equilbrium of acids and bases
DESCRIPTION
EQUILBRIUM OF Acids and Bases. Chapter 17. Water. AUTOIONIZATION. H 2 O can function as both an ACID and a BASE. Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C. Water. K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C - PowerPoint PPT PresentationTRANSCRIPT
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EQUILBRIUM OF Acids and BasesEQUILBRIUM OF Acids and Bases
Chapter 17Chapter 17
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WaterWater
H2O can function as both an ACID
and a BASE.Equilibrium constant for water
= Kw
Kw = [H3O+] [OH-] =
1.00 x 10-14 at 25 oC
AUTOIONIZATION
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WaterWater
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution
[H3O+] = [OH-]
soso
[H3O+] = [OH-] = 1.00 x 10-7 M
OH-
H3O+
OH-
H3O+
44Calculating [HCalculating [H33OO++] & [OH] & [OH--]]given pOH = 3.00given pOH = 3.00
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
[OH-] = 10-pOH = 0.0010
[H3O+] = Kw / 0.0010
= 1.0 x 10-11 M
If [H3O+] < [OH-]
This solution is _________
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Equilibria Involving Equilibria Involving Weak Acids and Weak Acids and
BasesBases
Equilibria Involving Equilibria Involving Weak Acids and Weak Acids and
BasesBases
Aspirin is a good
example of a
weak acid,
Ka = 3.2 x 10-4
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Weak Acids and Weak Acids and BasesBases
Weak Acids and Weak Acids and BasesBases
Acid Conjugate Base
acetic, CH3CO2H CH3CO2-, acetate
ammonium, NH4+ NH3, ammonia
bicarbonate, HCO3- CO3
2-, carbonate
A weak acid (or base) is one that ionizes to a VERY small extent (<
5%).
99Weak Acids and Weak Acids and BasesBases
acetic acid, CH3CO2H (HOAc)
HOAc + H2O H3O+ + OAc-
Acid Conj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated Ka for ACID)
[H3O+] and [OAc-] are SMALL, Ka << 1.
1010
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
1111Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH
Step 1. ICE table.
[HOAc] [H3O+] [OAc-]
I
C
E
1212Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
[HOAc] [H3O+] [OAc-]
I 1.00 0 0
C -x +x +x
E 1.00-x x x
Note that we neglect [H3O+] from H2O.
1313Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 2. Write Ka expression
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
This is a quadratic. Use quadratic formula or method of approximations
(see Appendix A).
HOWEVER
1414Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
Assume x is very small because Ka is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00Now we can more easily solve this Now we can more easily solve this
approximate expression.approximate expression.
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Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3. Solve Ka approximate expression
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = -log (4.2 x 10-3) = 2.37
1616Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of formic acid, HCO2H.
HCO2H + H2O HCO2- + H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.4
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.5
© 2011 Pearson Education, Inc.
Chapter 17 – Acids and Bases
What is the [H+] of a 1.0 M acetic acid if
Ka = 1.8 x 10-5 for the solution:
A. 1.0 M
B. 0.3 M
C. 0.0042 M
D. 1.8 x 10-5 M
© 2011 Pearson Education, Inc.
Chapter 17 – Acids and Bases
A 0.100 M solution of HX has a pH of is 4.5 Determine the Ka for the acid, HX.
A. 1.0 x 10-8
B. 3.0 x 10-7
C. 2.5 x 10-8
D. 4.5 x 10-9
Weak BasesWeak Bases
2020
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
2121Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 1; ICE table
[NH3] [NH4+] [OH-]
I
C
E
2222
Weak BaseWeak BaseStep 1. ICE table
[NH3] [NH4+] [OH-]
I 0.010 0 0
C -x +x +x
E 0.010 - x x x
2323
Weak BaseWeak BaseStep 2. Solve the equilibrium expression
Kb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
[NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid !
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AcidsAcidsAcidsAcids
ConjugatConjugatee
BasesBases
ConjugatConjugatee
BasesBases
Kw = Ka* Kb
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Relation Relation
of Kof Kaa, K, Kbb, ,
[H[H33OO++] ]
and pHand pH
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