chapter 15 applications of aqueous equilibria addition of base: normal human blood ph is 7.4 and has...
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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
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• 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 Solutions Buffer 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
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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 Solutions Understanding Buffer Solutions
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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:
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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 ).
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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
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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
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Blood Plasma pH = 7.4
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Buffer Solutions 02Buffer Solutions 02
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= 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
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Titrations
Slowly add baseto unknown acid
UNTIL
The indicatorchanges color
(pink)
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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) the equivalence point
Slowly add baseto unknown acid
UNTIL
The indicatorchanges color
(pink)
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Acid–Base Titration Curves 01Acid–Base Titration Curves 01
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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.
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Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
1. Before Addition of Any NaOH
0.100 M HCl
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2. Before the Equivalence Point
Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
2H2O(l)H3O1+(aq) + OH1-(aq)100%
Excess H3O1+
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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
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4. Beyond the Equivalence Point
Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
Excess OH1-
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Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
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Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
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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)
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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
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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
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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-
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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)
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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 ?
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
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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
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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
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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
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Measuring Ksp Measuring 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)
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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 = Ksps = 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
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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
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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.
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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
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The Common-Ion Effect and Solubility 01The Common-Ion Effect and Solubility 01
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Factors That Affect SolubilityFactors That Affect Solubility
Mg2+(aq) + 2F1-(aq)MgF2(s)
Solubility and the Common-Ion Effect
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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)
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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)
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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 =
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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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Factors That Affect SolubilityFactors That Affect Solubility
Solubility and Amphoterism
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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
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• 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?
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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
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Qualitative AnalysisQualitative Analysis
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Qualitative Analysis of
Cations
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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
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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 .
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Fractional Precipitation 01Fractional Precipitation 01
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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?
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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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EndEnd