Chapter 6 Problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 Comments on “Lab Report & Pop Rocks”
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Transcript of Chapter 6 Problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 Comments on “Lab Report & Pop Rocks”
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Chapter 6 Problems
6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24
Comments on “Lab Report & Pop Rocks”
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6.6
From the equationsHOCl H+ + OCl- K = 3.0 x 10-8
HOCl + OBr- HOBr + OCl- K= 15
Find K for
HOBr H+ + OBr—
HOBr + OCl- HOCl + OBr-
Flip
K’ = 1/K= 1/15
K= ?K= 3.0 x 10-8/15
K= 0.2 x 10-8
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6.9 The formation of Tetrafluorethlene
from its elements is highly exothermic. 2 F2 (g) + 2 C (s) F2C=CF2 (g)
(a) if a mixture of F2, graphite, and C2F4 is at equilibrium in a closed container, will the reaction go to the right or to the left if F2 is added?
(b) Rare bacteria … eat C2F4 and make teflon for cell walls. Will the reaction go to the right or to the left if these bacteria are added?
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6.9
The formation of Tetrafluorethlene from its elements is highly exothermic. 2 F2 (g) + 2 C (s) F2C=CF2 (g)
(C) will the reaction go to the right or to the left if graphite is added?
(d) will reaction go left or right if container is crushed to one-eighths of original volume?
(e) Does “Q” get larger or smaller if vessel is Heated?
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6-15. What concentration of Fe(CN)6
4- is in equilibrium with 1.0 uM Ag+ and Ag4Fe(CN)6 (s).
Ag4Fe(CN)6 4Ag+ + Fe(CN)64-
Ksp = [Ag+]4 [Fe(CN)64-]
8.5 x 10-45 = [1.0 x 10-6]4 [Fe(CN)64-]
[Fe(CN)64-] = 8.5 x 10-21 M = 8.5 zM
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6-16.
Cu4(OH)6(SO4) 4 Cu2+ + 6OH- + SO42-
I’d first set up an ICE table:
Cu4(OH)6(SO4) 4 Cu2+ + 6 OH + SO42-
I Some - 1.0 x 10-6 M -C -x + 4x Fixed at 1.0 x 10-6 M + xE Some –x + 4x Fixed at 1.0 x 10-6 M + x
Ksp = [Cu2+]4 [OH-]6 [SO42-] = 2.3 x 10-69
Ksp = [4x]4 [1.0 x 10-6]6 [x] = 2.3 x 10-69
X = 9.75 x 10-8 M Cu2+ = 4x = 3.90 x 10-7 M
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Chapter 6
Chemical Equilibrium
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Chemical Equilibrium
Equilibrium Constant Solubility product (Ksp) Common Ion Effect Separation by precipitation Complex formation
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Separation by Precipitation
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Separation by Precipitation
Complete separation can mean a lot … we should define complete.
Complete means that the concentration of the less soluble material has decreased to 1 X 10-
6M or lower before the more soluble material begins to precipitate
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Separation by Precipitation
EXAMPLE: Can Fe+3 and Mg+2 be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.
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Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Fe3+
Mg2+
Mg2+
Fe3+Fe3+
Mg2+
Mg2+
Add OHAdd OH--
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Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Mg2+
Mg2+ Mg2+
Mg2+
Fe(OH)Fe(OH)33(s)(s)
What is the [OH-] when this happens
^
@ equilibrium
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EXAMPLE: Separate Iron and Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume [Fe+3]eq = 1.0 X 10-6M when “completely” precipitated.
What will be the [OH-] @ equilibrium required to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ?Ksp = [Fe+3][OH-]3 = 2 X 10-39
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EXAMPLE: Separate Iron and Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
333 102][ OH
11103.1][ OH
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Dealing with Mg2+
Find [OH-] to start precipitating Mg2+
• Conceptually – •Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH-
Evaluate Q• If
•Q>K•Q<K•Q=K
“Left”“Right”“Equilibrium”
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Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Mg2+
Mg2+ Mg2+
Mg2+
Fe(OH)Fe(OH)33(s)(s)
[OH-]̂@ equilibrium
Is this [OH-] (that is in solution) great enough to start precipitating Mg2+?
= 1.3 x 10-11
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Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Mg2+
Mg2+ Mg2+
Mg2+
Fe(OH)Fe(OH)33(s)(s)
[OH-]̂@ equilibrium
Is this [OH-] (that is in solution) great enough to start precipitating Mg2+?
= 1.3 x 10-11
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EXAMPLE: Separate Iron and Magnesium?
What [OH-] is required to begin the precipitation of Mg(OH)2?
[Mg+2] = 0.10 M
Ksp =
[OH-] = 8.4 X 10-6M
[Mg2+]eq = 0.09999999999999999 M
Really, Really close to 0.1 M
(0.10 M)[OH-]2 = 7.1 X 10-12
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EXAMPLE: Separate Iron and Magnesium?
[OH-] to ‘completely’ remove Fe3+ = 1.3 X 10-11 M
[OH-] to start removing Mg2+ = 8.4 X 10-6M
“All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!!
^̂
@ equilibrium@ equilibrium
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EXAMPLE: Separate Iron and Magnesium?
Q vs. K
Ksp = [Mg2+][OH-]2 = 7.1 X 10-12
Q = [0.10 M ][1.3 x 10-11 ]2 = 1.69 x 10-23
Q<K Reaction will proceed to “Right”
Mg(OH)2(s) Mg2+ + 2OH-
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Dealing with Mg2+
Find [OH-] to start precipitating Mg2+
• Conceptually – •Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH-
Evaluate Q• If
•Q>K•Q<K•Q=K
“Left”“Right”“Equilibrium”
NO PPT
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“Real Example” Consider a 1 liter solution that
contains 0.3 M Ca2+ and 0.5 M Ba2+. Can you separate the ions by adding
Sodium Carbonate? Sodium Chromate ? Sodium Fluoride? Sodium Hydroxide? Sodium Iodate? Sodium Oxylate?
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An example
Consider Lead Iodide PbI2 (s) Pb2+ + 2I- Ksp
= 7.9 x 10-9
What should happen if I- is added to a solution?
Should the solubility go up or down?
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Complex Ion Formation
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Complex Formation
complex ions (also called coordination ions)
Lewis Acids and Basesacid => electron pair acceptor (metal)base => electron pair donor (ligand)
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Pb2+ I- I- Pb2+
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Pb2+ I- I- Pb2+
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Pb2+ I- I- Pb2+
I- I-
Pb2+ I-
I-
I-
I-
I-
I-
Pb2+
Pb2+
I-
I- I-
I-
I- I-
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Pb2+ I- I- Pb2+
I- I-
Pb2+
I-
I-
I- I-
I- I-
I-
I- I-
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
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Pb2+ I- I- Pb2+
I- I-
Pb2+
I-
I-
I- I-
I- I-
I-
I- I-
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
I-
I- I-
I-
I- I-
I-
I-
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Pb2+ I- I- Pb2+
I- I-
Pb2+
I-
I-
I-
I-
I- I-
I-
I-
I-
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
Pb2+ I- I- Pb2+
I-
I-
I-
I-
I- I-
I-
I-
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Effects of Complex Ion Formation on Solubility
Consider the addition of I- to a solution of Pb+2 ions
Pb2+ + I- <=> PbI+
PbI+ + I- <=> PbI2 K2 = 1.4 x 101
PbI2 + I- <=> PbI3- K3 =5.9
PbI3+ I- <=> PbI42- K4 = 3.6
221 100.1
]][[
][x
IPb
PbIK
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Effects of Complex Ion Formation on Solubility
Consider the addition of I- to a solution of Pb+2 ions
Pb2+ + I- <=> PbI+
PbI+ + I- <=> PbI2 K2 = 1.4 x 101
Pb2+ + 2I- <=> PbI2 K’ =?
221 100.1
]][[
][x
IPb
PbIK
Overall constants are designated with
This one is
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Protic Acids and Bases
Section 6-7
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Question
Can you think of a salt that when dissolved in water is not an acid nor a base?
Can you think of a salt that when dissolved in water IS an acid or base?
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Protic Acids and Bases - Salts
Consider Ammonium chloride Can ‘generally be thought of as the
product of an acid-base reaction.
NH4+Cl- (s) NH4
+ + Cl-
From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution
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Protic Acids and Bases
Conjugate Acids and Bases in the B-L concept
CH3COOH + H2O CH3COO- + H3O+ acid + base <=> conjugate base + conjugate acid
conjugate base => what remains after a B-L acid donates its protonconjugate acid => what is formed when a B-L base accepts a proton
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Question: Question: Calculate the Concentration of H+ and OH- in Pure
water at 250C.
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EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.
H2O H+ + OH-
Initial liquid - -
Change -x +x +x
Equilibrium Liquid-x +x +x
KW=(X)(X) = 1.01 X 10-14
Kw = [H+][OH-] = 1.01 X 10-14
(X) = 1.00 X 10-7
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Example
Concentration of OH-
if [H+] is 1.0 x 10-3 M @ 25 oC?
Kw = [H+][OH-]
1 x 10-14 = [1 x 10-3][OH-]
1 x 10-11 = [OH-]
“From now on, assume the temperature to be 25oC unless otherwise stated.”
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pH
~ -3 -----> ~ +16pH + pOH = - log Kw = pKw = 14.00
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Is there such a thing as Pure Water? In most labs the answer is NO Why?
A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value.
CO2 + H2O HCO3- + H+
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6-9 Strengths of Acids and 6-9 Strengths of Acids and BasesBases
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Strong Bronsted-Lowry Acid
A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). HCl as example
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Strong Bronsted-Lowry Base
Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added.
Example: NH2- (the amide ion)
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Weak Bronsted-Lowry acid
One that DOES not donate all of its acidic protons to water molecules in aqueous solution.
Example? Use of double arrows! Said to reach
equilibrium.
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Weak Bronsted-Lowry base
Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added.
example: NH3
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Common Classes of Weak Acids and Bases
Weak Acids carboxylic acids ammonium ions
Weak Bases amines carboxylate anion
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Weak Acids and Bases
HA H+ + A-
HA + H2O(l) H3O+ + A-
Ka
][
]][[
HA
AHKa
][
]][[ 3
HA
AOHKa
Ka’s ARE THE SAME
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Weak Acids and Bases
B + H2O BH+ + OH-
][
]][[
B
OHBHKb
Kb
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Relation Between KRelation Between Kaa and and KKbb
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Relation between Ka and Kb
Consider Ammonia and its conjugate base.
][
]][[
4
33
NH
OHNHKb
][
]][[
3
4
NH
OHNHKa
NH3 + H2O NH4+ + OH-
Ka
NH4+ + H2O NH3 + H3O+
Kb
H2O + H2O OH- + H3O+
][
]][[
][
]][[
3
4
4
33
NH
OHNH
NH
OHNHK
][][ 3 OHOHKw
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Example
The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base.
Kw = Ka x Kb
a
wb K
KK
105
14
107.51075.1
100.1
bK
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1st Insurance Problem
Challenge on page 120Challenge on page 120