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EQUILBRIUM OF Acids and BasesEQUILBRIUM OF Acids and Bases
Chapter 17Chapter 17
22
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
33
WaterWater
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a neutral solution [H3O+] = [OH-]
so [H3O+] = [OH-] = 1.00 x 10-7 M
OH-
H3O+
OH-
H3O+
Autoionization
44
[H[H33OO++], [OH], [OH--] and pH] and pH
A common way to express acidity and basicity is with pH
pH = - log [HpH = - log [H33OO++]]
In a neutral solution, [H3O+] = [OH-] =
1.00 x 10-7 at 25 oC
pH = -log (1.00 x 10-7) = - (-7) = 7
55[H[H33OO++], [OH], [OH--] and pH] and pH
What is the pH of the 0.0010 M NaOH solution?
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
General conclusion —
Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7
66
[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [OH], [OH--] and pH] and pH
If the pH of Coke is 3.12, it is ________.
log [H3O+] = - pH
Take antilog and get
[H3O+] = 10-pH
[H3O+] = 10-3.12 =
7.6 x 10-4 M
77
ppXX Scales ScalesppXX Scales Scales
In general
pX = -log X
pOH = - log [OH-]
pH = - log [H+]
pKw = 14 = pH + pOH
88
Weak BaseWeak BaseStep 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
99
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
1010
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%).
1111Weak 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.
1212
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
1313
Calculations Calculations with with
Equilibrium Equilibrium ConstantsConstants
pH of an acetic
acid solution.
What are your
observations?
pH of an acetic
acid solution.
What are your
observations?
0.0001 M
0.003 M
0.06 M
2.0 M
a pH meter, Screen 17.9a pH meter, Screen 17.9
1414Equilibria 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
1515Equilibria 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.
1616Equilibria 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
1717Equilibria 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.
1818
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
1919Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
For many weak acids
[H3O+] = [conj. base] = [Ka •
Co]1/2
where C0 = initial conc. of acid
Useful Rule of Thumb:
If 100•Ka < Co,
then [H3O+] = [Ka•Co]1/2
2020Equilibria 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
Weak BasesWeak Bases
2222
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
2323Equilibria 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
2424
Weak BaseWeak BaseStep 1. ICE table
[NH3] [NH4+] [OH-]
I 0.010 0 0
C -x +x +x
E 0.010 - x x x
2525
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 !
2626
AcidsAcidsAcidsAcids
ConjugatConjugatee
BasesBases
ConjugatConjugatee
BasesBases
2727
Relation Relation
of Kof Kaa, K, Kbb, ,
[H[H33OO++] ]
and pHand pH
2828
