] [H3O+] < [OH-]
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H+ H+ H+ H+ H+
OH-OH-OH-OH-OH-
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
[OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100
Relationship between [H+] & [OH-]
Kw =[H3O+][OH-]=1.0 x 10-14
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pH as a Measure of Acidity/Basicity
pH = -log [H3O+]
pH < 7 , acidic
pH > 7, basic
pH = 7, neutral
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pH of Common Substances
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H+ H+ H+ H+ H+
OH-OH-OH-OH-OH-
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
[OH-] 10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100
pH vs pOH
pH + pOH =14
pOH 14 13 11 9 7 5 3 1 0
pH 0 1 3 5 7 9 11 13 14
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1) What are the [OH–] and pH of household ammonia, an aqueous solution that has an [H3O+] of 1.99 x 10-12 M?
Kw =[H3O+][OH-]=1.0 x 10-14
(1.99 x10-12)[OH-]=1.0 x 10-14
[OH-]=(1.0 x 10-14)/(1.99 x10-12)
[OH-]=5.02 x 10-3
pH = -log[H3O+] =11.70
pOH = -log[OH-] = 2.30
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pKa = -logKa
pKa = -logKa
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16.3
Calculations Involving pH, Ka and Kb
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2) Calculate and compare [H+] and pH for a 0.100 M solution of HClO4 and a 0.100 M solution of HClO (Ka = 2.9 × 10–8).
HClO4 → ClO- + H+
HClO ⇄ ClO- + H+
For a strong acid, [H+] = 0.100 M, pH = -log(0.10) = 1
For the weak acid, a little more work is needed.
x2 0.100-x = 2.9 x 10-8
[H+] = x = 5.4 x 10-5
pH = 4.3
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3) The pH of a 0.050 M solution of a weak organic acid is 2.23 . Calculate [H+], Ka and % ionization for the acid.
[H+] = 10-pH = 10-2.23 = 5.89 x 10-3
[A-][H+] [HA]Ka = = [5.89 x 10-3]2
[0.044]= 7.9 x 10-4
% ionization = (5.9 x 10-3/0.050) x 100 = 12%
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4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 × 10–4).
[A-][H+] [HA]Ka = = [x][x]
[1.28-x] = 4.0 × 10–4
[x]2 [1.28] = 4.0 × 10–4
x = 2.3 × 10–2
% ionization = (2.3 x 10-2/1.28) x 100 = 1.8%
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4) Calculate the percent ionization for 1.28 M HNO2 and 0.0150 M HNO2 solutions (Ka = 4.0 × 10–4).
[A-][H+] [HA]Ka = = [x][x]
[0.0150-x] = 4.0 × 10–4
[x]2 [0.0150] = 4.0 × 10–4
x = 2.3 × 10–3
% ionization = (2.4 x 10-3/0.0150) x 100 = 16%
(2.6 x 10–3)
(17%)
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5) Calculate and compare [OH-] and pH for 0.200 M LiOH and a 0.200 M solution of methylamine, CH3-NH2 (Kb = 4.4 × 10–4).
LiOH → Li+ + OH-
CH3-NH2 + H2O ⇄ CH3-[NH3]+ + OH-
For a strong base, [OH-] = 0.200 M, pOH = -log(0.20) = 0.7, pH=13.3
For the weak base, once again, a little more work is needed.
[BH+][OH-] [B]Kb = =
[x][x] [0.200-x] = 4.4 × 10–4
[x][x] [0.200] = 4.4 × 10–4
x = 9.4 × 10–3 = [OH-]
pOH = -log(9.4 × 10–3) = 2.03, pH = 11.97
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6) Calculate [H+] and pH for 1.0 × 10–8 M HCl. HCl → H+ + Cl-
For a strong acid, [H+] = 1.0 × 10–8 M, pH = 8.0, BUT
THIS DOES NOT MAKE SENSE !!!
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H2O + H2O ⇄ H3O+ + OH-
THERE IS ANOTHER EQUILIBRIUM TO CONSIDER
[H3O+][OH-]KW = = 1.0 x 10-14
(x + 1.00 x 10-8) (x) = 1.0 x 10-14
x2 + (1.00 x 10-8)(x)-1.0 x 10-14= 0solve quadratic x = 9.5 x 10-8 = [OH-]
[H+] = 9.5 x 10-8 + 1.0 x 10-8 = 10.5 x 10-8 = 1.05 x 10-7
IF “x” REPRESENTS [H+] AND [OH-] FROM AUTOIONIZATION
pH = 6.98
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16.4 Polyprotic Acids
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Polyprotic Acids
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7) Calculate the pH of 0.034 M carbonic acid solution (Ka1 = 4.3 × 10–7 and Ka2 = 4.7 × 10–11).
[HCO3-][H+] [HA]Ka1 =
[x][x] [0.034]4.3 x 10-7 =
1.2 x 10-4 = x = [H+]
1.2 x 10-4 = x = [HCO3-]
0.034 = [H2CO3]
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7) Calculate the pH of 0.034 M carbonic acid solution (Ka1 = 4.3 × 10–7 and Ka2 = 4.7 × 10–11).
[CO32-][H+] [HCO3-]Ka2 =
[x][x +1.2 x 10-4] [1.2 x 10-4]4.7 x 10-11 =
4.7 x 10-11 = x = [CO32-]
1.2 x 10-4 = [HCO3-]
✔
pH = -log [H+] = -log (1.2 x 10-4 + 4.7 x 10-11) = 3.92
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Polyprotic Acids
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Polyprotic Acids