Transcript Acid-base titration Titration • In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration.

Acid-base titration
Titration
• In an acid-base titration, a solution of unknown
concentration (titrant) is slowly added to a solution of
known concentration from a burette until the reaction
is complete
– when the reaction is complete we have reached the
endpoint of the titration
• An indicator may be added to determine the endpoint
– an indicator is a chemical that changes color when the pH
changes
• When the moles of H3O+ = moles of OH−, the titration
has reached its equivalence point
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Titration
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Titration Curve
• A plot of pH vs. amount of added titrant
• The inflection point of the curve is the equivalence point of the
titration
• Prior to the equivalence point, the known solution in the flask is
in excess, so the pH is closest to its pH
• The pH of the equivalence point depends on the pH of the salt
solution
– equivalence point of neutral salt, pH = 7
– equivalence point of acidic salt, pH < 7
– equivalence point of basic salt, pH > 7
• Beyond the equivalence point, the unknown solution in the
burette is in excess, so the pH approaches its pH
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Titration Curve:
Unknown Strong Base Added to Strong
Acid
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
Because the solutions are
equal concentration, and 1:1
stoichiometry, the
equivalence point is at equal
volumes
After Equivalence
(excess base)
Equivalence Point
equal moles of
HCl and NaOH
pH = 7.00

Before Equivalence
(excess acid)
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
added 5.0 mL NaOH
Before equivalence point
5.0 x 10−4 mole NaOH added
mols before
HCl
NaCl
NaOH
2.50E-3
0
5.0E-4
mols change −5.0E-4 +5.0E-4 −5.0E-4
mols end
2.00E-3 5.0E-4
molarity, new 0.0667
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7
0.017
0
0
Continued...
Table
HCl
NaOH
NaCl
mols before
2.50E-3
5.0E-4
0
mols change
−5.0E-4
−5.0E-4
+5.0E-4
2.00E-3
0
5.0E-4
0.0667M
0
0.017M
mols end
molarity, new
Titration of 25 mL of 0.100 M HCl with
0.100 M NaOH
• HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq)
• To reach equivalence, the added moles NaOH =
initial moles of HCl = 2.50 x 10−3 moles
• At equivalence, we have 0.00 mol HCl and 0.00
mol NaOH left over
• Because the NaCl is a neutral salt, the pH at
equivalence = 7.00
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
At equivalence point
added 25.0 mL NaOH
2.5 x 10−3 mole NaOH added
HCl
NaCl
NaOH
mols before 2.50E-3
0
2.5E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
0
molarity, new
0
0.050
0
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
added 30.0 mL NaOH
After equivalence point
3.0 x 10−3 mole NaOH added
mols before
mols change
HCl
NaCl
NaOH
2.50E-3
0
3.0E-3
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
5.0E-4
molarity, new
0
0.045
0.0091
0.100 mol NaOH
L of added NaOH 
1L
 moles added NaOH
Adding 0.100 M NaOH to 0.100 M
HCl
added 10.0
5.0 mL
mLNaOH
NaOH
added 25.0 mL NaOH
0.00200 mol HCl
0.00150
equivalence point
pH = 1.37
1.18
pH = 7.00
added 15.0
40.0 mL NaOH
0.00150 mol HCl
0.00100
NaOH
pH = 1.60
12.36
added 20.0
50.0 mL NaOH
0.00250 mol HCl
0.00050
NaOH
pH = 1.95
12.52
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Practice – Calculate the pH of the solution that results
when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
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•
•
•
•
Practice – Calculate the pH of the solution that results
when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
Before equivalence point added 10.0 mL NaOH
mols before
HNO3
NaNO3
NaOH
1.25E-2
0
1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
mols end
1.1E-3
1.5E-3
0
molarity, new
0.018
0.025
0
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Practice – Calculate the amount of 0.15 M NaOH
solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence
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Practice – Calculate the amount of 0.15 M NaOH
solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence
• HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
• Initial pH = −log(0.250) = 0.60
• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
• At equivalence point: moles of NaOH = 1.25 x 10−2
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Practice – Calculate the pH of the solution that results
when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
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•
•
•
•
Practice – Calculate the pH of the solution that results
when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
After equivalence point
added 100.0 mL NaOH
HNO3
NaNO3
NaOH
mols before
1.25E-2
0
1.5E-2
mols change
−1.25E-2 +1.25E-2 −1.25E-2
mols end
0
1.25E-2
0.0025
molarity, new
0
0.0833
0.017
•
•
•
•
Practice – Calculate the pH of the solution that results
when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
After equivalence point
added 100.0 mL NaOH
HNO3
NaNO3
NaOH
mols before
1.25E-2
0
1.5E-2
mols change
−1.25E-2 +1.25E-2 −1.25E-2
mols end
0
1.25E-2
0.0025
molarity, new
0
0.0833
0.017
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Titration of a Strong Base with a Strong Acid
• If the titration is run
so that the acid is in
the burette and the
base is in the flask,
the titration curve
will be the reflection
of the one just
shown
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Titration of a Weak Acid with a Strong
Base
• Titrating a weak acid with
a strong base results in
•
•
•
•
differences in the titration curve at the equivalence
point and excess acid region
The initial pH is determined using the Ka of the weak
acid
The pH in the excess acid region is determined as you
would determine the pH of a buffer
The pH at the equivalence point is determined using the
Kb of the conjugate base of the weak acid
The pH after equivalence is dominated by the excess
strong base
– the basicity from the conjugate base anion is negligible
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Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
• Initial pH
Ka = 1.8 x 10−4
[HCHO2] [CHO2−] [H3O+]
initial
change
0.100
0.000
≈0
−x
+x
+x
x
x
equilibrium 0.100 − x
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Titration of 25 mL of 0.100 M HCHO2 with 0.100 M
NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence
added 5.0 mL NaOH
HA
A−
OH−
mols before
2.50E-3
0
0
mols added
–
–
5.0E-4
mols after
2.00E-3 5.0E-4
≈0
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Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
•
•
•
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
At equivalence
CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq)
added 25.0 mL NaOH
HA
A−
OH−
mols before
2.50E-3
0
0
mols added
–
–
2.50E-3
mols after
0
2.50E-3
≈0
[OH−] = 1.7 x 10−6 M
Kb = 5.6 x 10−11
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence
added 30.0 mL NaOH
3.0 x 10−3 mole NaOH added
HA
A−
NaOH
mols before 2.50E-3
0
3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3 5.0E-4
molarity, new
0
0.045
35
0.0091
Adding NaOH to HCHO2
25.0
added 30.0
5.0 mL
mLNaOH
NaOH
10.0
initial HCHO2 solution
equivalence
point
0.00200
0.00050
molmL
NaOH
0.00150
HCHO
2xs
added
35.0
NaOH
0.00250 mol HCHO
− 2
0.00250
mol
CHO
pH
=
3.14
11.96
3.56
2
0.00100
mol NaOH xs
pH = 2.37
−
[CHO
2 ]init = 0.0500 M
pH =− 12.22
[OH ]eq = 1.7 x 10−6
added
12.5 mL NaOH
pH
= 8.23
added
40.0 mL NaOH
0.00125 mol HCHO2
0.00150 mol NaOH xs
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
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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
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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 as you did in Example 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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Example 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
HNO2 + KOH  NO2 + H2O
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Example 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
HNO2 + KOH  NO2 + H2O
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
mols before
mols added
mols after
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HNO2
NO2−
OH−
0.00400
0
≈0
–
–
0.00100
0.00300
0.00100
≈0
Example 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.
 + H O+
HNO
+
H
O

NO
2
2
2
3
write an
Table 15.5 Ka = 4.6 x 10−4
equation for the
reaction of HA
with H2O
determine Ka
and pKa for HA
use the
HendersonHasselbalch
equation to
determine the
pH
HNO2
NO2−
OH−
0
≈0
mols before 0.00400
0.00100
mols added
–
–
0.00300 0.00100
mols after
≈0
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Example 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
HNO2 + KOH  NO2 + H2O
at half-equivalence, moles KOH = ½ mole HNO2
HNO2
mols before 0.00400
mols added
–
0.00200
mols after
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44
NO2−
OH−
0
≈0
–
0.00200
0.00200
≈0
Example 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.
HNO2 + H2O  NO2 + H3O+
write an equation
for the reaction of
HA with H2O
Table 15.5 Ka = 4.6 x 10-4
determine Ka and
pKa for HA
use the
HendersonHasselbalch
equation to
determine the pH
HNO2
NO2−
OH−
0
≈0
mols before 0.00400
0.00200
mols added
–
–
0.00200 0.00200
mols after
≈0
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Titration Curve of a Weak Base with
a Strong Acid
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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the initial pH of the
NH3 solution.
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Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M
HCl. Calculate the initial pH of the NH3(aq)
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
[HCl] [NH4+]
[NH3]
initial
0
0
0.10
change
+x
+x
−x
equilibrium
x
x
0.10−x
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Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M
HCl. Calculate the initial pH of the NH3(aq)
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
[HCl] [NH4+]
[NH3]
initial
0
0
0.10
change
+x
+x
−x
equilibrium
x
x
0.10−x
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Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution after
adding 5.0 mL of HCl.
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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution after
adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
NH4+(aq) + H2O(l)  NH4+(aq) + H2O(l)
pKb = 4.75
pKa = 14.00 − 4.75 = 9.25
NH3
NH4Cl
HCl
mols before
2.50E-3
0
5.0E-4
mols change
−5.0E-4 −5.0E-4 −5.0E-4
mols end
2.00E-3
5.0E-4
0
molarity, new
0.0667
0.017
0
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Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution after
adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
NH3
NH4Cl
HCl
mols before
2.50E-3
0
5.0E-4
mols change
−5.0E-4 −5.0E-4 −5.0E-4
mols end
2.00E-3
5.0E-4
0
molarity, new
0.0667
0.017
0
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Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the solution
at equivalence.
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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution at
equivalence.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence mol NH3 = mol HCl = 2.50 x 10−3
NH3
NH4Cl
HCl
mols before
2.50E-3
0
2.5E-3
mols change
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
0
molarity, new
0
0.050
0
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Tro: Chemistry: A Molecular Approach, 2/e
added 25.0 mL HCl
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution at
equivalence.
NH3(aq) + HCl(aq)  NH4Cl(aq)
at equivalence [NH4Cl] = 0.050 M
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
[NH3]
[NH4+] [H3O+]
initial
0
0.050
≈0
change
+x
−x
+x
equilibrium
x
0.050−x
x
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Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the solution
after adding 30.0 mL of HCl.
57
Tro: Chemistry: A Molecular Approach, 2/e
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)
with 0.10 M HCl. Calculate the pH of the solution after
adding 30.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence: after adding 30.0 mL HCl
NH3
NH4Cl
HCl
mols before
2.50E-3
0
3.0E-3
mols change
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
5.0E-4
molarity, new
0
0.045
0.0091
when you mix a strong acid, HCl,
with a weak acid, NH4+, you only
need to consider the strong acid
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Tro: Chemistry: A Molecular Approach, 2/e
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
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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
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Monitoring pH During a Titration
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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
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Phenolphthalein
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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
64
Tro: Chemistry: A Molecular Approach, 2/e
Monitoring a Titration with
an Indicator
• For most titrations, the titration curve shows a
very large change in pH for very small
additions of titrant 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
65
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Acid-Base Indicators
66
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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
67
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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
68
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69
Tro: Chemistry: A Molecular Approach, 2/e
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)
70
Tro: Chemistry: A Molecular Approach, 2/e
Example 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
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Initial
71
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Example 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
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
[Pb2+]
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Initial
72
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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
73
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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
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)
74
Tro: Chemistry: A Molecular Approach, 2/e
(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
initial
[Pb2+]
[Br−]
0
0
change
+(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium
(1.05 x 10−2)
75
Tro: Chemistry: A Molecular Approach, 2/e
(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
• To compare Ksps, the compounds must have the
same dissociation stoichiometry
76
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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
77
Tro: Chemistry: A Molecular Approach, 2/e
Example 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
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
78
Tro: Chemistry: A Molecular Approach, 2/e
Example 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
[Ca2+]
[F−]
0
0.100
change
+S
+2S
equilibrium
S
0.100 + 2S
initial
79
Tro: Chemistry: A Molecular Approach, 2/e
Practice – Determine the concentration of Ag+ ions
in seawater that has a [Cl−] of 0.55 M
Ksp of AgCl = 1.77 x 10−10
81
Tro: Chemistry: A Molecular Approach, 2/e
Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
write the
dissociation
reaction and Ksp
expression
AgCl(s)  Ag+(aq) + Cl−(aq)
create an ICE
table defining
the change in
terms of the
solubility of the
solid
[Ag+]
[Cl−]
0
0.55
change
+S
+S
equilibrium
S
0.55 + S
Ksp = [Ag+][Cl−]
initial
82
Tro: Chemistry: A Molecular Approach, 2/e
Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
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 = [Ag+][Cl−]
Ksp = (S)(0.55 + S)
Ksp = (S)(0.55)
[Ag+]
[Cl−]
0
0.55
Change
+S
+S
Equilibrium
S
0.55 + S
Initial
83
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[Ag+]
[Cl−]
0
0.55
Change
+S
+S
Equilibrium
S
0.55 + S
Initial
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)
85
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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
86
Tro: Chemistry: A Molecular Approach, 2/e
precipitation occurs if Q >
Ksp
a supersaturated solution will precipitate if a
seed crystal is added
87
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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
88
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Example 16.12: Will a precipitate form when we mix
Pb(NO3)2(aq) with NaBr(aq) if the concentrations after
mixing are 0.0150 M and 0.0350 M respectively?
write the equation for
the reaction
Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)
determine the ion
concentrations of the
original salts
determine the Ksp for
any “insoluble”
product
write the dissociation
reaction for the
insoluble product
Pb(NO3)2 = 0.0150 M
Pb2+ = 0.0150 M,
NO3− = 2(0.0150 M)
Ksp of PbBr2 = 4.67 x 10–6
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
Tro: Chemistry: A Molecular Approach, 2/e
NaBr = 0.0350 M
Na+ = 0.0350 M,
Br− = 0.0350 M
Q < Ksp, so no precipitation
89
Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after
mixing are both 0.0175 M?
Ksp of Ca(OH)2 = 4.68 x 10−6
90
Tro: Chemistry: A Molecular Approach, 2/e
Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after
mixing are both 0.0175 M?
write the equation for
the reaction
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
determine the ion
concentrations of the
original salts
determine the Ksp for
any “insoluble”
product
write the dissociation
reaction for the
insoluble product
Ca(NO3)2 = 0.0175 M
Ca2+ = 0.0175 M,
NO3− = 2(0.0175 M)
Ksp of Ca(OH)2 = 4.68 x 10–6
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
Tro: Chemistry: A Molecular Approach, 2/e
NaOH = 0.0175 M
Na+ = 0.0175 M,
OH− = 0.0175 M
Q > Ksp, so precipitation
91
Example 16.13: What is the minimum [OH−] necessary to
just begin to precipitate Mg2+ (with [0.059]) from
seawater?
Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
92
Tro: Chemistry: A Molecular Approach, 2/e
Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175 M
NaOH(aq)?
Ksp of Ca(OH)2 = 4.68 x 10−6
93
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Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175 M
NaOH(aq)?
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
[Ca(NO3)2] = [Ca2+] = 0.0153 M
94
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Example 16.14: What is the [Mg2+] when Ca2+
(with [0.011]) just begins to precipitate from
seawater?
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
95
Tro: Chemistry: A Molecular Approach, 2/e
Example 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
Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)
when Ca2+ just begins to
precipitate out, the [Mg2+]
has dropped from 0.059 M
to 4.8 x 10−10 M
96
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Practice – A solution is made by mixing Pb(NO3)2(aq) with
AgNO3(aq) so both compounds have a concentration of 0.0010
M. NaCl(s) is added to precipitate out both AgCl(s) and
PbCl2(aq). What is the [Ag+] concentration when the Pb2+ just
begins to precipitate?
97
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Practice – What is the [Ag+] concentration when the
Pb2+(0.0010 M) just begins to precipitate?
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
AgCl(s)  Ag+(aq) + Cl−(aq)
precipitating may just occur when Q = Ksp
98
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Practice – What is the [Ag+] concentration when the
Pb2+(0.0010 M) just begins to precipitate?
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
precipitating may just occur when Q = Ksp
99
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Practice – What is the [Ag+] concentration when the
Pb2+(0.0010 M) just begins to precipitate
precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M
precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M
AgCl(s)  Ag+(aq) + Cl−(aq)
when Pb2+ just begins to
precipitate out, the [Ag+]
has dropped from 0.0010 M
to 1.6 x 10−9 M
100
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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
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Qualitative Analysis
102
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103
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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
104
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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
105
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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
106
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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
107
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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
108
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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
109
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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, because its complex ion
is always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
110
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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
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Formation Constants
112
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Example 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
113
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Example 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)
Create an ICE
table. Because
Kf is large,
assume all the
Cu2+ is
converted into
complex ion,
then the system
returns to
equilibrium.
initial
change
[Cu2+]
[NH3]
[Cu(NH3)22+]
6.7E-4
0.11
0
≈−6.7E-4 ≈−4(6.7E-4)
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
x
114
0.11
≈+6.7E-4
6.7E-4
Example 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
confirm the
“x is small”
approximation
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
initial
change
equilibrium
[Cu2+]
[NH3]
[Cu(NH3)22+]
6.7E-4
0.11
0
≈−6.7E-4 ≈−4(6.7E-4)
x
0.11
2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid
115
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≈+6.7E-4
6.7E-4
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI
is reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
117
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Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is
reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Write the
formation
reaction and Kf
expression.
Look up Kf value
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
determine the
concentration of
ions in the
diluted solutions
118
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Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is
reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
Create an ICE
table. Because
Kf is large,
assume all the
lim. rgt. is
converted into
complex ion,
then the system
returns to
equilibrium.
I− is the limiting reagent
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
[Hg2+]
[I−]
[HgI42−]
3.75E-4
6.25E-4
0
≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
2.19E-4
119
x
1.56E-4
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is
reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
substitute in
and solve for
x
confirm the
“x is small”
approximation
initial
[Hg2+]
[I−]
[HgI42−]
3.75E-4
6.25E-4
0
≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
change
equilibrium
2.19E-4
x
2 x 10−8 << 1.6 x 10−4, so the approximation is valid
[HgI42−] = 1.6 x 10−4
120
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1.56E-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+
122
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123
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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+
124
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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)
125
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126
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