Chapter 6 Problems

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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 equations
• HOCl ?H OCl- K 3.0 x 10-8
• HOCl OBr- ? HOBr OCl- K 15
• Find K for
• HOBr ? H OBr

Flip
HOBr OCl- ? HOCl OBr-
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) ? F2CCF2 (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) ? F2CCF2 (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)64- is in equilibrium
  with 1.0 uM Ag and Ag4Fe(CN)6 (s).
• Ag4Fe(CN)6 ? 4Ag Fe(CN)64-
• Ksp Ag4 Fe(CN)64-
• 8.5 x 10-45 1.0 x 10-64 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- Id 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
x E Some x 4x Fixed at 1.0 x 10-6 M x
Ksp Cu24 OH-6 SO42- 2.3 x 10-69
Ksp 4x4 1.0 x 10-66 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 Fe3 and Mg2 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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Add OH-
Mg2
Mg2
Fe3
Fe3
Fe3
Fe3
Mg2
Mg2
Mg2
Fe3
Mg2
Fe3
Mg2
Fe3
Mg2
Fe3
Fe3
Fe3
Mg2
Mg2
Mg2
Fe3
Fe3
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Mg2
Mg2
Mg2
Mg2
Mg2
_at_ equilibrium
Fe3
Mg2
What is the OH- when this happens

Mg2
Mg2
Mg2
Mg2
Mg2
Fe(OH)3(s)
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EXAMPLE Separate Iron and Magnesium?

• Ksp Fe3OH-3 2 X 10-39
• Ksp Mg2OH-2 7.1 X 10-12

Assume Fe3eq 1.0 X 10-6M when completely
precipitated.
What will be the OH- _at_ equilibrium required to
reduce the Fe3 to Fe3 1.0 X 10-6M ?

• Ksp Fe3OH-3 2 X 10-39

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EXAMPLE Separate Iron and Magnesium?

• Ksp Fe3OH-3 2 X 10-39

(1.0 X 10-6M)OH-3 2 X 10-39
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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
• QgtK
• QltK
• QK

Left Right Equilibrium
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Mg2
Mg2
Mg2
Mg2
Mg2
_at_ equilibrium
Fe3
Mg2
1.3 x 10-11
OH-

Mg2
Mg2
Is this OH- (that is in solution) great enough
to start precipitating Mg2?
Mg2
Mg2
Mg2
Fe(OH)3(s)
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Mg2
Mg2
Mg2
Mg2
Mg2
_at_ equilibrium
Fe3
Mg2
1.3 x 10-11
OH-

Mg2
Mg2
Is this OH- (that is in solution) great enough
to start precipitating Mg2?
Mg2
Mg2
Mg2
Fe(OH)3(s)
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EXAMPLE Separate Iron and Magnesium?

• What OH- is required to begin the precipitation
  of Mg(OH)2?
• Mg2 0.10 M

Really, Really close to 0.1 M
Mg2eq 0.09999999999999999 M

• Ksp

(0.10 M)OH-2 7.1 X 10-12

• OH- 8.4 X 10-6M

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EXAMPLE Separate Iron and Magnesium?
_at_ equilibrium

• 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!!
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EXAMPLE Separate Iron and Magnesium?

• Q vs. K
• Ksp Mg2OH-2 7.1 X 10-12
• Q 0.10 M 1.3 x 10-11 2 1.69 x 10-23

Mg(OH)2(s) ? Mg2 2OH-
QltK
Reaction will proceed to Right
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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
• QgtK
• QltK
• QK

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 Bases
• acid gt electron pair acceptor (metal)
• base gt electron pair donor (ligand)

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Pb2
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Effects of Complex Ion Formation on Solubility

• Consider the addition of I- to a solution of Pb2
  ions

Pb2 I- ltgt PbI
PbI I- ltgt PbI2 K2 1.4 x 101
PbI2 I- ltgt PbI3- K3 5.9
PbI3 I- ltgt PbI42- K4 3.6
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Effects of Complex Ion Formation on Solubility

• Consider the addition of I- to a solution of Pb2
  ions

Pb2 I- ltgt PbI
PbI I- ltgt PbI2 K2 1.4 x 101
Pb2 2I- ltgt PbI2 K ?
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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.

NH4Cl- (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 ltgt conjugate
  base conjugate acid
• conjugate base gt what remains after a B-L acid
  donates its proton
• conjugate acid gt what is formed when a B-L base
  accepts a proton

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• 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 HOH- 1.01 X 10-14
KW(X)(X) 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 _at_ 25 oC?

From now on, assume the temperature to be 25oC
unless otherwise stated.
Kw HOH-
1 x 10-14 1 x 10-3OH-
1 x 10-11 OH-
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pH

• -3 -----gt 16
• pH 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 Bases
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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
Ka

• HA H A-

Kas ARE THE SAME
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Weak Acids and Bases
Kb

• B H2O BH OH-

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Relation Between Ka and Kb
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Relation between Ka and Kb

• Consider Ammonia and its conjugate base.

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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
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1st Insurance Problem

• Challenge on page 120