Transcript Applications of Aqueous Equilibria

Chapter 15
Applications of Aqueous Equilibria
Addition of base:
Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2
Buffer Solutions
•
•
When an acid or
base is added to
water, the pH
changes drastically.
A buffer solution
resists a change in
pH when an acid or
base is added.
Buffers:
•Mixture of a weak acid and its salt
•Mixture of a weak base and its salt
pH = 4
pH = 10.5
pH = 6.9
pH = 7.1
Understanding Buffer Solutions
The presence of a common ion suppresses
the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
CH3COOH (aq)
Na+ (aq) + CH3COO- (aq)
H+ (aq) + CH3COO- (aq)
common
ion
The common ion effect is the shift in equilibrium caused by the
addition of a compound having an ion in common with the
dissolved substance.
pH of a Buffer:
Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
[H+][A-]
Ka =
+
[HA]
HA (aq)
H (aq) + A (aq)
[H+]
Ka [HA]
=
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[A-]
pH = pKa + log
[HA]
Henderson-Hasselbalch
equation
[conjugate base]
pH = pKa + log
[acid]
pKa = -log Ka
Using Henderson-Hasselbalch
equation
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).
HCOOH : Ka= 1.67 X 10-4 , pKa = 3.78
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.02
pH = 3.77 + log
[0.30]
A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon
the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COOH (aq)
CH3COO- (aq) + H2O (l)
Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution
Blood Plasma pH = 7.4
Buffer Solutions
02
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
(Ka= 5.62 X 10-10 ).
NH4+ (aq)
[NH3]
pH = pKa + log
[NH4+]
H+ (aq) + NH3 (aq)
pKa = 9.25
[0.30]
pH = 9.25 + log
= 9.17
[0.36]
NH4+ (aq) + OH- (aq)
Initial (moles)
Change
0.36 X0.0 800 =0.029
-0.001
End (moles)
0.029-0.001 = 0.028
H2O (l) + NH3 (aq)
0.050 x 0.0200 = 0.001
-0.001
0.001-0.001 =0.0
0.30 X .0800 = 0.024
+0.001
0.024 + 0.001 = 0.025
final volume = 80.0 mL + 20.0 mL = 100 mL , 0.1 L
[NH4
+]
0.028
0.025
=
[NH3] =
0.10
0.10
[0.25]
pH = 9.25 + log
= 9.20
[0.28]
Titrations
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Acid–Base Titration Curves
01
Strong Acid-Strong Base Titrations
Equivalence Point: The point at which stoichiometrically
equivalent quantities of acid and base have been mixed
together.
Strong Acid-Strong Base Titrations
1. Before Addition of Any NaOH
0.100 M HCl
Strong Acid-Strong Base Titrations
2. Before the Equivalence Point
H3
O1+(aq)
+
OH1-(aq)
Excess H3O1+
100%
2H2O(l)
Strong Acid-Strong Base Titrations
3. At the Equivalence Point
The quantity of H3O1+ is
equal to the quantity of
added OH1-.
pH = 7
Strong Acid-Strong Base Titrations
4. Beyond the Equivalence Point
Excess OH1-
Strong Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
1. Before Addition of Any NaOH
CH3CO2H(aq) + H2O(l)
0.100 M CH3CO2H
H3O1+(aq) + CH3CO21-(aq)
Weak Acid-Strong Base Titrations
2. Before the Equivalence Point
CH3CO2H(aq) +
OH1-(aq)
Excess CH3CO2H
Buffer region
100%
H2O(l) + CH3CO21-(aq)
Weak Acid-Strong Base Titrations
3. At the Equivalence Point
CH3CO21-(aq) + H2O(l)
pH > 7
Notice how the pH at the
equivalence point is
shifted up versus the
strong acid titration.
OH1-(aq) + CH3CO2H(aq)
Weak Acid-Strong Base Titrations
4. Beyond the Equivalence Point
Excess OH1-
Notice that the strong
and weak acid titration
curves correspond after
the equivalence point.
Titration of a Weak Acid with a
Strong Base
With weaker acids, the initial
pH is higher and pH changes
near the equivalence point
are more subtle.
Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq)
H+ (aq) + NH3 (aq)
NH4Cl (aq)
NH4+ (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l)
NH3 (aq) + H+ (aq)
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M
NaOH solution. What is the pH at the equivalence point ?
start (moles)
end (moles)
0.01
0.01
HNO2 (aq) + OH- (aq)
0.0
0.0
Final volume = 200 mL
NO2- (aq) + H2O (l)
0.01
NO2- (aq) + H2O (l)
Initial (M)
Change (M)
0.01
= 0.05 M
0.200
OH- (aq) + HNO2 (aq)
[NO2-] =
0.05
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.05 - x
[OH-][HNO2]
x2
-11
=
2.2
x
10
Kb =
=
[NO2-]
0.05-x
pOH = 5.98
0.05 – x  0.05 x  1.05 x 10-6 = [OH-]
pH = 14 – pOH = 8.02
Polyprotic Acids:
Titration of Diprotic Acid + Strong Base
Titration of a Triprotic Acid:
H3PO4 + Strong Base
Fractional Precipitation
01
•
Fractional precipitation is a method of removing one ion
type while leaving others in solution.
•
Ions are added that will form an insoluble product with one
ion and a soluble one with others.
•
When both products are insoluble, their relative Ksp values
can be used for separation.
Solubility Equilibria
Ag+ (aq) + Cl- (aq)
AgCl (s)
Ksp = [Ag+][Cl-]
MgF2 (s)
Ag2CO3 (s)
Ksp = 1.8 x 10-10
Ksp is the solubility product constant
Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
2Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
3Ca2+ (aq) + 2PO43- (aq)
Ca3(PO4)2 (s)
Ksp = [Ca2+]3[PO43-]2
Dissolution of an ionic solid in aqueous solution:
Q (IP) < Ksp
Q (IP) = Ksp
Unsaturated solution
Q (IP) > Ksp
Supersaturated solution
No precipitate
Saturated solution
(IP) : Ion product
Precipitate will form
Ksp : Solubility Product
Measuring Ksp and Calculating
Solubility from Ksp
Molar solubility (mol/L): Number of moles of a solute that
dissolve to produce a litre of saturated solution
Solubility (g/L) is the number of grams of solute that dissolve
to produce a litre of saturated solution
Measuring Ksp
If the concentrations of Ca2+(aq) and F1-(aq) in a
saturated solution of calcium fluoride are known,
Ksp may be calculated.
CaF2(s)
Ca2+(aq) + 2F1-(aq)
[Ca2+] = 3.3 x 10-4 M
[F1-] = 6.7 x 10-4 M
Ksp = [Ca2+][F1-]2 = (3.3 x 10-4)(6.7 x 10-4)2 = 1.5 x 10-10
(at 25 °C)
Calculating Solubility from Ksp
What is the solubility of silver chloride in g/L ?
AgCl (s)
Initial (M)
Change (M)
Equilibrium (M)
[Ag+] = 1.3 x 10-5 M
Ag+ (aq) + Cl- (aq)
0.00
0.00
+s
+s
s
s
[Cl-] = 1.3 x 10-5 M
Ksp = 1.8 x 10-10
Ksp = [Ag+][Cl-]
Ksp = s2
s = Ksp
s = 1.3 x 10-5
1.3 x 10-5 mol AgCl 143.35 g AgCl
Solubility of AgCl =
x
= 1.9 x 10-3 g/L
1 L soln
1 mol AgCl
Fractional Precipitation
01
•
Fractional precipitation is a method of removing one ion
type while leaving others in solution.
•
Ions are added that will form an insoluble product with one
ion and a soluble one with others.
•
When both products are insoluble, their relative Ksp values
can be used for separation.
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]02 = 0.100 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 4.7 x 10-6
Q < Ksp
No precipitate will form
If this solution also contained 0.100 M AlCl3, Aluminum could be precipitated,
with Calcium remaining in the solution.
Fractional Precipitation
1)
Use the difference between Ksp’s to separate one
precipitate at a time.
2)
Use Common Ion effect to Precipitate one Ion at a
time
01
The Common-Ion Effect and
Solubility
01
Factors That Affect Solubility
Solubility and the Common-Ion Effect
MgF2(s)
Mg2+(aq) + 2F1-(aq)
Fractional Precipitation
01
1.
Use the difference between Ksp’s to separate one
precipitate at a time.
2.
Use Common Ion effect to Precipitate one Ion at a time
3.
Use pH to Increase or decrease solubility and
separate one ion at a time
Factors That Affect Solubility
Solubility and the pH of the Solution
CaCO3(s) + H3O1+(aq)
Ca2+(aq) + HCO31-(aq) + H2O(l)
Fractional Precipitation
01
1.
Use the difference between Ksp’s to separate one
precipitate at a time.
2.
Use Common Ion effect to Precipitate one Ion at a time
3.
Use pH to Increase or decrease solubility and separate
one ion at a time.
4.
Use complex formation to prevent certain Ions to
precipitate
Factors That Affect Solubility
Solubility and the Formation of Complex Ions
Ag1+(aq) + NH3(aq)
Ag(NH3)1+(aq)
K1 = 2.1 x 103
Ag(NH3)1+(aq) + NH3(aq)
Ag(NH3)21+(aq)
K2 = 8.1 x 103
Ag1+(aq) + 2NH3(aq)
Ag(NH3)21+(aq)
Kf = 1.7 x 107
(at 25 °C)
Kf is the formation constant.
How is it calculated?
Factors That Affect Solubility
Solubility and the Formation of Complex Ions
K1 =
Kf = K1K2 =
[Ag(NH3)1+]
[Ag1+][NH3]
[Ag(NH3)21+]
[Ag1+][NH3]2
K2 =
[Ag(NH3)21+]
[Ag(NH3)1+][NH3]
= (2.1 x 103)(8.1 x 103) = 1.7 x 107
Factors That Affect Solubility
Solubility and the Formation of Complex Ions
Ag1+(aq) + 2NH3(aq)
Ag(NH3)21+(aq)
Factors That Affect Solubility
Solubility and Amphoterism
Aluminum hydroxide is soluble both in strongly acidic
and in strongly basic solutions.
In acid:
In base:
Al(OH)3(s) + 3H3O1+(aq)
Al(OH)3(s) + OH1-(aq)
Al3+(aq) + 6H2O(l)
Al(OH)41-(aq)
Factors That Affect Solubility
Solubility and Amphoterism
The Common Ion Effect and Solubility
AgBr (s)
Ag+ (aq) + Br- (aq)
The presence of a common ion decreases
the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water
and (b) 0.0010 M NaBr?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 5.4 x 10-13
s2 = Ksp
s = 7.3 x 10-7
M
NaBr (s)
Na+ (aq) + Br- (aq)
[Br-] = 0.0010 M
AgBr (s)
Ag+ (aq) + Br- (aq)
[Ag+] = s
[Br-] = 0.0010 + s  0.0010
Ksp = 0.0010 x s
s = 5.4 x 10-10 M
pH and Solubility
• The presence of a common ion
decreases the solubility.
• Insoluble bases dissolve in acidic
solutions
• Insoluble acids dissolve in basic
solution.
pH and Solubility
What is the pH of a solution containing Mg(OH)2? what pH
Would reduce the solubility of this precipitate?
remove
add
Mg(OH)2 (s)
Mg2+ (aq) + 2OH- (aq)
At pH less than 10.35
Ksp = [Mg2+][OH-]2 = 5.6 x 10-12
Lower [OH-], Increase solubility of Mg(OH)2
Ksp = (s)(2s)2 = 4s3
- (aq) + H+ (aq)
OH
H2O (l)
3
-12
4s = 5.6 x 10
s = 1.1 x 10-4 M
At pH greater than 10.35
[OH-] = 2s = 2.2 x 10-4 M
pOH = 3.65 pH = 10.35
Raise [OH-]
Decrease solubility of Mg(OH)2
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
CoCl42- (aq)
Co2+ (aq) + 4Cl- (aq)
The formation constant or stability constant (Kf ) is the
equilibrium constant for the complex ion formation.
Co(H2O)2+
6
CoCl24
Kf =
[CoCl42- ]
[Co2+][Cl-]4
Kf
stability of
complex
--
Qualitative Analysis
Qualitative
Analysis of
Cations
Fractional Precipitation
01
•
Fractional precipitation is a method of removing one ion
type while leaving others in solution.
•
Ions are added that will form an insoluble product with one
ion and a soluble one with others.
•
When both products are insoluble, their relative Ksp values
can be used for separation.
Flame Test for Cations
lithium
sodium
potassium
copper
Nessler's is K2[HgI4] in a basic solution , [HgI4]2- reacts with the NH4+ to form
a yellow brown precipitate .
Fractional Precipitation
01
Separation of Ions by Selective Precipitation
Determine whether Cd+2 can be separated from Zn+2 by bubbling
H2S through ([H2S ] = 0.1 ) a 0.3 M HCl solutions that contains
0.005 M Cd+2 and 0.005 M Zn+2?
•
MS(s) + 2 H3O+(aq) ae M2+(aq) + H2S(aq) + 2 H2O(l)
(see page 631) : K spa = [Cd+2 ] [H2S ] / [H3O+ ]2
• Kspa = For ZnS, Kspa = 3 x 10-2;
• for CdS, Kspa = 8 x 10-7
• [Cd+2 ] = [Zn+2] = 0.005 M
• Because the two cation concentrations are equal,
• Qspa is :[Cd+2 ] [H2S ] / [H3O+ ]2 =(0.005).(0.1)/ (0.3)2
• Qspa = 6 x 10-3
• For CdS , Qspa > Kspa; CdS will precipitate.
• For ZnS , Qspa< Kspa; Zn+2 will remain in solution.
•
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
Change (M)
Equilibrium (M)
H+ (aq) + HCOO- (aq)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
X (0.52 + X) = 1.67 X 10-4
0.30 - X
X = 9.77 X 10-5 = [ H3O+]
pH = - log ( 9.77 X 10-5) = 4.01
End