sch 4u. ionize only partially in water, exist primarily in molecule form dynamic equilibrium...
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WEAK ACIDS & BASES
SCH 4U
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WEAK ACIDS Ionize only partially in water,
exist primarily in molecule form
Dynamic equilibrium established between unreacted molecules and ions formed from rxn with water
Like all equilibria, can be shifted by removal/ addition of reactants or products
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WEAK BASES Have a weak attraction for protons
Recall: the conjugate base of a strong acid is a weak base
HA + H2O A– + H3O+
Usually non-hydroxide bases(Recall Arrhenius vs. Bronsted-Lowry
definitions)
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PERCENT IONIZATIONFor weak acids:
p = [H3O+] x 100%
[HA]
For weak bases:p = [OH-] x 100%
[B]
Ex. The pH of a 0.10 mol/L methanoic acid solution is 2.38. What is the percent ionization of methanoic acid?
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PERCENT IONIZATION EXAMPLEEx. The pH of a 0.10 mol/L methanoic acid
solution is 2.38. What is the percent ionization of methanoic acid?
HCO2H(aq) H+(aq) + HCO2
-(aq)
[H+(aq)] = 10-pH
= 10-2.38
= 4.2 x 10-3 mol/L
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PERCENT IONIZATION EXAMPLE
p = conc. of acid ionized x 100% conc. of acid solute
p = 4.2 x 10-3 mol/L x 100% 0.10 mol/L
p = 4.2 %
Therefore methanoic acid ionizes 4.2% in a 0.10 mol/L solution
i.e. HCO2H(aq) H+(aq) + HCO2
–(aq)
4.2 %
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IONIZATION CONSTANTS (Ka) FOR WEAK ACIDS
Equilibrium constant found as before, called “acid ionization constant,” Ka
E.g. for acetic acid:HC2H3O2(aq) H+
(aq) + C2H3O2-(aq)
Ka = [H+][C2H3O2-]
[HC2H3O2]
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CALCULATING Ka FROM PERCENT IONIZATION Calculate the acid ionization constant, Ka,
of acetic acid if a 0.1000 M solution at equilibrium at SATP has a percent ionization of 1.3%
HC2H3O2(aq) H+(aq) + C2H3O2
-(aq)
Ka = [H+][C2H3O2-]
[HC2H3O2]
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A percent ionization of 1.3% means initial [HC2H3O2] is diminished by 1.3% by time system reaches equilibrium
Note in this example molar ratio is 1:1:1
Use an ICE table:
Solve for x:x = 0.1 mol/L X 0.013x = 0.0013 mol/L
HC2H3O2(aq)
H+(aq)
+ C2H3O2
-
(aq)
Initial 0.1000 0 0
Change - x + x + x
Equilibrium 0.1000 - x 0 + x 0 + x
HC2H3O2(aq)
H+(aq)
+ C2H3O2
-
(aq)
Initial 0.1000 0 0
Change - 0.0013 + 0.0013 + 0.0013
Equilibrium 0.0987 0.0013 0.0013
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Now we can calculate Ka:
Ka = [H+][C2H3O2-]
[HC2H3O2]
= (0.0013) (0.0013) (0.0987)
Ka = 1.7 x 10-5
HC2H3O2(aq)
H+(aq)
+ C2H3O2
-
(aq)
Initial 0.1000 0 0
Change - 0.0013 + 0.0013 + 0.0013
Equilibrium 0.0987 0.0013 0.0013
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PERCENT IONIZATION & CONCENTRATION Ka values provide a means of comparing
relative strengths of acids
Can also compare % ionization values, but only when acids are equal in initial conc.
Data for acetic acid
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PERCENT IONIZATION & CONCENTRATION More dilute the solution, greater the
degree of ionization
Can explain using Le Chatelier’s Principle:
HA(aq) A–(aq) + H+
(aq)
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IONIZATION CONSTANTS (Kb) FOR WEAK BASES
Equilibrium constant called “base ionization constant,” Kb
E.g. for ammonia:
NH3(aq) + H2O(l) OH-(aq) + NH4
+(aq)
Kb = [OH-][NH4+]
[NH3]
Note: many weak bases contain one or more N atom, others are conjugate bases of strong acids
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RELATIONSHIP BETWEEN Ka AND Kb
Consider an acetic acid solution at equilibrium:
HC2H3O2(aq) + H2O(l) H3O+(aq) +
C2H3O2-(aq)
As a base, the acetate ion also reacts with water, establishing an equilibrium:
C2H3O2-(aq) + H2O(l) OH-
(aq) + HC2H3O2(aq)
Ka = [H3O+][C2H3O2
-] [HC2H3O2]
Kb = [OH-][HC2H3O2] [C2H3O2
-]
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RELATIONSHIP BETWEEN Ka AND Kb
Ka x Kb =
Kw = Ka x Kb
[H3O+] [C2H3O2-] x [OH-]
[HC2H3O2] [HC2H3O2] [C2H3O2
-]
= [H3O+] [OH-]
Ka = Kw
Kb
Kb = Kw
Ka
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STRENGTH GENERALIZATIONS The conjugate base of a strong acid is a
______________ base
The conjugate base of a weak acid is a ______________ base
The conjugate base of a very weak acid is a ______________ base
* Go over page 562 together
very weak
weak
strong
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STRENGTH GENERALIZATIONS The conjugate acid of a strong base is a
______________ acid
The conjugate acid of a weak base is a ______________ acid
The conjugate acid of a very weak base is a ______________ acid
very weak
weak
strong
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CALCULATING CONC & pH OF A WEAK ACID GIVEN Ka
Calculate the hydrogen ion conc. and pH of a 0.10 mol/L acetic acid solution. Ka for acetic acid is 1.8 x 10-5.
*First need to compare Ka’s of all equilibria that may contribute H+ to the system...
Ka = 1.8 x 10-5 Kw = 1.0 x 10-14
HC2H3O2(aq) H+(aq) + C2H3O2
-
(aq)
I 0.10 0 0C - x +x +xE 0.10 – x x x
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Ka = [H+][C2H3O2-]
[HC2H3O2]
1.8 x 10-5 = x2
0.10 –x
Could solve quadratic or could make it simpler...
FINISH... on p.563!
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FINDING Kb, GIVEN CONC & pHFor diethylamine (CH3CH2)2NH, the pH of a 2.6 x 10-2 M solution is 11.56. What is Kb for diethylamine?
B(aq) + H2O(l) OH-(aq) + HB+
(aq)
Can use pH to det. [OH]...
pH = 11.56 therefore pOH = 14.00 – 11.56 = 2.44
pOH = -log[OH-][OH-] = 10-2.44
[OH-] = 3.6 x 10-3 M
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B(aq) + H2O(l) OH-(aq) + HB+
(aq)
I 2.6 x 10-2 0 0C -x +x +xE 2.6 x 10-2 – x x x
x = 3.6 x 10-3
Kb = [OH-][HB+]
[B]
Kb = (3.6 x 10-3)2
2.6 x 10-2 – 3.6 x 10-3
= 5.8 x 10-4 See summary of problem-solving steps p.574
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POLYPROTIC ACIDS E.g. Sulfuric acid H2SO4, boric acid H3BO3
Different Ka values (Ka1, Ka2, etc.)
In general,
Ka1 > Ka2 > Ka3 ...
See p. 574, 575
* Because Ka1 is usually >> Ka2, Ka3, etc., typically use just Ka1 to determine pH
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HOMEWORK
p. 579 # 3-6, 11-13