Transcript Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Adding NaOH to HCHO2
initial HCHO
added
30.0mL
35.0
5.0
10.0
25.0
mL
NaOH
NaOH
2 solution
0.00050 mol point
0.00100
0.00250
0.00200
0.00150
equivalence
NaOH2xs
HCHO
pH = 3.56
0.00250
11.96
12.22
2.37
3.14
mol CHO2−
[CHO2−]init = 0.0500 M
−] = 1.7 x 10-6
[OH
added
eq12.5 mL NaOH
added
40.0 mL NaOH
pH
= 8.23
0.00125
mol HCHO
0.00150 mol NaOH2xs
pH = 3.74 = pKa
pH = 12.36
half-neutralization
added 50.0
15.0 mL NaOH
0.00100 mol NaOH
0.00250
HCHO2xs
pH = 12.52
3.92
added 20.0 mL NaOH
0.00050 mol HCHO2
pH = 4.34
Tro, Chemistry: A Molecular
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2
Titrating Weak Acid with a Strong Base
• the initial pH is that of the weak acid solution
– calculate like a weak acid equilibrium problem
• e.g., 15.5 and 15.6
• before the equivalence point, the solution
becomes a buffer
– calculate mol HAinit and mol A−init using reaction
stoichiometry
– calculate pH with Henderson-Hasselbalch using mol
HAinit and mol A−init
• half-neutralization pH = pKa
Tro, Chemistry: A Molecular
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Titrating Weak Acid with a Strong Base
• at the equivalence point, the mole HA = mol Base, so
the resulting solution has only the conjugate base
anion in it before equilibrium is established
– mol A− = original mole HA
• calculate the volume of added base like Ex 4.8
– [A−]init = mol A−/total liters
– calculate like a weak base equilibrium problem
• e.g., 15.14
• beyond equivalence point, the OH is in excess
– [OH−] = mol MOH xs/total liters
– [H3O+][OH−]=1 x 10-14
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Approach
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Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the volume of KOH at the
equivalence point
Write an equation for
the reaction for B with
HA.
Use Stoichiometry to
determine the volume
of added B
0.0400L NO2 
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L
 0.0400 L
1 mL
0.100mol NO2 1 mol KOH
1 L KOH


1 L NO2
1 mol NO2 0.200mol KOH
 0.0200L KOH
0.0200 L 
Tro, Chemistry: A Molecular
Approach
8
1 mL
 20.0 mL
0.001 L
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH after adding 5.00 mL KOH
Write an equation
for the reaction for
B with HA.
Determine the
moles of HAbefore &
moles of added B
Make a
stoichiometry table
and determine the
moles of HA in
excess and moles
A made
Tro, Chemistry: A Molecular
Approach
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L 0.100 mol HNO 2

 0.00400 mol HNO 2
1 mL
1L
0.001 L 0.200 mol KOH
5.00 mL 

 0.00100 mol KOH
1 mL
1L
HNO2
mols Before 0.00400
mols added
mols After 0.00300
9
NO2-
OH−
0
≈0
-
0.00100
0.00100
≈0
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.
Write an equation for
the reaction of HA
with H2O
Determine Ka and pKa
for HA
Use the HendersonHasselbalch Equation
to determine the pH
Tro, Chemistry: A Molecular
Approach
HNO2 + H2O  NO2 + H3O+
Table 15.5 Ka = 4.6 x 10-4


pK a   log K a   log 4.6 104  3.15
 NO2  

pH  pK a  log
 HNO2 


HNO
NO2OH−
 02.00100
pH  3.15  log
  2.67
0
≈0
mols Before 0.00400
 0.00300
0.00100
mols added
mols After 0.00300 0.00100 ≈ 0
10
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH at
the half-equivalence point
Write an equation
for the reaction for
B with HA.
Determine the
moles of HAbefore &
moles of added B
Make a
stoichiometry table
and determine the
moles of HA in
excess and moles
A made
Tro, Chemistry: A Molecular
Approach
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L 0.100 mol HNO 2

 0.00400 mol HNO 2
1 mL
1L
at half-equivalence, moles KOH = ½ mole HNO2
HNO2
mols Before 0.00400
mols added
mols After 0.00200
11
NO2-
OH−
0
≈0
-
0.00200
0.00200
≈0
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is
titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence point.
Write an equation for
the reaction of HA
with H2O
Determine Ka and pKa
for HA
Use the HendersonHasselbalch Equation
to determine the pH
Tro, Chemistry: A Molecular
Approach
HNO2 + H2O  NO2 + H3O+
Table 15.5 Ka = 4.6 x 10-4


pK a   log K a   log 4.6 104  3.15
 NO2  

pH  pK a  log
 HNO2 


HNO
NO2OH−
 02.00200
pH  3.15  log
  3.15
0
≈0
mols Before 0.00400
 0.00200
0.00200
mols added
mols After 0.00200 0.00200 ≈ 0
12
Titration Curve of a Weak Base with
a Strong Acid
Tro, Chemistry: A Molecular
Approach
13
Titration of a Polyprotic Acid
• if Ka1 >> Ka2, there will be two equivalence
points in the titration
– the closer the Ka’s are to each other, the less
distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M
NaOH
Tro, Chemistry: A Molecular
Approach
14
Monitoring pH During a Titration
• the general method for monitoring the pH during the
course of a titration is to measure the conductivity of
the solution due to the [H3O+]
– using a probe that specifically measures just H3O+
• the endpoint of the titration is reached at the
equivalence point in the titration – at the inflection
point of the titration curve
• if you just need to know the amount of titrant added
to reach the endpoint, we often monitor the titration
with an indicator
Tro, Chemistry: A Molecular
Approach
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Monitoring pH During a Titration
Tro, Chemistry: A Molecular
Approach
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Indicators
• many dyes change color depending on the pH of the solution
• these dyes are weak acids, establishing an equilibrium with the
H2O and H3O+ in the solution
HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
• the color of the solution depends on the relative
concentrations of Ind:HInd
– when Ind:HInd ≈ 1, the color will be mix of the colors of Ind
and HInd
– when Ind:HInd > 10, the color will be mix of the colors of Ind
– when Ind:HInd < 0.1, the color will be mix of the colors of HInd
Tro, Chemistry: A Molecular
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Phenolphthalein
18
Methyl Red
H
C
(CH3)2N
H
C
C
H
C
C
C
H
N
(CH3)2N
C
OH-
C
N
N
CH
C
C
H
H
C
H
C
NaOOC
H
C
C
C
H
N
H
C
H
H3O+
H
C
N
H
C
N
C
C
H
CH
C
C
H
NaOOC
Tro, Chemistry: A Molecular
Approach
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Monitoring a Titration with
an Indicator
• for most titrations, the titration curve shows a
very large change in pH for very small
additions of base near the equivalence point
• an indicator can therefore be used to
determine the endpoint of the titration if it
changes color within the same range as the
rapid change in pH
– pKa of HInd ≈ pH at equivalence point
Tro, Chemistry: A Molecular
Approach
20
Acid-Base Indicators
21
Solubility Equilibria
• all ionic compounds dissolve in water to some
degree
– however, many compounds have such low
solubility in water that we classify them as
insoluble
• we can apply the concepts of equilibrium to
salts dissolving, and use the equilibrium
constant for the process to measure relative
solubilities in water
Tro, Chemistry: A Molecular
Approach
22
Solubility Product
• the equilibrium constant for the dissociation of a solid
salt into its aqueous ions is called the solubility
product, Ksp
• for an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
• the solubility product would be
Ksp = [Mm+]n[Xn−]m
• for example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
• and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular
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Tro, Chemistry: A Molecular
Approach
24
Molar Solubility
• solubility is the amount of solute that will dissolve in a
given amount of solution
– at a particular temperature
• the molar solubility is the number of moles of solute that
will dissolve in a liter of solution
– the molarity of the dissolved solute in a saturated solution
• for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
molar solubility  n  m 
Tro, Chemistry: A Molecular
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K sp
n
n
m
m
Ex 16.8 – Calculate the molar solubility of PbCl2
in pure water at 25C
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
Tro, Chemistry: A Molecular
Approach
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Initial
26
Ex 16.8 – Calculate the molar solubility of PbCl2
in pure water at 25C
Substitute into the
Ksp expression
Find the value of
Ksp from Table
16.2, plug into the
equation and
solve for S
Tro, Chemistry: A Molecular
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Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
3
2+]
[Pb
K sp  4 S
[Cl−]
Initial K
0
05
1.17  10
sp
3
3
 S +S

Change
+2S
4
4
Equilibrium
S  1.43S 10  2 M 2S
27
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Tro, Chemistry: A Molecular
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Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
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PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Initial
[Pb2+]
[Br−]
0
0
Change
+(1.05 x 10-2) +2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
29
(2.10 x 10-2)
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Substitute into the
Ksp expression
plug into the
equation and
solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10-2)(2.10 x 10-2)2


[Pb2+]
2
2
K

1
.
05

10
2
.
10

10
sp
Initial
0
0
6
K

4
.
63

10
Change
+(1.05 x 10-2) +2(1.05 x 10-2)
sp
Equilibrium
Tro, Chemistry: A Molecular
Approach

[Br−2]
30
(1.05 x 10-2)
(2.10 x 10-2)
Ksp and Relative Solubility
• molar solubility is related to Ksp
• but you cannot always compare solubilities of
compounds by comparing their Ksps
• in order to compare Ksps, the compounds must
have the same dissociation stoichiometry
Tro, Chemistry: A Molecular
Approach
31
The Effect of Common Ion on Solubility
• addition of a soluble salt that contains one of
the ions of the “insoluble” salt, decreases the
solubility of the “insoluble” salt
• for example, addition of NaCl to the solubility
equilibrium of solid PbCl2 decreases the
solubility of PbCl2
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the equilibrium to the left
Tro, Chemistry: A Molecular
Approach
32
Ex 16.10 – Calculate the molar solubility of CaF2
in 0.100 M NaF at 25C
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
Tro, Chemistry: A Molecular
Approach
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
[F−]
0
0.100
Change
+S
+2S
Equilibrium
S
0.100 + 2S
Initial
33
Ex 16.10 – Calculate the molar solubility of CaF2
in 0.100 M NaF at 25C
Substitute into the
Ksp expression
assume S is small
Find the value of
Ksp from Table
16.2, plug into the
equation and
solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
K sp
Initial
ChangeS

2+] 2
[Ca
 S 0.100 
0 10
1.46  10
+S 2
0.100 
Equilibrium
S 8
S  1.46  10
Tro, Chemistry: A Molecular
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[F−]
0.100
+2S
M
0.100 + 2S
The Effect of pH on Solubility
• for insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxide
– and the lower the pH, the higher the solubility
– higher pH = increased [OH−]
M(OH)n(s)  Mn+(aq) + nOH−(aq)
• for insoluble ionic compounds that contain anions of
weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l)
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Precipitation
• precipitation will occur when the concentrations of the
ions exceed the solubility of the ionic compound
• if we compare the reaction quotient, Q, for the current
solution concentrations to the value of Ksp, we can
determine if precipitation will occur
– Q = Ksp, the solution is saturated, no precipitation
– Q < Ksp, the solution is unsaturated, no precipitation
– Q > Ksp, the solution would be above saturation, the salt
above saturation will precipitate
• some solutions with Q > Ksp will not precipitate unless
disturbed – these are called supersaturated solutions
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Approach
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precipitation occurs if Q >
Ksp
Tro, Chemistry: A Molecular
Approach
a supersaturated solution will precipitate if a
seed crystal is added
37
Selective Precipitation
• a solution containing several different cations
can often be separated by addition of a
reagent that will form an insoluble salt with
one of the ions, but not the others
• a successful reagent can precipitate with more
than one of the cations, as long as their Ksp
values are significantly different
Tro, Chemistry: A Molecular
Approach
38
Ex 16.13 What is the minimum [OH−] necessary to just
begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
2
 2
Q  [Mg ][OH ]
Q  K sp
[0.059 ][OH  ]2  2.06  10 13

[OH ] 
Tro, Chemistry: A Molecular
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2.06 10   1.9 10
13
0.059 
39
6
Ex 16.14 What is the [Mg2+] when Ca2+ (with
[0.011]) just begins to precipitate from
seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
2
 2
Q  [Ca ][OH ]
Q  K sp
[0.011][OH  ]2  4.68  10  6

[OH ] 
Tro, Chemistry: A Molecular
Approach
4.68 10   2.06 10
6
0.011
40
2
Ex 16.14 What is the [Mg2+] when Ca2+ (with
[0.011]) just begins to precipitate from
seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
2
 2
Q  [Mg ][OH ]



when Q  K sp

[Mg2  ][ 2.06  10 2 ]2  2.06  1013
13
2.06  10
2
10
[Mg ] 

4
.
8

10
M
2
2.06  10 2
Tro, Chemistry: A Molecular
Approach


41
when Ca2+ just begins to
precipitate out, the
[Mg2+] has dropped from
0.059 M to 4.8 x 10-10 M
Qualitative Analysis
• an analytical scheme that utilizes selective
precipitation to identify the ions present in a
solution is called a qualitative analysis scheme
– wet chemistry
• a sample containing several ions is subjected to
the addition of several precipitating agents
• addition of each reagent causes one of the ions
present to precipitate out
Tro, Chemistry: A Molecular
Approach
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Qualitative Analysis
Tro, Chemistry: A Molecular
Approach
43
44
Group 1
• group one cations are Ag+, Pb2+, and Hg22+
• all these cations form compounds with Cl−
that are insoluble in water
– as long as the concentration is large enough
– PbCl2 may be borderline
• molar solubility of PbCl2 = 1.43 x 10-2 M
• precipitated by the addition of HCl
Tro, Chemistry: A Molecular
Approach
45
Group 2
• group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,
Pb2+, Sb3+, and Hg2+
• all these cations form compounds with HS− and
S2− that are insoluble in water at low pH
• precipitated by the addition of H2S in HCl
46
Group 3
• group three cations are Fe2+, Co2+, Zn2+, Mn2+,
Ni2+ precipitated as sulfides; as well as Cr3+,
Fe3+, and Al3+ precipitated as hydroxides
• all these cations form compounds with S2− that
are insoluble in water at high pH
• precipitated by the addition of H2S in NaOH
47
Group 4
• group four cations are Mg2+, Ca2+, Ba2+
• all these cations form compounds with PO43−
that are insoluble in water at high pH
• precipitated by the addition of (NH4)2HPO4
Tro, Chemistry: A Molecular
Approach
48
Group 5
• group five cations are Na+, K+, NH4+
• all these cations form compounds that are
soluble in water – they do not precipitate
• identified by the color of their flame
49
Complex Ion Formation
• transition metals tend to be good Lewis acids
• they often bond to one or more H2O molecules to
form a hydrated ion
– H2O is the Lewis base, donating electron pairs to form
coordinate covalent bonds
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
• ions that form by combining a cation with several
anions or neutral molecules are called complex ions
– e.g., Ag(H2O)2+
• the attached ions or molecules are called ligands
– e.g., H2O
Tro, Chemistry: A Molecular
Approach
50
Complex Ion Equilibria
• if a ligand is added to a solution that forms a
stronger bond than the current ligand, it will
replace the current ligand
Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l)
– generally H2O is not included, since its complex ion is
always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Tro, Chemistry: A Molecular
Approach
51
Formation Constant
• the reaction between an ion and ligands to
form a complex ion is called a complex ion
formation reaction
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
• the equilibrium constant for the formation
reaction is called the formation constant, Kf

[Ag(NH 3 )2 ]
Kf 

2
[Ag ][ NH3 ]
Tro, Chemistry: A Molecular
Approach
52
Formation Constants
Tro, Chemistry: A Molecular
Approach
53
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Write the
formation reaction
and Kf expression.
Look up Kf value
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Determine the
concentration of
ions in the diluted
solutions
1.5  10-3 mol
0.200 L 
1L
[Cu 2  ] 
 6.7  10  4 M
0.200 L  0.250 L 
Tro, Chemistry: A Molecular
Approach
[Cu(NH 3 )42 ]
13
Kf 

1
.
7

10
[Cu 2 ][ NH3 ]4
2.0  10-1 mol
0.250 L 
1L
[ NH3 ] 
 1.1  10 1 M
0.200 L  0.250 L 
54
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Kf 
[Cu 2 ][ NH3 ]4
[Cu(NH 3 )42 ]
 1.7  1013
Create an ICE
[Cu2+]
[NH3] [Cu(NH3)22+]
table. Since Kf is
large, assume all
Initial
6.7E-4
0.11
0
2+
the Cu is
Change
-≈6.7E-4 -4(6.7E-4)
+ 6.7E-4
converted into
complex ion, then Equilibrium
x
0.11
6.7E-4
the system returns
to equilibrium
55
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Substitute in
and solve for x
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
[Cu(NH 3 )42 ]
13
Kf 

1
.
7

10
[Cu 2 ][ NH3 ]4
confirm the
“x is small”
approximation

6.7  10 

4
13
1.7  10[Cu
2+]
x 0.11
[NH43]
6.7E-4
0.11

6.7  10 
x
 2.7  10
Change
-≈6.7E-4 -4(6.7E-4)
1.7 10 0.11
4
Initial
4
13
Equilibrium
x
0.11
since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid
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[Cu(NH3)22+]
13
0
+ 6.7E-4
6.7E-4
The Effect of Complex Ion Formation
on Solubility
• the solubility of an ionic compound that contains
a metal cation that forms a complex ion
increases in the presence of aqueous ligands
AgCl(s)  Ag+(aq) + Cl−(aq)
Ksp = 1.77 x 10-10
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Kf = 1.7 x 107
• adding NH3 to a solution in equilibrium with
AgCl(s) increases the solubility of Ag+
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58
Solubility of Amphoteric
Metal Hydroxides
• many metal hydroxides are insoluble
• all metal hydroxides become more soluble in acidic
solution
– shifting the equilibrium to the right by removing OH−
• some metal hydroxides also become more soluble in
basic solution
– acting as a Lewis base forming a complex ion
• substances that behave as both an acid and base are
said to be amphoteric
• some cations that form amphoteric hydroxides include
Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
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Al3+
• Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
• addition of OH− drives the equilibrium to the right
and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l)
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