Acids, Bases and Buffers

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Chapter 17

• Acids, Bases and Buffers

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Overview

• strong acid strong base
• strong acid weak base
• weak acid strong base
• weak acid weak base
• common ion effect
• buffers Henderson-Hasselbalch Equation
• titration curves

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• strong acid strong base at the equivalence
  point (OH - H) the pH 7
• strong acid weak base at the equivalence
  point, pH lt 7
• weak acid strong base at the equivalence
  point, pH gt 7
• weak acid weak base at the equivalence point,
  pH depends on the relative value of the Ka and
  the Kb of the acid and base

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example weak acid strong base solution
H2O HCN H3O CN -
Ka H3O OH -
2H2O 1/Kw
HCN OH - H2O CN -
Ka /Kw
WA
SB
Ka /Kw is large so rxn. will be, essentially,
complete
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example strong acid weak base solution
H2O CN - HCN OH -
Kb H3O OH -
2H2O 1/Kw
CN - H3O H2O HCN
Kb /Kw
SA
WB
Kb /Kw is large so rxn. will be, essentially,
complete
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example weak acid weak base solution
H2O HCN H3O CN -
Ka H2O NH3 OH
- NH4 Kb
pH depends upon relative magnitude of Ka and Kb
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Common Ion Effect
HCN H2O H3O CN -
Initial addition of CN - (as NaCN) shifts
equilibrium, decreasing H3O thereby increasing
the pH
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How does this affect the pH quantitatively?
HCN H2O H3O CN -
0.20 M 0 0.10 M initail
-x x x change
0.20 - x x 0.10 x equil.
4.9 x 10 -10 (x)(0.10 x) 0.20 - x
x 9.8 x 10 -10 M H3O pH 9.0 (vs.
9.9 x 10 - 6 M pH 5.0 with no added NaCN)
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Buffers

• Resist change in pH on addition of small
  amounts of strong acid or base
• Composed of A weak acid and the salt of its
  conj. base or A weak base and the salt of its
  conj. acid
• Most effective when pH is 1 of the pKa

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HX H X-
H Ka HX X-

• adding OH- causes inc. in X- dec. in HX OH-
  HX H2O X-
• adding H causes dec. in X- inc. in HX H
  X- HX
• as long as amt. of OH- or H is small compared
  to HX X-, ratio changes little

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H Ka HX X-
-log H - log Ka -log HX
X-
pH pKa - log HX X-
pH pKa log X- HX
Henderson-Hasselbalch Equation
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Henderson-Hasselbalch Equation

• Calculates pH of buffer solutions
• Can be used only for buffer solutions
• Can be used only when the equilibrium
  approximation can be used
• pH pKa log (X - / HX) where HX is
  the weak acid and X - is its conjugate base

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Titration Curves

• strong acid strong base

Calculation of pH after addition of aliquots
of base HCl NaOH H2O
NaCl pH 50 mL 0.10 0 mL 0.10
M 1.0 10 mL 1.2
20 mL 1.4 49 mL
3.0 50 mL 7.0 55
mL 11.7 80 mL 12.4 100
mL 12.5 Note large increase in pH near
equivalence pt.
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50 mL 0.10 M HCl titrated with 0.10 M NaOH
pH
Equivalence Pt.
Volume of Base, mL
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Titration Curves

• Weak acid strong base

Calculation of pH after addition of aliquots
of base HC2H3O2 NaOH H2O
NaC2H3O2 pH 50 mL 0.10 0 mL 0.10
M 2.9 10 mL 4.1
20 mL 4.6 25 mL
half equivalence point pH pKa 4.7
49 mL 6.4 50 mL equivalence
point 8.7 55 mL 11.7 80
mL 12.4 100 mL 12.5 Note
large increase in pH near equivalence pt.
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50 mL 0.10 M HC2H3O2 titrated with 0.10 M NaOH
12108 6 4 2
Equivalence Pt.
pH
half equivalence point pKa
50
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Volume of Strong Base, mL
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Strong Base Titrated with Strong Acid
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50 mL 0.10 M NH3 titrated with 0.10 M HCl
12108 6 4 2
half equivalence point pKb
pH
Equivalence Pt.
50
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Volume of Strong Acid, mL
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Diprotic acid, H2C2O4, titrated with 0.10 M NaOH
12108 6 4 2
second equivalence pt.
pH
first equivalence point pKa
200
100
Volume of Base, mL
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Indicators
Indicators are generally weak acids
HInd H2O H3O Ind -
Ka Ind H3OInd - HInd
H3O HInd this ratio controls color
Ka Ind -
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Solubility Equilibria

• Solubility and Solubility Product
• Precipitation of Insoluble Salts Ksp and Q
• Common Ion Effect and Solubility
• Simultaneous Equilibria
• Solubility and pH
• Solubility and Complex Ions
• Separations and Qualitative Analysis

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Solubility Product Constant
The equilibrium constant expression for the
solution of a solid
AgCl(s) Ag(aq) Cl -(aq)

Ksp AgCl -
Solubility Product
All rules for equilibrium constants and
expressions apply
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Examples
CaF2(s) Ca2(aq) 2F -(aq)
Ksp Ca2F - 2
Ag3PO4(s) 3Ag(aq) PO43- (aq)

Ksp Ag 3PO43-
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The quantity of CaF2(s) dissolved is reflected by
the quantity of Ca2(aq) ions in solution
Ca2
F -
The Molar Solubility of CaF2(s) is equal to the
Ca2 at the eq. point a saturated solution
F -
F -
Ca2
F -
solid CaF2(s)
dissolved Ca2(aq) F- (aq) ions
CaF2(s) Ca2(aq) 2F -(aq)
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Calculate the Ksp from experimental data
Prob Ba2 7.5 x 10 -3 M in saturated BaF2.
Calculate Ksp.
BaF2(s) Ba2(aq)
2F -(aq) x 2x 7.5 x
10 -3 2(7.5 x 10 -3)
Ksp (7.5 x 10 -3)(15.0 x 10 -3)2 1.6 x
10 -6
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Estimating Solubility
Much like doing equilibrium problems
Determine the molar solubility of CaCO3 if the
Ksp is 3.8 x 10 -9 at 25 C
CaCO3(s) Ca2(aq)
CO32-(aq) initial 0 0change
x xequil. x x
x2 3.8 x 10 -9 x 6.2 x 10 -5 M Ca2
CO32-
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Prob Calculate the molar solubility of Mg(OH)2
if the Ksp is 1.5 x 10 -11
Mg(OH)2(s) Mg2(aq) 2OH
-(aq) initial 0 0change x
2xequil. x 2x
Ksp Mg2OH -2 (x) (2x)
(x)(2x)2 4x3 1.5 x 10 -11 x 1.6 x 10
-4 M Mg2 1.6 x 10 -4 M OH - 3.2 x
10 -4 M
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Which is more soluble in water?
AgCl or Ag2CrO4
AgCl(s) Ag(aq) Cl -(aq)
Ksp 1.8 x 10 -10
molar solubility 1.3 x 10 -5 M
Ag2CrO4(s) 2Ag(aq) CrO4 2- (aq)
Ksp 9.0 x 10 -12
molar solubility 1.3 x 10 -4 M
Direct comparisons of Ksps can only be used if
the ion ratios are the same
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Which is more soluble in water?
AgCl Ksp 1.8 x 10 -10 orAgCN Ksp 1.2 x
10 -16 Mg(OH)2 Ksp 1.5 x 10 -11
orCa(OH)2 Ksp 7.9 x 10 -9
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Q, Reaction Quotient

• Q Ksp system is at equilibrium
• Q gt Ksp system not at equilibrium solid
  forms (rxn. shifts )
• Q lt Ksp system not at equilibrium solid
  dissolves (rxn. shifts )

Prob PbI2(s) (Ksp 8.7 x 10 -9) placed in
solution where Pb2 1.1 x 10 -3 M. Is the
solution saturated?
PbI2(s) Pb2(aq) 2I -(aq) Q
5.3 x 10 -9 lt Ksp
No, more will dissolve
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Concentrations required for precipitation
Prob What is the minimum conc. of I - that can
cause precipitation of PbI2 from a 0.050 M
solution of Pb(NO3)2? Ksp (PbI2) 8.7 x 10 -9.
Ksp Pb2I -2 8.7 x 10 -9 I -2
0.050 I - 4.2 x 10 -4 M
How much Pb2 remains when I - 0.0015 M
Pb2 8.7 x 10 -9 3.9 x 10 -3 M
(0.0015)2
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Ksp and Precipitation
Prob You have 100.0 mL of 0.0010 M AgNO3. Does
AgCl precipitate if you add 5.0 mL of 0.025 HCl?
AgNO3(aq) Ag(aq)
NO3-(aq) HCl(aq) H2O
H3O(aq) Cl -(aq)
AgCl(s) Ag(aq)
Cl -(aq)
1.0 x 10 -4 mol 1.25 x 10 -4 mol 0.105 L
0.105 L 9.5 x 10 -4 M 1.2 x 10 -3 M
Q (9.5 x 10 -4)(1.2 x 10 -3)Q 1.1 x 10 -6 gt
Ksp will precipitate
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Solubility and Common Ion Effect
CaF2(s) Ca2(aq) 2F
-(aq)
Adding extra Ca2 or F - shifts
equilibrium
shift toward solid CaF2
causing a decrease in solubility of CaF2(s)
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Molar Solubility of CaF2 (no added F -)
CaF2(s) Ca2(aq) 2F
-(aq)
initial 0 0change
x 2x equil. x 2x
x 2.1 x 10 -4 M Ca2 Ksp 3.9 x 10 -11
(x)(2x)2
molar solubility is 2.1 x 10 -4 moles CaF2 / L
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Molar Solubility of CaF2 (with 0.010 M NaF)
CaF2(s) Ca2(aq) 2F
-(aq)
initial 0 0.010change
x 2x equil. x 0.010
2x
Ksp 3.9 x 10 -11 (x)(0.010 2x)2 x 3.9 x
10 -7 M Ca2
molar solubility is 3.9 x 10 -7 moles CaF2 / L
with added NaF which suppresses solubility
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What happens if acid is added to CaF2?
CaF2(s) Ca2(aq) 2F
-(aq)
H addition shifts rxn, increasing solubility
H3O F - HF
H2O SA WB
complete reaction
The second rxn. has the effect of removing F -
from the first equilibrium, affecting the
solubility of CaF2
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Which of the following would be more soluble in
acid solution?
PbCl2 CaCO3 Mg(OH)2
the stronger the conj. base the more soluble
if the anion is a hydrolyzing conjugate base, the
stronger base it is, the more soluble the salt
is in acid solution
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Cl - H2O HCl OH -
K very small
CO32- H2O HCO3- OH - Kb
2.1 x 10 -4
OH - H2O H2O OH - K
very large
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For example
CaCO3(s) Ca2(aq) CO32-(aq)
Ksp 3.8 x 10 -9 CO32-(aq) H2O
HCO3- OH - Kb 2.1 x 10 -4 OH -
H3O 2H2O Kw-1 1.0 x 10 14
CaCO3(s) H3O Ca2 HCO3-
H2O
K 79.8
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Which of the following would be more soluble in
acid solution?
hydroxy apatite - tooth enamel
Ca10(PO4)6(OH)2
Ca10(PO4)6F2
fluoro apatite - fluoridated tooth enamel
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Formation of Complex Ions

• Complex ions are large, polyatomic ions
• Most complex ions have large K values
• Formation of complex ions can affect solubility
  of some salts
• Kf is the formation constant
• Most transition metals form stable complex ions-
  the transition metal is a Lewis Acid

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For example Ex. 17.14 Calc. Ag present at
eq. when conc. NH3 added to 0.010 M AgNO3 to give
eq. NH3 0.20 M. Neglect vol. change.
Ag 2NH3(aq) Ag(NH3)2 Kf
1.7 x 10 7 x M 0.20 M (0.010 - x) M
K 1.7 x 10 7 Ag(NH3)2 0.010
AgNH32 (x)(0.20)2
x 1.5 x 10 -8 M Ag
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Selective Precipitation
Prob 0.050 M Mg2 0.020 M Cu2. Which will
ppt first as OH - is added?
Mg(OH)2 Mg2 2OH- Ksp 1.8 x 10
-11
Cu(OH)2 Cu2 2OH- Ksp 2.2 x
10 -20
Cu(OH)2
OH- (Ksp / M2)1/2
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What concentration of OH- is necessary?
OH- (Ksp / Cu2)1/2 (2.2 x
10 -20 / 0.020)1/2 1.0 x 10 -9 M
takes less OH-
OH- (Ksp / Mg2)1/2 (1.8 x
10 -11 / 0.050)1/2 1.9 x 10 -5 M
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For example
AgCl(s) Ag(aq) Cl -(aq)
Ksp 1.8 x 10 -10 Ag 2NH3(aq)
Ag(NH3)2 Kf 1.7 x 10 7
formation
AgCl(s) 2NH3(aq) Ag(NH3)2 Cl
- K 3.1 x 10 -3 Ag(NH3)2Cl
- NH32
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Prob Does 100 mL of 4.0 M aqueous ammonia
completely dissolve 0.010 mol of AgCl suspended
in 1.0 L of solution?
AgCl(s) 2NH3(aq) Ag(NH3)2
Cl -
0.010 moles
0.020 moles
0.010 moles
0.010 moles
K 3.1 x 10 -3 Ag(NH3)2 Cl -
NH32
NH32 (1.0 x 10 -2)(1.0 x 10 -2) (0.032)1/
2 0.18 M (3.1 x 10 -3) we have 0.4
moles of NH3 available which is plenty to provide
the 0.020 mol necessary to form the complex plus
0.16 mol to achieve an equilibrium concentration
of 0.18 M NH3