Chapter 6 Problems

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Chapter 6 Problems

• 6.19, 6.21, 6.24
• 6-29, 6-31, 6-39, 6.41
• 6-42, 6-48,

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6-19.

• A solution contains 0.0500 M Ca2 and 0.0300 M
  Ag. Can 99 of Ca2 be precipitated by sulfate
  without precipitating Ag? What will be the
  concentration of Ca2 when Ag2SO4 begins to
  precipitate?

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6-19.

• A solution contains 0.0500 M Ca2 and 0.0300 M
  Ag. Can 99 of Ca2 be precipitated by sulfate
  without precipitating Ag? What will be the
  concentration of Ca2 when Ag2SO4 begins to
  precipitate?

Ca2 at equilibrium
CaSO4 ? Ca2 SO42- Ksp 2.4 x 10-5
0.000500SO42-
CaSO4 ? Ca2 SO42- Ksp 2.4 x 10-5
Ca2SO42-
SO42- 0.048 M
Sep. is NOT feasible
QgtK
Ag2SO4 ? 2Ag SO42- Ksp 1.5 x 10-5
Q Ag2SO42-
Q 0.0320.000500
Q 4.3 x 10-5
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6-19.

• A solution contains 0.0500 M Ca2 and 0.0300 M
  Ag. Can 99 of Ca2 be precipitated by sulfate
  without precipitating Ag? What will be the
  concentration of Ca2 when Ag2SO4 begins to
  precipitate?

Ag2SO4 ? 2Ag SO42- Ksp 1.5 x 10-5
K/Ag22 SO42-
1.5 x 10-5/0.03002 SO42-
1.67 x 10-2 SO42-
Find Ca2
CaSO4 ? Ca2 SO42- Ksp 2.4 x 10-5
Ca21.67 x 10-2
Ca2 0.0014 M
About 2.8 remains in solution
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6.21

• If a solution containing 0.10 M Cl-, Br-, I- and
  Cr2O42- is treated with Ag, in what order will
  the anions precipitate?
• AgCl ? Ag Cl- Ksp 1.8 x 10-10AgCl
• AgBr ? Ag Br- Ksp 5.0 x 10-13AgBr
• AgI ? Ag I- Ksp 8.3 x 10-17AgI
• Ag2CrO4 ? 2Ag CrO4- Ksp 1.2 x
  10-12Ag2Cl

• AgCl ? Ag Cl- Ksp 1.8 x 10-10Ag0.1
• AgBr ? Ag Br- Ksp 5.0 x 10-13Ag0.1
• AgI ? Ag I- Ksp 8.3 x 10-17Ag0.1
• Ag2CrO4 ? 2Ag CrO4- Ksp 1.2 x
  10-12Ag20.1

1.8 x 10-9Ag 5.0 x 10-12Ag 8.3 x
10-16Ag 3.5 x 10-6Ag
SOLVE for Ag required at equilibrium
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6-24.
SnCl2(aq)

• The cumulative formation constant for SnCl2(aq)
  in 1.0 M NaNO3 is b212. Find the concentration
  of SnCl2 for a solution in which the
  concentration of Sn2 and Cl- are both somehow
  fixed at 0.20 M.

Sn2 (aq) 2Cl- (aq) ? SnCl2 (aq) b212
b212
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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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Effects of Complex Ion Formation on Solubility

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

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

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

Pb2 I- PbI
PbI I- PbI2 K2 1.4 x 101
PbI2 I- PbI3- K3 5.9
PbI3 I- PbI42- K4 3.6
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Acids and Bases Equilibrium

• Section 6-7

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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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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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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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Equilibrium and Water

• 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- ?
KW(X)(X) ?
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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

• What is the concentration of OH- in a solution of
  water that is 1.0 x 10-3 M in H (_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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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 acid.

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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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Example

• Calculate the hydroxide ion concentration in a
  0.0100 M sodium hypochlorite solution.
• OCl- H2O ? HOCl OH-

The acid dissociation constant 3.0 x 10-8
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1st Insurance Problem

• Challenge on page 120

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Chapter 8

• Activity

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Write out the equilibrium constant for the
following expression

• Fe3 SCN- D Fe(SCN)2

Q What happens to K when we add, say KNO3 ?
A Nothing should happen based on our K, our K
is independent of K NO3-
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K decreases when an inert salt is added!!! Why?
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8-1 Effect of Ionic Strength on Solubility of
Salts

• Consider a saturated solution of Hg2(IO3)2 in
  pure water. Calculate the concentration of
  mercurous ions.
• Hg2(IO3)2(s) D Hg22 2IO3- Ksp1.3x10-18

some - - -x x 2x some-x x 2x
I C E
A seemingly strange effect is observed when a
salt such as KNO3 is added. As more KNO3 is added
to the solution, more solid dissolves until
Hg22 increases to 1.0 x 10-6 M. Why?
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Increased solubility

• Why?
• Complex Ion?
• No
• Hg22 and IO3- do not form complexes with K or
  NO3-.
• How else?

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The Explanation

• Consider Hg22 and the IO3-

Electrostatic attraction
-
2
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The Explanation

• Consider Hg22 and the IO3-

Electrostatic attraction
-
2
Hg2(IO3)2(s) The Precipitate!!
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The Explanation

• Consider Hg22 and the IO3-

NO3-
K
NO3-
K
NO3-
NO3-
Electrostatic attraction
K
-
K
2
NO3-
NO3-
NO3-
NO3-
K
K
NO3-
NO3-
Add KNO3
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The Explanation

• Consider Hg22 and the IO3-

NO3-
K
K
NO3-
NO3-
NO3-
K
-
K
2
NO3-
NO3-
NO3-
K
K
NO3-
NO3-
NO3-
Hg22 and IO3- cant get CLOSE ENOUGH to form
Crystal lattice
Or at least it is a lot Harder to form crystal
lattice
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The potassium hydrogen tartrate example
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Alright, what do we mean by Ionic strength?

• Ionic strength is dependent on the number of ions
  in solution and their charge.
• Ionic strength (m) ½ (c1z12 c2z22 )

Or Ionic strength (m) ½ S cizi2
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Examples

• Calculate the ionic strength of (a) 0.1 M
  solution of KNO3 and (b) a 0.1 M solution of
  Na2SO4 (c) a mixture containing 0.1 M KNO3 and
  0.1 M Na2SO4.

(m) ½ (c1z12 c2z22 )
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Alright, thats great but how does it affect the
equilibrium constant?

• Activity Ac Cgc
• AND

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Relationship between activity and ionic strength
Debye-Huckel Equation
m ionic strength of solution g activity
coefficient Z Charge on the species x a
effective diameter of ion (nm)
2 comments

1. What happens to g when m approaches zero?
2. Most singly charged ions have an effective radius
   of about 0.3 nm

Anyway we generally dont need to calculate g
can get it from a table
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Activity coefficients are related to the hydrated
radius of atoms in molecules
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Relationship between m and g
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Back to our original problem

• Consider a saturated solution of Hg2(IO3)2 in
  pure water. Calculate the concentration of
  mercurous ions.
• Hg2(IO3)2(s) D Hg22 2IO3- Ksp1.3x10-18

At low ionic strengths g -gt 1
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Back to our original problem

• Consider a saturated solution of Hg2(IO3)2 in
  pure water. Calculate the concentration of
  mercurous ions.
• Hg2(IO3)2(s) D Hg22 2IO3- Ksp1.3x10-18

In 0.1 M KNO3 - how much Hg22 will be dissolved?
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Back to our original problem

• Consider a saturated solution of Hg2(IO3)2 in
  pure water. Calculate the concentration of
  mercurous ions.
• Hg2(IO3)2(s) D Hg22 2IO3- Ksp1.3x10-18

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