CHM 1046 FINAL REVIEW - Palm Beach State CollegeCHM 1046 FINAL REVIEW Prepared & Presented By:...
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CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub
PART II
Chapter Description
14 Chemical Equilibrium
15 Acids and Bases
16 Acid-Base Equilibrium
17 Solubility and Complex-Ion Equilibrium
19 Electrochemistry
20 Nuclear Chemistry
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CH. 14 CHEMICAL EQUILIBRIUM
• DYNAMIC EQUILIBRIUM: WHEN FORWARD AND REVERSE REACTION RATES ARE THE SAME.
*PURE SOLIDS AND LIQUIDS DO NOT APPEAR IN THE EQUILIBRIUM CONSTANT EXPRESSION (ANY K OR Q)
• THE EQUILIBRIUM CONSTANT KC: THE VALUE OBTAINED FOR THE EQUILIBRIUM-CONSTANT
EXPRESSION WHEN EQUILIBRIUM CONCENTRATIONS ARE SUBSTITUTED.
• THE EQUILIBRIUM CONSTANT KP: THE EQUILIBRIUM CONSTANT EXPRESSION FOR A GASEOUS
REACTION IN TERMS OF PARTIAL PRESSURES.
FOR GASES: KP = K
C (RT)ΔN , WHERE ΔN = DIFFERENCE OF COEFFICIENT.
Ch. 14 Chemical Equilibrium
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CALCULATING VALUES OF K
• IF 2 OR MORE REACTIONS ARE ADDED TO ACHIEVE A GIVEN REACTION, THE
EQUILIBRIUM CONSTANT FOR THE GIVEN EQUATION EQUALS THE PRODUCT
OF THE EQUILIBRIUM CONSTANTS (K) OF THE ADDED EQUATIONS.
• IF REACTION IS REVERSED YOU TAKE THE INVERSE OF ORIGINAL K OF REACTION.
• IF MULTIPLIED OR DIVIDED, RAISE K TO THAT POWER.
EX: MULTIPLY BY 2
EX: DIVIDE BY 3
1/K
K1xK2
K2
K1/3
Ch. 14 Chemical Equilibrium
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USING QC TO DETERMINE THE DIRECTION OF EQUILIBRIUM
• REACTION QUOTIENT (QC): THE INITIAL REACTION RATE OF A REACTION.
ITS NOT A CONSTANT BUT DYNAMIC.
• THE RELATIONSHIP BETWEEN QC AND KC GIVES THE DIRECTION OF THE
EQUILIBRIUM.
• IF KC > QC EQUILIBRIUM IS IN THE FORWARD DIRECTION.
• IF KC = QC REACTION IS IN EQUILIBRIUM.
• IF KC < QC EQUILIBRIUM IS IN THE REVERSE DIRECTION.
• IF QC = 0 ONLY REACTANTS ARE PRESENT.
• IF QC = ∞ ONLY PRODUCTS ARE PRESENT.
Ch. 14 Chemical Equilibrium
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1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide:
N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K
a) What is the value of Kc for this reaction?
b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g)
c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)?
Ch. 14 Chemical Equilibrium
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1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide:
N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K
a) What is the value of Kc for this reaction?
b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g)
c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)?
Ch. 14 Chemical Equilibrium
a) Kp = K
c (RT)Δn
Kp
Kc = _______
(RT)Δn
0.660
Kc = ________________
(0.0821 x 319 )2-1
b) Reaction reversed then take the
inverse of Kp
1/ Kp = 1/ 0.660
c) (PNO2)2
Kp = _________________
(PN2O4 )
(0.223) 2
0.660 = ___________
(PN2O4 )
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2. If a 2.50 L vessel at 1000˚C containes 0.525 mol CO2, 1.25 mol CF4, and 0.75 mol COF2, in what
direction will a net reaction occur to reach the equilibrium?
CO2 (g) + CF4 (g) ↔ 2COF2 (g) Kc = 0.50 @ 1000 K
Ch. 14 Chemical Equilibrium
[CO2] = 0.525 mol / 2.5 L = 0.21 M
[CF4] = 1.25 mol / 2.5 L = 0.5 M
[COF2] = 0.75 mol / 2.5 L = 0.3 M
[COF2]2
Qc = ___________ [CO2] [CF4]
[0.3]2
Qc = ___________
[0.21] [0.5]
Kc = 0.50
Solution:
How do you determine the direction of a reaction in order to reach equilibrium?
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3. Starting with 0.100 mol each of CO and H2O in a 5.00 L flask, equilibrium is established in the
following reaction at 600K:
CO(g) + H2O(g) ↔ CO2 (g) + H2 (g) Kc = 23.2 at 600K
What is the concentration of hydrogen at equilibrium?
Ch. 14 Chemical Equilibrium
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Ch. 14 Chemical Equilibrium
[CO] = 0.100 mol/ 5 L = 0.02 M
[H2O] = 0.100 mol / 5 L = 0.02 M
[X] [X]
23.2 = ______________
[0.02-X] [0.02-X]
[X]2
23.2 = _________________
[0.02-X]2
*Take the square root of both sides:
[X]
4.82 = _________________
[0.02-X]
0.096 -4.82X = X
0.096 = X + 4.82X
5.82X = 0.096
X = 0.096/ 5.82
X = 0.0165
[CO2] [H2]
Kc = _________________
[CO] [H2O]
SO [H2]= X at equilibrium
Balanced Equation
CO(g) + H2O(g) ↔
CO2 (g) H2 (g)
Initial
0.02
0.02
0
0
Change
- x
-x
+x
+x
Equilibrium
0.02-x
0.02-x
x
x
Solution
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4. The Keq of a reaction is 4x10-7. At equilibrium_____.
a) The products are favored
b) The reactants are favored
c) The reactants and products are present in equal amounts
d) The rate of the forward and reverse reaction are the same
Ch. 14 Chemical Equilibrium
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4. The Keq of a reaction is 4x10-7. At equilibrium_____.
a) The products are favored
c) The reactants and products are present in equal amounts
d) The rate of the forward and reverse reaction are the same
Explanation: Here Keq= 0.0000004
Keq<< 1 Therefore Reactants are favored!!
What if Keq >> 1?????
Ch. 14 Chemical Equilibrium
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4. The Keq of a reaction is 4x10-7. At equilibrium_____.
a) The products are favored
c) The reactants and products are present in equal amounts
d) The rate of the forward and reverse reaction are the same
Explanation: Here Keq= 0.0000004
Keq<< 1 Therefore Reactants are favored!!
What if Keq >> 1?????
Products would be favored
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5
Remove Cl2
Add Ar
Decrease V (or increase P)
Increase T
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2
Add Ar
Decrease V (or increase P)
Increase T
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar
Decrease V (or increase P)
Increase T
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar No effect
Decrease V (or increase P)
Increase T
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar No effect
Decrease V (or increase P) Left
High to Low #of moles of Gases
1 mol ← 2 mol
Increase T
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar No effect
Decrease V (or increase P) Left High to Low #of moles of Gases
1 mol ← 2 mol
Increase T Right
Add Catalyst
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar No effect
Decrease V (or increase P) Left High to Low #of moles of Gases
1 mol ← 2 mol
Increase T Right
Add Catalyst No effect on equilibrium
*Catalysts Lowers Activation Energy only
Ch. 14 Chemical Equilibrium
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5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill
in the following table:
Changes Shifts RX'N which way?
Add PCl5 Right
Remove Cl2 Right
Add Ar No effect
Decrease V (or increase P) Left High to Low #of moles of Gases
1 mol ← 2 mol
Increase T Right
Add Catalyst No effect on equilibrium
*Catalysts Lowers Activation Energy only
Ch. 14 Chemical Equilibrium
Which of the top changes would cause a change in K?
• ONLY a change in Temperature can cause a change in K
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CH. 15 ACIDS AND BASES
Arrhenius Definitions:
1) Acid: substance that when dissolved in water increases [H+]
2) Base: substance that when dissolved in water increases [OH-]
Example: Identify following as Arrhenius Acid/Base
HCl NaOH
Ch. 15 Acids and Bases
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CH. 15 ACIDS AND BASES
Arrhenius Definitions:
1) Acid: substance that when dissolved in water increases [H+]
2) Base: substance that when dissolved in water increases [OH-]
Example: HCl (Acid) NaOH (Base)
Brønsted-Lowry Definitions:
1) Acid: substance that donates a proton, H+, in a reaction
2) Base: substance that accepts a proton, H+, in a reaction
Example: Identify following as Bronsted-Lowry Acid/Base
:NH3 HCl
Ch. 15 Acids and Bases
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CH. 15 ACIDS AND BASES
Arrhenius Definitions:
1) Acid: substance that when dissolved in water increases [H+]
2) Base: substance that when dissolved in water increases [OH-]
Example: HCl (Acid) NaOH (Base)
Brønsted-Lowry Definitions:
1) Acid: substance that donates a proton, H+, in a reaction
2) Base: substance that accepts a proton, H+, in a reaction
Example: HCl (Acid) :NH3 (Base)
Lewis Definitions:
1) Acid: is an electron pair acceptor.
2) Base: is an electron pair donor.
Example: Identify following as Lewis Acid/Base
Ag+ :NH3
Ch. 15 Acids and Bases
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CH. 15 ACIDS AND BASES
Arrhenius Definitions:
1) Acid: substance that when dissolved in water increases [H+]
2) Base: substance that when dissolved in water increases [OH-]
Example: HCl (Acid) NaOH (Base)
Brønsted-Lowry Definitions:
1) Acid: substance that donates a proton, H+, in a reaction
2) Base: substance that accepts a proton, H+, in a reaction
Example: HCl (Acid) :NH3 (Base)
Lewis Definitions:
1) Acid: is an electron pair acceptor.
2) Base: is an electron pair donor.
Example: Ag+ (Acid) :NH3(Base)
Ch. 15 Acids and Bases
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CH. 15 ACIDS AND BASES
Arrhenius Definitions:
1) Acid: substance that when dissolved in water increases [H+]
2) Base: substance that when dissolved in water increases [OH-]
Example: HCl (Acid) NaOH (Base)
Brønsted-Lowry Definitions:
1) Acid: substance that donates a proton, H+, in a reaction
2) Base: substance that accepts a proton, H+, in a reaction
Example: HCl (Acid) :NH3 (Base)
Lewis Definitions:
1) Acid: is an electron pair acceptor.
2) Base: is an electron pair donor.
Example: Ag+ (Acid) :NH3(Base)
Ch. 15 Acids and Bases
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STRONG ACIDS AND BASES
• HNO3
• H2SO4
• HClO4
• HCl
• HBr
• HI
• H3O+ or H +
• GROUP 1A & 2A HYDROXIDE
• EX: NaOH, Ca(OH)2
• EXCEPT Be(OH)2
Ch. 15 Acids and Bases
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COMPARING THE STRENGTHS OF GIVEN ACIDS
• Binary Acids (H-NONMETAL):
A. Electronegativity if same period (↑ EN = ↑ strength)
EN increases Left to Right
B. Atomic size if same group (↑ atomic size = ↑ strength)
Atomic size increases Top to Bottom
• Oxacids (H-O-nonmetal): • Electronegativity (↑ EN = ↑ Strength)
EN increases Left to Right and Bottom to Top
• Other Groups (H-O-NM-On): • More Oxygen = Stronger Acid
• Carboxylic Acids (RCOOH): A. Resonance (EN)
B. Inductive Effect: distance of EN element from COOH group.
(Closer = More Acidic)
Ch. 15 Acids and Bases
EX: H2Te vs HI
EX: HI vs HF
EX: HOCl vs HOBr
EX: H3AsO4 vs H3AsO3
EX: FCH3COOH
vs
FCH3CH2CH2COOH
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Strong base gives weak CA and weak base gives strong CA
Strong acid gives weak CB and weak acid gives strong CB
Reaction is favored in the direction which leads to the formation
of weaker acid or base (or CB and CA).
Ex: Which side of the following reaction is favored?
CH3COOH + H2O ↔ CH3COO- + H3O+
Strength of Conjugate Acid and Conjugate Bases
Ch. 15 Acids and Bases
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Strong base gives weak CA and weak base gives strong CA
Strong acid gives weak CB and weak acid gives strong CB
Reaction is favored in the direction which leads to the formation
of weaker acid or base (or CB and CA).
Ex: Which side of the following reaction is favored?
CH3COOH + H2O ↔ CH3COO- + H3O+
Answer: Weak Acid Strong C.B.
Strength of Conjugate Acid and Conjugate Bases
Ch. 15 Acids and Bases
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Strong base gives weak CA and weak base gives strong CA
Strong acid gives weak CB and weak acid gives strong CB
Reaction is favored in the direction which leads to the formation
of weaker acid or base (or CB and CA).
Ex: Which side of the following reaction is favored?
CH3COOH + H2O ↔ CH3COO- + H3O+
Answer: Weak Acid Strong C.B.
Weak Base Strong C.A.
Therefore the Reactant side (left) is Favored
Strength of Conjugate Acid and Conjugate Bases
Ch. 15 Acids and Bases
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Ch. 15 Acids and Bases
1. Identify the acid and base on each side of the following equation:
H2S + NH3 ↔ NH4+ + HS-
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Ch. 15 Acids and Bases
1. Identify the acid and base on each side of the following equation:
H2S + NH3 ↔ NH4+ + HS-
ANSWER: Acid + Base → C.A. + C.B.
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2. Which of the following is the strongest Acid?
A) HOI B) HOCl C) HOBr
Ch. 15 Acids and Bases
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2. Which of the following is the strongest Acid?
A) HOI C) HOBr
Answer: HOCl
With Oxacids (H-O-nonmetal)
We use Electronegativity trend (increases bottom to top in a group)
↑EN= ↑Strength of Acid.
Ch. 15 Acids and Bases
↑EN
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3. What is the pH of a solution prepared by dissolving
0.025 mol Ba(OH)2 in water to give 455 mL of solution?
Ch. 15 Acids and Bases
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3. What is the pH of a solution prepared by dissolving
0.025 mol Ba(OH)2 in water to give 455 mL of solution?
Ch. 15 Acids and Bases
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3. What is the pH of a solution prepared by dissolving
0.025 mol Ba(OH)2 in water to give 455 mL of solution?
Ch. 15 Acids and Bases
Solution: There is more than one way to solve this..
0.025 mol x 2 = 0.05 mol OH-
0.05 mol/0.455 L = 0.1099 M OH-
pOH = -log [OH-]
pOH = -log [0.1099]
pOH = 0.959
pOH + pH = 14
pH = 14 - pOH
pH = 14 – 0.959
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3. What is the pH of a solution prepared by dissolving
0.025 mol Ba(OH)2 in water to give 455 mL of solution?
Ch. 15 Acids and Bases
Solution: There is more than one way to solve this..
0.025 mol x 2 = 0.05 mol OH-
0.05 mol/0.455 L = 0.1099 M OH-
pOH = -log [OH-]
pOH = -log [0.1099]
pOH = 0.959
pOH + pH = 14
pH = 14 - pOH
pH = 14 – 0.959
OR
0.1099 M OH-
Kw= [H+] [OH-] = 1x10-14
1x10-14
[H+] = __________ = 9.1x10-14 M H+
0.1099
pH = -log [H+]
pH = -log [9.1x10-14 ]
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4. A sample of milk is found to have a pH of 6.25. What is the
[OH-] in this milk?
Ch. 15 Acids and Bases
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4. A sample of milk is found to have a pH of 6.25. What is the
[OH-] in this milk?
Ch. 15 Acids and Bases
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4. A sample of milk is found to have a pH of 6.25. What is the
[OH-] in this milk?
Ch. 15 Acids and Bases
Solution: There is more than one way to solve this..
pH + pOH = 14
pOH = 14 - 6.25
pOH = 7.75
[OH-] = 10-pOH
[OH-] = 10-7.75
OR
[H+] = 10-pH
[H+] = 10-6.25
[H+] = 5.62x10-7
Kw= [H+] [OH-] = 1x10-14
1x10-14
[OH-] = __________ =
5.62x10-7
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CH. 16 ACID-BASE EQUILIBRIUM
Acid ionization constant:
*Lower value of pKa = strong acid
*Higher value of pKa = weak acid
Base ionization constant:
*Lower value of pKb = strong Base
*Higher value of pKb = weak Base
Polyprotic acids: more than one ionizable proton e.g. H2SO4 is diprotic and H3PO4 is triprotic.
In a solution there will be more than just the one conjugate acid/base pair.
Ch. 16 Acid-Bases Equilibrium
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• Strong Acid + Strong Base → Neutral (ex. NaCl, KNO3)
• Strong Acid + Weak Base → Acidic Solution (ex. NH4Cl )
• Weak Acid + Strong Base → Basic Solution (ex. Na2CO3)
• Weak Acid + Weak Base → Depends on K and Kb
Hydrolysis
Ch. 16 Acid-Bases Equilibrium
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The common ion in an equilibrium can shift the equilibrium to the opposite
side.
Ex: adding sodium acetate CH3COONa to the following reaction will
increase the concentration of acetate ion thus moving the equilibrium to the
left.
Common Ion Effect
Ch. 16 Acid-Bases Equilibrium
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• Buffers are solutions containing weak acid and its conjugate base
OR weak base and its conjugate acid.
• The pH changes slightly with the addition of a little acid or base.
•Acid Base Indicators: chemicals that change color with pH.
• pH of buffers can be calculated using
The traditional ICE method
or by
Henderson-Hasselbalch equation
Buffers
Ch. 16 Acid-Bases Equilibrium
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Buffer Calculation:
1. What is the [H3O
+] for a buffer solution that is 0.250 M in acid and 0.600
M in the corresponding salt if the weak acid Ka = 5.80 x 10
−7?
Ch. 16 Acid-Bases Equilibrium
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Buffer Calculation:
1. What is the [H3O
+] for a buffer solution that is 0.250 M in acid and 0.600
M in the corresponding salt if the weak acid Ka = 5.80 x 10
−7?
Solution:
HA(aq) + H2O(l) ↔ H3O
+ (aq) + A
− (aq)
pKa = −log(5.80 x 10−7
) = 6.237
[base] = 0.600 M and [acid] = 0.250 M
Ch. 16 Acid-Bases Equilibrium
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Buffer Calculation:
1. What is the [H3O
+] for a buffer solution that is 0.250 M in acid and 0.600
M in the corresponding salt if the weak acid Ka = 5.80 x 10
−7?
Solution:
HA(aq) + H2O(l) ↔ H3O
+ (aq) + A
− (aq)
pKa = −log(5.80 x 10−7
) = 6.237
[base] = 0.600 M and [acid] = 0.250 M
Ch. 16 Acid-Bases Equilibrium
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Buffer Calculation:
1. What is the [H3O
+] for a buffer solution that is 0.250 M in acid and 0.600
M in the corresponding salt if the weak acid Ka = 5.80 x 10
−7?
Solution:
HA(aq) + H2O(l) ↔ H3O
+ (aq) + A
− (aq)
pKa = −log(5.80 x 10−7
) = 6.237
[base] = 0.600 M and [acid] = 0.250 M
Ch. 16 Acid-Bases Equilibrium
6.237 [0.600]
[0.250]
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Buffer Calculation:
1. What is the [H3O
+] for a buffer solution that is 0.250 M in acid and 0.600
M in the corresponding salt if the weak acid Ka = 5.80 x 10
−7?
Solution:
HA(aq) + H2O(l) ↔ H3O
+ (aq) + A
− (aq)
pKa = −log(5.80 x 10−7
) = 6.237
[base] = 0.600 M and [acid] = 0.250 M
Ch. 16 Acid-Bases Equilibrium
6.237 [0.600]
[0.250]
pH = 6.237 + log 2.40
pH = 6.237 + 0.380 = 6.617
[H3O
+] = 10
–pH
[H3O
+] = 10
–6.617
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• The reactant in a solubility equilibrium is a slightly soluble salt and the
equilibrium constant for the reaction is the Solubility Product Constant, Ksp.
• Solubility problems are equilibrium problems.
Ex: the solubility equilibrium and Ksp for the salt SrF2(s) are
SrF2(s) Sr2+(aq) + 2 F – (aq) Ksp = [Sr2+][F –]2 = 2.0 x 10-10
*Note that since the reactant is a Solid, its concentration Does Not appear in the Ksp expression.
•The Molar Solubility of a salt in water can be formed by setting up an
equilibrium table and solving for x.
•The Solubility is the same quantity expressed in g/L (rather than M = mol/L).
CH. 17 Solubility and Complex-Ion Equilibrium
Ch. 17 Solubility Equilibrium
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Solubility Quotient (Qip): determines if precipitation will occur
Qip > Ksp precipitation occurs (Supersaturated)
Qip = Ksp saturation of solution (Saturated)
Qip < Ksp precipitation will not occur (Unsaturated)
*In general very small value of Ksp indicates complete precipitation)
Solubility Quotient (Qip)
Ch. 17 Solubility Equilibrium
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• Common ion effect: Addition of an ion “common” to a solubility equilibrium will reduce
solubility. *Le Châtelier’s Principle.
Ex: SrF2(s) Sr2+(aq) + 2 F –
(aq)
Adding fluoride ion, F–(aq), to the SrF2(s) equilibrium above will shift it left.
The shift will increase the amount of SrF2(s) in solid form, and thus decrease solubility.
• Effect of pH on solubility: pH of a solution will affect solubility if the conjugate ion (acid
or base) is acidic or basic.
Ex: Cl- is very weak conjugate base (CB) and is not considered basic,
whereas HCO3 - is a weak CB whose solubility is affected by the pH.
*An Anion (CB) of a weak Acid is more soluble in Acidic solutions.
Factors Affecting Solubility
Ch. 17 Solubility Equilibrium
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• If two precipitates are possible, the least soluble salt will precipitate first (the one
with the smaller Ksp).
Ex: If you add NaOH(aq) to a solution containing equal amounts of Ca2+and Mg2+
*Mg(OH)2(s) will precipitate before Ca(OH)2(s) since
Ksp(Mg(OH)2) = 1.2 x 10-11 < Ksp(Ca(OH)2) = 8.0 x 10-6
Comparing Precipitation Of Given Salts
Ch. 17 Solubility Equilibrium
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Common Ion Effect:
1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Ksp = 2.0 x 10-10
Ch. 17 Solubility Equilibrium
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We set up an equilibrium (ICE) table, but now we have an initial concentration of fluoride ion:
Ch. 17 Solubility Equilibrium
Common Ion Effect:
1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Ksp = 2.0 x 10-10
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Balanced Equation
SrF2(s)
Sr2+(aq) +
2 F – (aq)
Initial (M)
0
0.10
Change (M)
+x
+2x
Equilibrium (M)
x
(0.10 + 2x)
Ch. 17 Solubility Equilibrium
Common Ion Effect:
1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Ksp = 2.0 x 10-10
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Balanced Equation
SrF2(s)
Sr2+(aq) +
2 F – (aq)
Initial (M)
0
0.10
Change (M)
+x
+2x
Equilibrium (M)
x
(0.10 + 2x)
We next use the equilibrium concentrations in the table and the Ksp value given above, and solve for x.
[0.10/ 2.0 x 10-10] = 500,000,000 >>>>1000
Note that since x is small, (0.10 + 2x) ~ 0.10
Ch. 17 Solubility Equilibrium
Common Ion Effect:
1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Ksp = 2.0 x 10-10
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Balanced Equation
SrF2(s)
Sr2+(aq) +
2 F – (aq)
Initial (M)
0
0.10
Change (M)
+x
+2x
Equilibrium (M)
x
(0.10 + 2x)
We next use the equilibrium concentrations in the table and the Ksp value given above, and solve for x.
[0.10/ 2.0 x 10-10] = 500,000,000 >>>>1000
Note that since x is small, (0.10 + 2x) ~ 0.10
Ksp = [Sr2+][F –]2
2.0 x 10-10 = (x)( 0.10) 2
2.0 x 10-10 = (x)( 0.10)2
2.0 x 10-10 = 0.010 x
Ch. 17 Solubility Equilibrium
Common Ion Effect:
1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Ksp = 2.0 x 10-10
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2. Which of the following salts is most soluble in pure water?
A) HgS, Ksp = 2.0 x 10-53
B) AgI, Ksp = 8.5 x 10-17
C) PbI2, Ksp = 7.1 x 10-9
D) CuS, Ksp = 8.7 x 10-36
E) ZnS, Ksp =1.6 x 10-24
Ch. 17 Solubility Equilibrium
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2. Which of the following salts is most soluble in pure water?
A) HgS, Ksp = 2.0 x 10-53
B) AgI, Ksp = 8.5 x 10-17
C) PbI2, Ksp = 7.1 x 10-9
D) CuS, Ksp = 8.7 x 10-36
E) ZnS, Ksp =1.6 x 10-24
The higher the Ksp , the more soluble it would be.
Ch. 17 Solubility Equilibrium
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2. Which of the following salts is most soluble in pure water?
A) HgS, Ksp = 2.0 x 10-53
B) AgI, Ksp = 8.5 x 10-17
D) CuS, Ksp = 8.7 x 10-36
E) ZnS, Ksp =1.6 x 10-24
The higher the Ksp , the more soluble it would be.
Ch. 17 Solubility Equilibrium
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CH. 19 Electrochemistry
Ch. 19 Electrochemistry
Galvanic cell is Spontaneous
Ecell > 0
∆G < 0
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Galvanic/Voltaic Cells: A cell that has a spontaneous redox reaction.
The terminology of a cell are:
Electrode : a metal piece at which the electrochemical reaction takes place.
Anode: where oxidation (loss of e- ) occurs. (AN OX)
Cathode: where reduction (gain of e- ) occurs. (RED CAT)
Half-cell: the reduction or the oxidation part of the cell.
Coulomb (C): the unit of electric charge.
Volt (V): one joule per coulomb.
Voltmeter: measures volts.
Cell potential (Ecell): electro-potential difference that moves the electrons from the
anode to cathode.
*Electrons flow from Anode to Cathode (Mnemonic: A to C in alphabetical order)
Galvanic/Voltaic Cells
Ch. 19 Electrochemistry
Galvanic cell is Spontaneous
Ecell > 0
∆G < 0
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Standard Electrode Potential: the tendency for reduction to occur at an
electrode under the conditions of SHE.
More Reduction Potential = Oxidizing Agent = Reduction
More Reduction Potential = Reducing Agent = Oxidation
+
-
Standard Cell Potential: E
o cell = E
o (cathode) – E
o (anode)
Eo cell = E
o (right) – E
o (left)
*Reduction potential = potential of the reduction
half reaction.
*Oxidation potential = potential of the oxidation
half reaction = reverse the sign of the reduction
potential.
Ch. 19 Electrochemistry
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Oxidation-Reduction
Oxidation
• Losing Electrons
• Reducing Agent
• Increases In Charge
Reduction
• Gaining Electrons
• Oxidizing Agent
• Decreases In Charge
Ch. 19 Electrochemistry
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1. Identify the following two reactions:
A) 1 is oxidation, 2 is reduction
B) 1 is reduction, 2 is oxidation
C) Neither
D) Both oxidation
E) Both reduction
Ch. 19 Electrochemistry
1) Cu(s) → Cu2+ (aq) + 2e-
2) 2 Ag+(aq) + 2e- → 2 Ag(s)
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1. Identify the following two reactions:
B) 1 is reduction, 2 is oxidation
C) Neither
D) Both oxidation
E) Both reduction
LeO GeR
Ch. 19 Electrochemistry
1) Cu(s) → Cu2+ (aq) + 2e-
2) 2 Ag+(aq) + 2e- → 2 Ag(s)
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2. The following questions are about this cell:
Al|Al+3
||Pb+2
|Pb
a. Identify the anode and the cathode.
b. Write the balanced overall reaction.
c. What is the potential of this cell under standard conditions?
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2. The following questions are about this cell:
Al|Al+3
||Pb+2
|Pb
a. Identify the anode and the cathode.
b. Write the balanced overall reaction.
c. What is the potential of this cell under standard conditions?
Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation)
Al loses e- and it is on the left, therefore it is the Anode
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2. The following questions are about this cell:
Al|Al+3
||Pb+2
|Pb
a. Identify the anode and the cathode.
b. Write the balanced overall reaction.
c. What is the potential of this cell under standard conditions?
Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation)
Al loses e- and it is on the left, therefore it is the Anode
Red Cat, GeR (Reduction Cathode, Gain e- Reduction)
Pb gain e- and it is on the right, therefore it is the cathode
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2. The following questions are about this cell:
Al|Al+3
||Pb+2
|Pb
a. Identify the anode and the cathode.
b. Write the balanced overall reaction.
c. What is the potential of this cell under standard conditions?
Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation)
Al loses e- and it is on the left, therefore it is the Anode
Red Cat, GeR (Reduction Cathode, Gain e- Reduction)
Pb gain e- and it is on the right, therefore it is the cathode
b. 2Al + 3Pb+2
→ 2Al+3
+ 3Pb
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2. The following questions are about this cell:
Al|Al+3
||Pb+2
|Pb
a. Identify the anode and the cathode.
b. Write the balanced overall reaction.
c. What is the potential of this cell under standard conditions?
Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation)
Al loses e- and it is on the left, therefore it is the Anode
Red Cat, GeR (Reduction Cathode, Gain e- Reduction)
Pb gain e- and it is on the right, therefore it is the cathode
b. 2Al + 3Pb+2
→ 2Al+3
+ 3Pb
c.
Eo cell = E
o (cathode) – E
o (anode)
Eo cell = E
o (right) – E
o (left)
Eo cell = (-0.13) – (-1.66)
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Ch. 19 Electrochemistry
Balance the following redox reactions • (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. • (b) give the balanced net reaction. • (c) identify the oxidizing agent and the reducing agent.
Cl2(g) + S
2O
3
-2 (aq) Cl
-(aq) + SO
4-2 (aq) in acid solution.
Oxidation- Reduction Reactions in
Acidic/Basic Solutions
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Ch. 19 Electrochemistry
Balance the following redox reactions • (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. • (b) give the balanced net reaction. • (c) identify the oxidizing agent and the reducing agent.
Cl2(g) + S
2O
3
-2 (aq) Cl
-(aq) + SO
4-2 (aq) in acid solution.
1. Assign Oxidation numbers and determine which is oxidized and which is reduced Cl2(g) + S
2O
3
-2 (aq) Cl (aq) + SO4(aq)
Cl : 0 → -1 (reduced)
S : +2→ +6 (Oxidized)
S2O
3
-2
2S + 3O = -2
2S + 3(-2) = -2
2S - 6 = -2
2S = +4
S = +2
SO4
-2
S + 4O = -2
S + 4(-2) = -2
S - 8 = -2
S = +6
S = +6
2. Split into two half reactions (Oxidation and Reduction) S
2O
3
-2(aq) SO4-2 (aq)
Cl2(g) Cl
-(aq)
Oxidation- Reduction Reactions in
Acidic/Basic Solutions
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3. Balance both half Reactions
1) Balance all atoms except H and O S
2O
3
-2(aq) 2 SO4-2 (aq)
Cl2(g) 2 Cl-(aq)
2) Balance O atoms by adding H20 to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq)
3) Balance H atoms by adding H+ ions to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+
4) Balance electric charge by adding electrons (e-) to the more
positive side
S
2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
Cl2(g) + 2 e- 2 Cl
-(aq)
Ch. 19 Electrochemistry
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3. Balance both half Reactions
1) Balance all atoms except H and O S
2O
3
-2(aq) 2 SO4-2 (aq)
Cl2(g) 2 Cl-(aq)
2) Balance O atoms by adding H20 to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq)
3) Balance H atoms by adding H+ ions to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+
4) Balance electric charge by adding electrons (e-) to the more
positive side
S
2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
Cl2(g) + 2 e- 2 Cl
-(aq)
4. Combine the two half reactions and obtain a final balanced equation
a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
x4 (Cl2(g) + 2 e- 2 Cl-(aq))
4 Cl2(g) + 8 e- 8 Cl-(aq)
b. Simplify the balanced equation
S2O
3
-2(aq) + 5 H20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H+(aq) + 8 Cl
-(aq)
Ch. 19 Electrochemistry
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3. Balance both half Reactions
1) Balance all atoms except H and O S
2O
3
-2(aq) 2 SO4-2 (aq)
Cl2(g) 2 Cl-(aq)
2) Balance O atoms by adding H20 to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq)
3) Balance H atoms by adding H+ ions to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+
4) Balance electric charge by adding electrons (e-) to the more
positive side
S
2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
Cl2(g) + 2 e- 2 Cl
-(aq)
4. Combine the two half reactions and obtain a final balanced equation
a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
x4 (Cl2(g) + 2 e- 2 Cl-(aq))
4 Cl2(g) + 8 e- 8 Cl-(aq)
b. Simplify the balanced equation
S2O
3
-2(aq) + 5 H20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H+(aq) + 8 Cl
-(aq)
Oxidizing Agent?
Reducing Agent?
Ch. 19 Electrochemistry
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3. Balance both half Reactions
1) Balance all atoms except H and O S
2O
3
-2(aq) 2 SO4-2 (aq)
Cl2(g) 2 Cl-(aq)
2) Balance O atoms by adding H20 to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq)
3) Balance H atoms by adding H+ ions to one side of the equation
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+
4) Balance electric charge by adding electrons (e-) to the more
positive side
S
2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
Cl2(g) + 2 e- 2 Cl
-(aq)
4. Combine the two half reactions and obtain a final balanced equation
a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out
S2O
3
-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-
x4 (Cl2(g) + 2 e- 2 Cl-(aq))
4 Cl2(g) + 8 e- 8 Cl-(aq)
b. Simplify the balanced equation
S2O
3
-2(aq) + 5 H20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H+(aq) + 8 Cl
-(aq)
Oxidizing Agent? Cl2(g)
Reducing Agent? S2O
3
-2(aq)
Ch. 19 Electrochemistry
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If this was asking to balance in Basic solution, then there are two more steps.
5. Add as many OH- ions to both sides of the equation as there H
+ ions.
S2O
3
-2(aq) + 5 H20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H+(aq) + 8 Cl
-(aq) + 10 OH
-(aq)
6. each added OH- will react with H
+ to form H
2O
S
2O
3
-2(aq) + 5 H
20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H
20 + 8 Cl
-(aq)
*simplify the equation
S2O
3
-2(aq) + 4 Cl2(g) 2 SO4-2 (aq) + 5 H
20 + 8 Cl
-(aq)
Ch. 19 Electrochemistry
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Bibliography
Ebbing, D. D., & Gammon, S. D. (2013). General Chemistry. Cengage Learning.
General Chemistry Course Information. (2014). Retrieved from Dr. Sapna Gupta: http://drsapnag.manusadventures.com/index.php/general-chemistry