Chapter 6 Problems
description
Transcript of 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?
MM 000500.001.0*0500.0 Ca2+ at equilibrium
CaSO4 Ca2+ + SO42- Ksp
= 2.4 x 10-5 = [Ca2+][SO42-]CaSO4 Ca2+ + SO4
2- Ksp= 2.4 x 10-5 = [0.000500][SO4
2-]
[SO42-] = 0.048 M
Ag2SO4 2Ag+ + SO42- Ksp = 1.5 x 10-5
Q = [Ag+]2[SO42-] Q = [0.03]2[0.000500] Q = 4.3 x 10-5
Q>K
Sep.
is N
OT fe
asib
le
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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/[Ag2+]2 = [SO42-] 1.5 x 10-5/[0.0300]2 = [SO4
2-]
1.67 x 10-2= [SO42-]
Find Ca2+
CaSO4 Ca2+ + SO42- Ksp
= 2.4 x 10-5
[Ca2+] = 0.0014 M
= [Ca2+][1.67 x 10-2]
About 2.8 % remains in solution
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6.21 If a solution containing 0.10 M Cl-, Br-, I- and Cr2O4
2- is treated with Ag+, in what order will the anions precipitate?
AgCl Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][Cl]
AgBr Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][Br]
AgI Ag+ + I- Ksp = 8.3 x 10-17=[Ag][I]
Ag2CrO4 2Ag+ + CrO4
- Ksp = 1.2 x 10-12=[Ag]2[Cl]
AgCl Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][0.1]
AgBr Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][0.1]
AgI Ag+ + I- Ksp = 8.3 x 10-17=[Ag][0.1]
Ag2CrO4 2Ag+ + CrO4
- Ksp = 1.2 x 10-12=[Ag]2[0.1]
SOLVE for Ag+ required at equilibrium
1.8 x 10-9=[Ag]5.0 x 10-12=[Ag]8.3 x 10-16=[Ag]3.5 x 10-6=[Ag]
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6-24. The cumulative formation constant for SnCl2(aq) in
1.0 M NaNO3 is 2=12. Find the concentration of SnCl2 for a solution in which the concentration of Sn2+ and Cl- are both somehow fixed at 0.20 M.
SnCl2(aq)
Sn2+ (aq) + 2Cl- (aq) SnCl2 (aq) 2=12
2=12 222
]][[)]([
ClSnaqSnCl
22
]20.0][20.0[)]([ aqSnCl
MaqSnCl 096.0)]([ 2
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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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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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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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Acids and Bases & EquilibriumEquilibrium
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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Question: Question: Calculate the Concentration of H+ and OH- in Pure
water at 250C.
Equilibrium and Equilibrium and WaterWater
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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) = ?
Kw = [H+][OH-] = ?
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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
What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 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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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 acid.
][
]][[
4
33
NH
OHNHKa
][
]][[
3
4
NH
OHNHKb
NH3 + H2O NH4+ + OH-
Kb
NH4+ + H2O NH3 + H3O+
Ka
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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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
][
]][[
OCl
OHHOClKb
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1st Insurance Problem
Challenge on page 120Challenge on page 120
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Chapter 8
ActivityActivity
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Write out the equilibrium constant for the following expression
Fe3+ + SCN- 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
-
]][[
])([3
2
SCNFe
SCNFeK
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K decreases when an inert salt is added!!! Why?
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8-1 Effect of Ionic Strength 8-1 Effect of Ionic Strength on Solubility of Saltson Solubility of Salts
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) Hg22+ + 2IO3
- Ksp=1.3x10-18
A seemingly strange effectseemingly 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?
1823
22 103.1]][[ IOHgK sp
IICCEE
somesome -- - --x-x +x+x +2x+2xsome-xsome-x +x+x +2x+2x
182 103.1]2][[ xxK sp7109.6][ x
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Increased solubility
Why? Complex Ion?
No Hg2
2+ and IO3- do not form complexes
with K+ or NO3-.
How else?
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The Explanation
Consider Hg22+ and the IO3
-
-2+
Electrostatic attraction
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The Explanation
Consider Hg22+ and the IO3
-
2+
Electrostatic attraction
-
Hg2(IO3)2(s) The Precipitate!!The Precipitate!!
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The Explanation
Consider Hg22+ and the IO3
-
-2+
Electrostatic attraction
Add KNO3
K+ K+
K+K+
K+K+
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
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The Explanation
Consider Hg22+ and the IO3
-
-2+
K+ K+
K+K+
K+K+
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
Hg22+ and IO3
- can’t getCLOSE ENOUGH to form Crystal latticeOr at least it is a lot “Harder” to form crystal lattice
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The potassium hydrogen tartrate example
K+-O
O
OH
OH
O
OH
potassium hydrogen tartrate
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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 () = ½ (c1z12+ c2z2
2 + …)
Or Ionic strength (m) = ½ 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.() = ½ (c1z1
2+ c2z22 + …)
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Alright, that’s great but how does it affect the equilibrium constant?
Activity = Ac = [C]c
AND
bB
baA
a
dD
dcC
c
bB
aA
dD
cC
BA
DC
AA
AAK
][][
][][
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Relationship between activity and ionic strength
x
x
xz
3.31
51.0log
2
Debye-Huckel Equation
2 comments
= ionic strength of solution = activity coefficientZ = Charge on the species x = effective diameter of ion (nm)
(1)What happens to when approaches zero?(2)Most singly charged ions have an effective radius of about 0.3 nm
Anyway … we generally don’t need to calculate – can get it from a table
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Activity coefficients are Activity coefficients are related to the hydrated related to the hydrated radius of atoms in radius of atoms in moleculesmolecules
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Relationship between and
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Back to our original problem
223
22
2
3223
22
][][ IOHgIOHgsp IOHgAAK
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) Hg22+ + 2IO3
- Ksp=1.3x10-18
At low ionic strengths -> 1
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Back to our original problem
223
22
2
3223
22
][][ IOHgIOHgsp IOHgAAK
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) Hg22+ + 2IO3
- Ksp=1.3x10-18
In 0.1 M KNO3 - how much Hg22+ will be dissolved?
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Back to our original problem
223
22
2
3223
22
][][ IOHgIOHgsp IOHgAAK
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) Hg22+ + 2IO3
- Ksp=1.3x10-18
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