Chapter 15Chapter 15
Applications of Aqueous EquilibriaApplications of Aqueous Equilibria
Addition of base:
Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2
• 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.
Buffer SolutionsBuffer Solutions
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
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.
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) Na+ (aq) + CH3COO- (aq)
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
common ion
Understanding Buffer SolutionsUnderstanding Buffer Solutions
Consider mixture of salt NaA and weak acid HA.
HA (aq) H+ (aq) + A- (aq)
NaA (s) Na+ (aq) + A- (aq)
Ka =[H+][A-]
[HA]
[H+] =Ka [HA]
[A-]
-log [H+] = -log Ka - log[HA][A-]
-log [H+] = -log Ka + log[A-][HA]
pH = pKa + log[A-][HA]
pKa = -log Ka
Henderson-Hasselbalch equation
pH = pKa + log[conjugate base]
[acid]
pH of a Buffer:
Using Henderson-Hasselbalch equationUsing Henderson-Hasselbalch equation
pH = pKa + log[HCOO-][HCOOH]
HCOOH : Ka= 1.67 X 10-4 , pKa = 3.78
pH = 3.77 + log[0.52][0.30]
= 4.02
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).
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.
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)
Consider an equal molar mixture of CH3COOH and CH3COONa
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 basebuffer solution
(b) HBr is a strong acidnot a buffer solution
(c) CO32- is a weak base and HCO3
- is it conjugate acidbuffer solution
= 9.20
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) H+ (aq) + NH3 (aq)
pH = pKa + log[NH3][NH4
+]pKa = 9.25 pH = 9.25 + log
[0.30][0.36]
= 9.17
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
Initial (moles)
End (moles)
0.36 X0.0 800 =0.029 0.050 x 0.0200 = 0.001 0.30 X .0800 = 0.024
0.029-0.001 = 0.028 0.001-0.001 =0.0 0.024 + 0.001 = 0.025
pH = 9.25 + log[0.25][0.28]
[NH4+] =
0.0280.10
final volume = 80.0 mL + 20.0 mL = 100 mL , 0.1 L
[NH3] = 0.0250.10
-0.001 -0.001 +0.001Change
TitrationsIn 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) theequivalence point
Slowly add baseto unknown acid
UNTIL
The indicatorchanges color
(pink)
Strong Acid-Strong Base TitrationsStrong 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 TitrationsStrong Acid-Strong Base Titrations
1. Before Addition of Any NaOH
0.100 M HCl
2. Before the Equivalence Point
Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
2H2O(l)H3O1+(aq) + OH1-(aq)100%
Excess H3O1+
3. At the Equivalence Point
Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
The quantity of H3O1+ is equal to the quantity of added OH1-.
pH = 7
4. Beyond the Equivalence Point
Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
Excess OH1-
Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
1. Before Addition of Any NaOH
0.100 M CH3CO2H
H3O1+(aq) + CH3CO21-(aq)CH3CO2H(aq) + H2O(l)
Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
2. Before the Equivalence Point
H2O(l) + CH3CO21-(aq)CH3CO2H(aq) + OH1-(aq)
100%
Excess CH3CO2H
Buffer region
췠ѥ
Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
3. At the Equivalence Point
OH1-(aq) + CH3CO2H(aq)CH3CO21-(aq) + H2O(l)
Notice how the pH at the equivalence point is shifted up versus the strong acid titration.
pH > 7
Notice that the strong and weak acid titration curves correspond after
the equivalence point.
Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
4. Beyond the Equivalence Point
Excess OH1-
Titration of a Weak Acid with a Strong BaseTitration 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.
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Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)
At equivalence point (pH < 7):
H+ (aq) + NH3 (aq) NH4+ (aq)
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 MNaOH solution. What is the pH at the equivalence point ?
HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)
start (moles)
end (moles)
0.01 0.01
0.0 0.0 0.01
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.05 0.00
-x +x
0.05 - x
0.00
+x
x x
[NO2-] =
0.010.200 = 0.05 MFinal volume = 200 mL
Kb =[OH-][HNO2]
[NO2-]
=x2
0.05-x= 2.2 x 10-11
0.05 – x ≈ 0.05 x ≈ 1.05 x 10-6 = [OH-]
pOH = 5.98
pH = 14 – pOH = 8.02
Polyprotic Acids:Titration of Diprotic Acid + Strong Base
Polyprotic Acids:Titration of Diprotic Acid + Strong Base
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Titration of a Triprotic Acid: H3PO4 + Strong BaseTitration of a Triprotic Acid: H3PO4 + Strong Base
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• 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.
Fractional Precipitation 01Fractional Precipitation 01
터
Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-] Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2
Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3
2-]
Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO4
3-]2
Dissolution of an ionic solid in aqueous solution:
Q (IP) = Ksp Saturated solution
Q (IP) < Ksp Unsaturated solution No precipitate
Q (IP) > KspSupersaturated solution Precipitate will form
Ksp = 1.8 x 10-10
Ksp : Solubility Product(IP) : Ion product
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Measuring Ksp and Calculating Solubility from Ksp
Measuring Ksp and Calculating Solubility from Ksp
터
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
터
Measuring KspMeasuring Ksp
If the concentrations of Ca2+(aq) and F1-(aq) in a saturated solution of calcium fluoride are known, Ksp may be calculated.
Ca2+(aq) + 2F1-(aq)CaF2(s)
Ksp = [Ca2+][F1-]2 = (3.3 x 10-4)(6.7 x 10-4)2 = 1.5 x 10-10
[F1-] = 6.7 x 10-4 M[Ca2+] = 3.3 x 10-4 M
(at 25 °C)
터
What is the solubility of silver chloride in g/L ?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]Initial (M)
Change (M)
Equilibrium (M)
0.00
+s
0.00
+s
s s
Ksp = s2
s = Ksp√s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M
Solubility of AgCl = 1.3 x 10-5 mol AgCl
1 L soln143.35 g AgCl
1 mol AgClx = 1.9 x 10-3 g/L
Ksp = 1.8 x 10-10
Calculating Solubility from Ksp
터
• 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.
Fractional Precipitation 01Fractional Precipitation 01
터
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
Ksp = [Ca2+][OH-]2 = 4.7 x 10-6
Q = [Ca2+]0[OH-]02 = 0.100 x (4.0 x 10-4)2 = 1.6 x 10-8
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.
터
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
Fractional Precipitation 01Fractional Precipitation 01
터
Factors That Affect SolubilityFactors That Affect Solubility
Mg2+(aq) + 2F1-(aq)MgF2(s)
Solubility and the Common-Ion Effect
터
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
Fractional Precipitation 01Fractional Precipitation 01
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Factors That Affect SolubilityFactors That Affect Solubility
Solubility and the pH of the Solution
Ca2+(aq) + HCO31-(aq) + H2O(l)CaCO3(s) + H3O1+(aq)
터
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
Fractional Precipitation 01Fractional Precipitation 01
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Factors That Affect SolubilityFactors That Affect Solubility
Solubility and the Formation of Complex Ions
Ag(NH3)21+(aq)Ag(NH3)1+(aq) + NH3(aq)
Ag(NH3)21+(aq)Ag1+(aq) + 2NH3(aq) Kf = 1.7 x 107
K1 = 2.1 x 103Ag(NH3)1+(aq)Ag1+(aq) + NH3(aq)
K2 = 8.1 x 103
Kf is the formation constant.
How is it calculated?
(at 25 °C)
ѫ
Factors That Affect SolubilityFactors That Affect Solubility
Solubility and the Formation of Complex Ions
= (2.1 x 103)(8.1 x 103) = 1.7 x 107[Ag(NH3)2
1+]
[Ag1+][NH3]2
K1 =[Ag(NH3)1+]
[Ag1+][NH3]K2 =
[Ag(NH3)21+]
[Ag(NH3)1+][NH3]
Kf = K1K2 =
Factors That Affect SolubilityFactors That Affect Solubility
Solubility and the Formation of Complex Ions
Ag(NH3)21+(aq)Ag1+(aq) + 2NH3(aq)
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Factors That Affect SolubilityFactors That Affect Solubility
Solubility and Amphoterism
Al(OH)41-(aq)Al(OH)3(s) + OH1-(aq)
Al3+(aq) + 6H2O(l)Al(OH)3(s) + 3H3O1+(aq)
In base:
In acid:
Aluminum hydroxide is soluble both in strongly acidic and in strongly basic solutions.
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The Common Ion Effect and Solubility
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
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
AgBr (s) Ag+ (aq) + Br- (aq)
M
M
터
• The presence of a common ion decreases the solubility.
• Insoluble bases dissolve in acidic solutions
• Insoluble acids dissolve in basic solution.
pH and SolubilitypH and Solubility
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pH and Solubility
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
Ksp = [Mg2+][OH-]2 = 5.6 x 10-12
Ksp = (s)(2s)2 = 4s3
4s3 = 5.6 x 10-12
s = 1.1 x 10-4 M
[OH-] = 2s = 2.2 x 10-4 M
pOH = 3.65 pH = 10.35
At pH less than 10.35
Lower [OH-], Increase solubility of Mg(OH)2
OH- (aq) + H+ (aq) H2O (l)
remove
At pH greater than 10.35
Raise [OH-]
add
Decrease solubility of Mg(OH)2
What is the pH of a solution containing Mg(OH)2? what pH
Would reduce the solubility of this precipitate?
터
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq) CoCl4 (aq)2-
Kf =[CoCl4 ]
[Co2+][Cl-]4
2-
The formation constant or stability constant (Kf ) is the equilibrium constant for the complex ion formation.
Co(H2O)62+ CoCl4
2-
Kf
stability of complex
터
• 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.
Fractional Precipitation 01Fractional Precipitation 01
辠Ѯ
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 .
ѫ
Separation of Ions by Selective PrecipitationSeparation of Ions by Selective Precipitation
• 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.
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?
犰
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.52
+x
x 0.52 + x
Common ion effect
0.30 – x ≈ 0.30
0.52 + x ≈ 0.52
Mixture of weak acid and conjugate base!
X (0.52 + X)0.30 - X
= 1.67 X 10-4
X = 9.77 X 10-5 = [ H3O+]
pH = - log ( 9.77 X 10-5) = 4.01
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