Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
Chapter 9
9-1
(a) A weak electrolyte only partially ionizes when dissolved in water. NaHCO3 is an
example of a weak electrolyte.
(b) A Brønsted-Lowry acid is a molecule that donates a proton when it encounters a base
(proton acceptor). By this definition, NH4+ can be a Brønsted-Lowry acid.
(c) The conjugate acid of a Brønsted-Lowry base is the species formed when a BrønstedLowry base accepts a proton. For example, the NH4+ is the conjugate acid of NH3.
(d) Neutralization, according to the Brønsted-Lowry concept, occurs when a reaction
involving an acid and its conjugate base is combined with a second reaction involving a
base and its conjugate acid. Thus,
+
NH 3 + H 2 O →
NH 4 + OH −
←
In the example above, NH3 acts as a base with NH4+ as its conjugate acid. H2O acts as an
acid with OH- as its conjugate base.
(e) An amphiprotic solvent can act either as an acid or a base depending on the solute.
Water is an example of an amphiprotic chemical species.
(f) A zwitterion is a chemical species that bears both positive and negative charges. Free
amino acids, such as glycine, can exist as zwitterions in solution.
NH 2CH 2COOH → NH 3+ CH 2COO −
glycine
← zwitterion
(g) Autoprotolysis is the act of self-ionization of a solvent to produce both a conjugate
acid and a conjugate base.
(h) A strong acid dissociates completely such that no undissociated molecules are left in
aqueous solution. Hydrochloric acid, HCl, is an example of a strong acid.
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(i) The Le Châtelier principle states that the position of an equilibrium always shifts in
such a direction to relieve a stress applied to the system.
(j) The common-ion effect is responsible for the reduced solubility of an ionic precipitate
when one of the soluble components reacting to form the precipitate is added to the
solution in equilibrium with the precipitate.
9-2
(a) An amphiprotic solute is a chemical species that possesses both acidic and basic
properties. The dihydrogen phosphate ion, H2PO4-, is an example of an amphiprotic
solute.
(b) A differentiating solvent reveals different strengths of acids. By this definition,
anhydrous acetic acid is a differentiating solvent because perchloric acid dissociates 5000
times more than hydrochloric acid.
(c) A leveling solvent is one in which a series of acids (or bases) all dissociate completely.
Water is an example, since strong acids like HCl and HClO4 ionize completely.
(d) A mass-action effect arises when a shift in the chemical equilibrium occurs due to the
introduction of one of the participating chemical species (i.e., addition of one of the
reactants or products.
9-3
For an aqueous equilibrium in which water is a participant, the concentration of water is
normally so much larger than the concentrations of the other reactants or products that it
can be assumed to be constant and independent of the equilibrium position. Thus, its
concentration is included within the equilibrium constant. For a pure solid, the
concentration of the chemical species in the solid phase is constant. As long as some
solid exists as a second phase, its effect on the equilibrium is constant and is included
within the equilibrium constant.
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
9-4
Acid
Conjugate Base
(a)
HOCl
OCl-
(b)
H2 O
OH-
(c)
NH4+
NH3
-
CO32-
(d)
HCO3
(e)
H2PO4-
HPO42-
Base
Conjugate Acid
(a)
H2 O
H3 O+
(b)
HONH2
HONH3+
(c)
H2 O
H3 O+
(d)
HCO3-
H2CO3
(e)
PO43-
HPO42-
9-5
9-6
(a) 2 H 2 O →
H 3O + + OH −
←
(b) 2CH 3COOH
(c) 2CH 3 NH 2
(d) 2CH 3OH
9-7
→
←
→
←
→
←
+
CH 3COOH 2 + CH 3COO −
+
CH 3 NH 3 + CH 3 NH −
+
CH 3OH 2 + CH 3O −
+
(a) C 2 H 5 NH 2 + H 2 O →
C 2 H 5 NH 2 + OH −
←
+
Kb =
K w 1.00 × 10 −14 [C 2 H 5 NH 2 ][OH − ]
=
=
= 4.33 × 10 −4
K a 2.31 × 10 −11
[C 2 H 5 NH 2 ]
(b) HCN + H 2O →
CN − + H 3O +
←
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
K w [CN − ][H 3O + ]
Ka =
=
= 6.2 × 10 −10
Kb
[HCN]
C 5 H 5 N + H 3O +
(c) C 5 H 5 NH + + H 2 O →
←
K w [C5 H 5 N][H 3O + ]
Ka =
=
= 5.90 × 10 −6
+
Kb
[C 5 H 5 NH ]
(d) CN − + H 2 O →
HCN + OH −
←
K
1.00 × 10
Kb = w =
K a 6.2 × 10 −10
−14
=
[HCN][OH − ]
= 1.6 × 10 −5
−
[CN ]
(e)
−
H 3 AsO 4 + H 2 O →
H 2 AsO 4 + H 3O +
←
−
H 2 AsO 4 + H 2 O →
HAsO 4
←
HAsO 4
2−
2−
+ H 3O +
3−
+ H 2O →
AsO 4 + H 3O +
←
3−
H 3 AsO 4 + 3H 2 O →
AsO 4 + 3H 3O +
←
−
[H AsO 4 ][H 3O + ]
K a1 = 2
[H 3 AsO 4 ]
2−
Ka2
[HAsO 4 ][H 3O + ]
=
−
[H 2 AsO 4 ]
3−
β3 =
3−
Ka3
2−
(f)
2−
−
+ H 2O →
HCO 3 + OH −
←
−
HCO 3 + H 2 O ←→ H 2 CO 3 + OH −
CO 3
−
[AsO 4 ][H 3O + ]3 K a 3 [HAsO 4 ][H 3O + ]2 K a 3 K a 2 [H 2 AsO 4 ][H 3O + ]
=
=
= K a 3 K a 2 K a1
[H 3 AsO 4 ]
[ H 3 AsO 4 ]
[H 3 AsO 4 ]
β 3 = (3.2 × 10 −12 ) × (1.1 × 10 −7 ) × (5.8 × 10 −3 ) = 2.0 × 10 − 21
CO 3
[AsO 4 ][H 3O + ]
=
2−
[HAsO 4 ]
2−
+ 2H 2O ←
H 2 CO 3 + 2OH −
→
Fundamentals of Analytical Chemistry: 8th ed.
−
Chapter 9
−
[HCO 3 ][OH − ]
[ HCO 3 ]K w
K
=
= w
K b1 =
2−
2−
+
[CO 3 ]
[ H 3O ][CO3 ] K a 2
K b2 =
[H 2 CO 3 ][OH − ]
[ H 2 CO 3 ]K w
K
=
= w
−
−
+
[HCO 3 ]
[ H 3O ][HCO 3 ] K a1
−
2
[H CO ][OH − ]2 K b 2 [ HCO 3 ][OH − ]
Kw
β 2 = 2 3 2−
=
= K b 2 K b1 =
2−
K a1 K a 2
[CO 3 ]
[CO 3 ]
β2 =
9-8
(1.0 × 10 −14 ) 2
= 1.4 × 10 −14
(1.5 × 10 −4 ) × ( 4.69 × 10 −11 )
Cu + + I −
(a) CuI( s ) →
←
K sp = [Cu + ][I − ]
→
(b) PbClF( s ) ←
Pb 2+ + Cl − + F −
K sp = [Pb 2+ ][Cl − ][F − ]
(c) PbI 2 ( s ) →
Pb 2+ + 2I −
←
K sp = [Pb 2+ ][I − ]2
(d) BiI 3 ( s ) →
Bi 3+ + 3I −
←
K sp = [Bi 3+ ][I − ]3
+
(e) MgNH 4 PO 4 ( s ) →
Mg 2+ + NH 4 + PO 4
←
9-9
3−
+
(a) S = [Cu + ] = [I − ] K sp = [Cu + ][I − ] = S 2
(b) S = [Pb2+] = [Cl-] = [F-]
Ksp = [Pb2+][Cl-][F-] = S3
(c) S = [Pb2+] = 1/2[I-]
Ksp = [Pb2+][I-]2 = (S)(2S)2 = 4S3
(d) S = [Bi3+] = 1/3[I-]
Ksp = [Bi2+][I-]3 = (S)(3S)3 = 27S4
(e) S = [Mg2+] = [NH4+] = [PO43-]
3−
K sp = [ Mg 2+ ][NH 4 ][PO 4 ]
Ksp = [Mg2+][NH4+][PO43-] = S3
Fundamentals of Analytical Chemistry: 8th ed.
9-10
Chapter 9
(a)
[Cu 2+ ] = [SeO 3 ] = 1.42 × 10 −4 mole
2−
→
CuSeO 3 ( s ) ← Cu
2+
+ SeO 3
2−
L
2−
K sp = [Cu 2+ ][SeO 3 ] = (1.42 × 10 −4 ) × (1.42 × 10 −4 )
= 2.02 × 10 −8
(b)
[Pb 2+ ] = 4.3 × 10 −5 mole
[IO 3 ] = 2[Pb 2+ ] = 2 × (4.3 × 10 −5 mole ) = 8.6 × 10 −5 mole
L
L
−
L
Pb(IO 3 ) 2 ( s ) →
Pb 2+ + 2IO 3
←
−
−
K sp = [ Pb 2+ ][IO 3 ]2 = ( 4.3 × 10 −5 ) × (8.6 × 10 −5 ) 2 = 3.2 × 10 −13
(c)
[Sr 2+ ] = 8.6 × 10 −4 mole
SrF2 ( s ) ←→ Sr 2+ + 2F −
L
[F − ] = 2[Sr 2+ ] = 2 × (8.6 × 10 −4 mole ) = 1.7 × 10 −3 mole
L
L
K sp = [Sr 2+ ][F − ]2 = (8.6 × 10 − 4 ) × (1.7 × 10 −3 ) 2 = 2.5 × 10 −9
(d)
[Th 4+ ] = 3.3 × 10 −4 mole
L
[OH − ] = 4[Th 4+ ] = 4 × (3.3 × 10 −4 mole ) = 1.3 × 10 −3 mole
L
L
Th(OH) 4 ( s ) →
Th 4+ + 4OH −
←
9-11
K sp = [Th 4+ ][OH − ]4 = (3.3 × 10 −4 ) × (1.3 × 10 −3 ) 4 = 1.0 × 10 −15
(a)
Ksp = 2.02×10-8 = [Cu2+][SeO32-] = (0.050 M)[SeO32-]
S=
[SeO32-]
2.0 × 10 −8
=
= 4.0 × 10 −7 M
0.050
(b)
Ksp = 3.2×10-13 = [Pb2+][IO3-]2 = (0.050 M)[IO3-]2
[IO3-] =
3.2 × 10 −13
= 2.5 × 10 −6 M
0.050
S = ½[IO3-] = ½(2.5×10-6 M) = 1.3×10-6 M
Fundamentals of Analytical Chemistry: 8th ed.
(c)
K sp = 2.5 × 10 −9 = [Sr 2+ ][F − ]2 = (0.050 M)[F − ]2
2.5 × 10 −9
= 2.2 × 10 −4 M
0.050
1
1
S = [ F − ] = × ( 2.2 × 10 − 4 M ) = 1.1 × 10 − 4 M
2
2
[F− ] =
(d)
K sp = 1.0 × 10 −15 = [Th 4+ ][OH − ]4 = (0.050 M)[OH − ]4
1.0 × 10 −15
= 3.8 × 10 −4 M
0.050
1
1
S = [OH − ] = × (3.8 × 10 −4 M ) = 9.4 × 10 −5 M
4
4
[OH − ] = 4
9-12
(a)
2−
K sp = 2.02 × 10 −8 = [Cu 2+ ][SeO 3 ] = [Cu 2+ ](0.050M )
S = [Cu 2+ ] =
2.02 × 10 −8
= 4.0 × 10 −7 M
0.050
(b)
−
K sp = 3.2 × 10 −13 = [ Pb 2+ ][IO 3 ]2 = [ Pb 2+ ](0.050M ) 2
S = [ Pb 2+ ] =
3.2 × 10 −13
= 1.3 × 10 −10 M
(0.050) 2
(c)
Ksp = 2.5×10-9 = [Sr2+][F-]2 = [Sr2+](0.050 M)2
2.5 × 10 −9
S = [Sr ] =
= 1.0 × 10 −6 M
2
(0.050)
2+
Chapter 9
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(d)
Ksp = 1.0×10-15 = [Th4+][OH-]4 = [Th4+](0.050 M)4
S = [Th4+] =
9-13
1.0 × 10 −15
= 1.6 × 10 −10 M
(0.050) 4
Ag 2CrO4 ( s ) →
2Ag + + CrO4
←
(a) [CrO 4
2−
2−
Ksp = 1.2 × 10−12 = [ Ag + ]2 [CrO4 ]
1.2 × 10 −12
]=
= 1.0 × 10 −9 M
−2 2
(3.41 × 10 )
2−
(b) [CrO 4 ] =
9-14
2−
1.2 × 10 −12
= 0.30 M
(2.00 × 10 −6 ) 2
Al(OH) 3 ( s ) →
Al 3+ + 3OH −
←
K sp = 3.0 × 10− 34 = [Al 3+ ][OH − ]3
1
Molar solubility of Al(OH) 3 = [OH − ]
3
1
[Al 3+ ] = 5.0 × 10 − 2 M + [OH − ]
3
1


K sp =  5.0 × 10 − 2 M + [OH − ] [OH − ]3 = 3.0 × 10 − 34
3


(a) Because the Ksp is so small, we can assume the solubility of Al(OH)3 is not large.
Therefore, it is reasonable that the [Al3+] ≈ 5.0 X 10-2 M. The Ksp equation then
simplifies to
Ksp = (5.0×10-2 M)[OH-]3 = 3.0×10-34
1
 3.0 × 10 −34  3
 = 2.0 × 10 −11 M
[OH-] = 
−2
 5.0 × 10 M 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(b) As in part (a),
1
3
 3.0 × 10

 = 1.1 × 10 −9 M
[OH − ] = 
−7
 2.00 × 10 M 
−34
9-15
−
K sp = [Ce3+ ][IO 3 ]3 = S(3S) 3 = 3.2 × 10−10
(a)
0.0250 M Ce 3+ ≡ 2.5 × 10 −2
mole
L
×
× 50.0 mL = 1.25 × 10 −3 mole
L
1000 mL
1.25 × 10 −3 mole
1000 mL
×
= 0.0125 M
(50.0 mL + 50.0 mL)
L
−
(b) 0.040 M IO 3 ≡ 4.0 × 10 − 2
mole
L
×
× 50.0 mL = 2.0 × 10 −3 mole
L
1000 mL
moles Ce 3+ unreacted = 1.25 × 10 −3 mole −
1
(
2.0 × 10 −3 mole) = 5.833 × 10 −4 mole
3
 5.833 × 10 −4 mole
1000 mL 
 + S = 5.833 × 10 −3 M + S
[Ce 3+ ] = 
×
L
 (50.0 mL + 50.0 mL)

where S is derived iteratively using the equation
K sp = (5.833 × 10 −3 + S) (3S) 3
We start by solving for S assuming no contribution to [Ce3+] other than from the
dissociation of Ce(IO3)3. In this case, S equals 1.855 × 10-3. Now, we substitute 1.855 ×
10-3 back into the Ksp equation above and solve for Kiterative. Kiterative equals 1.3259 × 10-9.
S is too large; we choose a smaller S (i.e., 1 × 10-3) and recalculate Kiterative. Iteration
continues until Kiterative ≈ Ksp. The results of this approach are shown in the Table below.
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
Iteration
S
Kiterative
1
1.855×10-3
1.3259×10-9
2
1×10-3
1.8449×10-10
3
1.2×10-3
3.2813×10-10
4
1.19×10-3
3.1954×10-10
We now substitute S = 1.19 × 10-3 into the equation below,
[Ce 3+ ] = 5.833 × 10 −3 M + 1.19 × 10 −3 = 7.0 × 10 −3 M
(c)
−
0.250 M IO 3 ≡ 2.5 × 10 −1
mole
L
×
× 50.0 mL = 0.0125 mole
L
1000 mL
The Ce3+ is completely consumed; thus,
−
moles IO 3 unreacted = 0.0125 mole − 3(1.25 × 10 −3 mole) = 8.75 × 10 −3 mole
 8.75 × 10 −3 mole 1000 mL 
−
 + 3S = 0.0875 M + 3S
[ IO 3 ] = 
×
100
.
0
mL
L


3
−10
K sp = S(0.0875 M + 3S) = 3.2 × 10
Make the assumption that 3S << 0.0375 M.
K sp = S(0.0875 M ) 3 = 3.2 × 10 −10
 3.2 × 10 −10 
 = 4.8 × 10 −7 M
[Ce 3+ ] = S = 
3 
 (0.0875 M ) 
(d)
−
0.150 M IO 3 ≡ 1.5 × 10 −1
mole
L
×
× 50.0 mL = 7.5 × 10 −3 mole
L
1000 mL
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
The Ce3+ is completely consumed; thus,
−
moles IO 3 unreacted = 7.5 × 10 −3 mole − 3(1.25 × 10 −3 mole) = 3.75 × 10 −3 mole
 3.75 × 10 −3 mole 1000 mL 
−
 + 3S = 0.0375 M + 3S
[ IO 3 ] = 
×
L
 100.0 mL

3
−10
K sp = S(0.0375 M + 3S) = 3.2 × 10
Make the assumption that 3S << 0.0375 M.
K sp = S(0.0375 M ) 3 = 3.2 × 10 −10
 3.2 × 10 −10 
 = 6.1 × 10 −6 M
[Ce 3+ ] = S = 
3 
(
0
.
0375
M
)


9-16
2−
K sp = [ K + ]2 [PdCl 6 ] = ( 2S) 2 (S) = 6.0 × 10 −6
0.200 M K + ≡ 2.0 × 10 −1
mole
L
×
× 50.0 mL = 0.0100 mole
L
1000 mL
1.00 × 10 −2 mole
1000 mL
×
= 0.100 M
(50.0 mL + 50.0 mL)
L
(a) 0.0500 M PdCl 6
2−
≡ 5.0 × 10 −2
mole
L
×
× 50.0 mL = 2.5 × 10 −3 mole
L
1000 mL
moles K + unreacted = 1.00 × 10 −2 mole − 2(2.5 × 10 −3 mole) = 5.0 × 10 −3 mole
5.0 × 10 −3 mole
1000 mL
×
= 0.050 M
(50.0 mL + 50.0 mL)
L
The [K+] is 0.05 M plus the contribution from the dissociation of the precipitate, x, or
[K+] = 0.05 M + 2x
Substituting the equation above into the equilibrium expression, we find
2−
K sp = [ K + ]2 [ PdCl 6 ] = (0.05 + 2 x ) 2 x
4 x 3 + 0.20 x 2 + ( 2.5 × 10 −3 ) x = 6.0 × 10 −6
We use an iterative approach to solve for x. We begin by ignoring the contribution to
[K+] from the dissociation of the precipitate.
Fundamentals of Analytical Chemistry: 8th ed.
x=3
Chapter 9
6.0 × 10 −6
= 0.011447
4
Substituting this value back into the Ksp equation above, we obtain Kiterative equal to
6.0824 × 10-5, which is too large. We now proceed by choosing a smaller value of x equal
1 × 10-3 and obtain Kiterative equals to 2.7040× 10-6. We continue the iterative approach
until Kiterative ≈ Ksp as shown in the table below.
Iteration
x
Kiterative
1
0.011447
6.0814 × 10-5
2
1 × 10-3
2.7040 ×10-6
3
2 × 10-3
5.8320 ×10-6
Substituting x equals 2 × 10-3 into the [K+] equation, we find
[K+] = 0.05 M + 2(2 × 10-3M) = 0.054 M
(b)
0.100 M PdCl 6
2−
≡ 1.0 × 10 −1
mole
L
×
× 50.0 mL = 5.00 × 10 −3 mole
L
1000 mL
The [K+] is determined directly from Ksp; thus,
2−
[ K + ] = 2S [ PdCl 6 ] = S
K sp = ( 2S) 2 S = 4S3 = 6.0 × 10 −6
1
 6.0 × 10 −6  3
 = 0.011
S = 
4


+
[ K ] = 2S = 2(0.011 M ) = 0.022 M
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(c)
0.200 M PdCl 6
2−
≡ 2.0 × 10 −1
mole
L
×
× 50.0 mL = 0.010 mole
L
1000 mL
The K+ is completely consumed; thus,
2−
moles PdCl 6 unreacted = 1.0 × 10 −2 mole −
1
(1.00 × 10 −2 mole) = 5.0 × 10 -3 mole
2
 5.0 × 10 −3 mole 1000 mL 
 + S = 0.05 M + S
×
[PdCl 6 ] = 
L
 100.0 mL

Assume that S << 0.05 M
2−
2−
[ PdCl 6 ] = 0.05 M
K sp = 4S 2 (0.05 M ) = 6.0 × 10 −6
1
 6.0 × 10 −6  2
 = 5.477 × 10 −3 M
S = 
 4(0.05) 
[ K + ] = 2S = 2(5.477 × 10 −3 M ) = 0.011 M
9-17
CuI(s) Cu+ + I-
Ksp = [Cu+][I-] = 1×10-12
AgI(s) Ag+ + I-
Ksp = [Ag+][I-] = 8.3×10-17
PbI2(s) Pb2+ + 2I- Ksp = [Pb2+][I-]2 = 7.1×10-9
= S(2S)2 = 4S3
BiI3(s) Bi3+ + 3I-
= S(3S)3 = 27S4
Ksp = [Bi3+][I-]3 = 8.1×10-19
(a)
S CuI = [Cu + ] = [I − ] S = 1 × 10 −12 = 1 × 10 −6
S AgI = [Ag + ] = [I − ] S = 8.3 × 10 −17 = 9.1 × 10 −9
S PbI 2 = [ Pb 2+ ] =
1 −
(7.1 × 10 −9 )
[I ] S = 3
= 1.2 × 10 −3
2
4
1
(8.1 × 10 −19 )
S BiI 3 = [ Bi 3+ ] = [I − ] S = 4
= 1.3 × 10 −5
3
27
PbI 2 > BiI 3 > CuI > AgI in water
Fundamentals of Analytical Chemistry: 8th ed.
(b)
S CuI =
1 × 10 −12
= 1 × 10 −11
0.10 M
S AgI
8.3 × 10 −17
=
= 8.3 × 10 −16
0.10 M
S PbI 2
7.1 × 10 −9
=
= 7.1 × 10 −7
2
(0.10 M )
S BiI 3 =
8.1 × 10 −19
= 8.1 × 10 −16
3
(0.10 M )
PbI 2 > CuI > AgI > BiI 3 in 0.10 M NaI
(c)
S CuI =
1 × 10 −12
= 1 × 10 −10
0.010 M
S AgI =
8.3 × 10 −17
= 8.3 × 10 −15
0.010 M
S PbI 2 =
1 7.1 × 10 −9
= 4.2 × 10 − 4
2 0.010 M
S BiI 3 =
1 8.1 × 10 −19
3
= 1.4 × 10 − 6
3 0.010 M
PbI 2 > BiI 3 > CuI > AgI in a 0.010 M solution of the solute cation
9-18
BiOOH(s) BiO+ + OH-
Ksp = [BiO+][OH-] = 4.0×10-10
Be(OH)2(s) Be2+ + 2OH- Ksp = [Be2+][OH-]2 = 7.0×10-22 = S(2S)2 = 4S3
Tm(OH)3(s) Tm3+ + 3OH- Ksp = [Tm3+][OH-]3 = 3.0×10-24 = S(3S)3 = 27S4
Hf(OH)4(s) Hf4+ + 4OH-
Ksp = [Hf4+][OH-]4 = 4.0×10-26 = S(4S)4 = 256S5
Chapter 9
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(a)
S BiOOH = [ BiO + ] = [OH − ] S = 4.0 × 10 −10 = 2.0 × 10 −5
S Be ( OH )2
1
7.0 × 10 −22
−
3
= [ Be ] = [OH ] S =
= 5.6 × 10 −8
2
4
2+
1
3.0 × 10 −24
S Tm ( OH )3 = [Tm 3+ ] = [OH − ] S = 4
= 5.8 × 10 −7
3
27
1
4.0 × 10 − 26
[OH − ] S = 5
= 2.7 × 10 −6
4
256
Be(OH) 2 has the lowest solubility in water.
S Hf ( OH )4 = [ Hf 4+ ] =
(b)
S BiOOH =
4.0 × 10 −10
= 4.0 × 10 −9
0.10 M
S Be ( OH )2 =
7.0 × 10 − 22
= 7.0 × 10 −20
2
(0.10 M )
S Tm ( OH )3 =
3.0 × 10 − 24
= 3.0 × 10 −21
3
(0.10 M )
S Hf ( OH )4
4.0 × 10 −26
=
= 4.0 × 10 −22
4
(0.10 M )
Hf(OH) 4 has the lowest solubility in 0.10 M NaOH.
9-19
At 0ºC,
K sp = [ H 3O + ][OH − ] = 0.114 × 10 −14
[H 3O + ] = [OH − ] = 0.144 × 10 −14 = 3.38 × 10 −8 M
pH = −log[H 3O + ] = −log(3.38 × 10 −8 M) = 7.472
At 100ºC,
K sp = [ H 3O + ][OH − ] = 49 × 10 −14
[H 3O + ] = [OH − ] = 49 × 10 −14 = 7.00 × 10 −7 M
pH = −log[H 3O + ] = −log(7.00 × 10 −7 M) = 6.155
Fundamentals of Analytical Chemistry: 8th ed.
9-20
Chapter 9
(a)
→
−
HOCl + H 2 O ← OCl + H 3O
+
[OCl − ][H 3O + ]
Ka =
= 3.0 × 10 −8
[HOCl]
[ H 3O + ] = [OCl − ] [HOCl] = 0.0300 M − [H 3O + ]
[H 3O + ]2
= 3.0 × 10 −8
+
(0.0300 M − [H 3O ])
[H 3O + ]2 = 3.0 × 10 −8 (0.0300 M − [H 3O + ])
[H 3O + ]2 + 3.0 × 10 −8 [H 3O + ] − 9.0 × 10 −10 = 0
3.0 × 10 −8 + (3.0 × 10 −8 ) 2 + 4(9.0 × 10 −10 )
= 3.0 × 10 −5 M
2
−14
1.0 × 10
[OH − ] =
= 3.3 × 10 −10 M
−5
3.0 × 10 M
[H 3O + ] = −
(b)
CH 3CH 2 CH 2 COOH + H 2 O →
CH 3CH 2 CH 2 COO − + H 3O +
←
[CH 3CH 2 CH 2 COO − ][H 3O + ]
Ka =
= 1.52 × 10 −5
[CH 3CH 2 CH 2 COOH]
[H 3O + ] = [CH 3CH 2 CH 2 COO − ] [CH 3CH 2 CH 2 COOH] = 0.0600 M − [H 3O + ]
[H 3O + ]2
= 1.52 × 10 −5
+
(0.0600 M − [H 3O ])
[H 3O + ]2 = 1.52 × 10 −5 (0.0600 M − [H 3O + ])
[H 3O + ]2 + 1.52 × 10 −5 [H 3O + ] − 9.12 × 10 −7 = 0
1.52 × 10 −5 + (1.52 × 10 −5 ) 2 + 4(9.12 × 10 −7 )
[H 3O ] = −
= 9.47 × 10 −4 M
2
−14
1.0 × 10
[OH − ] =
= 1.06 × 10 −11 M
9.47 × 10 −4 M
+
Fundamentals of Analytical Chemistry: 8th ed.
(c)
+
C 2 H 5 NH 2 + H 2 O →
C 2 H 5 NH 3 + OH −
←
+
[C H NH 3 ][OH − ] K w
1.0 × 10 −14
=
=
= 4.33 × 10 −4
Kb = 2 5
[C 2 H 5 NH 2 ]
K a 2.31 × 10 −11
+
[OH − ] = [C 2 H 5 NH 3 ] [C 2 H 5 NH 2 ] = 0.100 M − [OH − ]
[OH − ]2
= 4.33 × 10 − 4
(0.100 M − [OH − ])
[OH − ]2 = 4.33 × 10 − 4 (0.100 M − [OH − ])
[OH − ]2 + 4.33 × 10 −4 [OH − ] − 4.33 × 10 −5 = 0
4.33 × 10 −4 + (4.33 × 10 −4 ) 2 + 4(4.33 × 10 −5 )
= 6.4 × 10 −3 M
2
1.0 × 10 −14
[H 3O + ] =
= 1.6 × 10 −12 M
−3
6.37 × 10 M
[OH − ] = −
(d)
(CH 3 ) 3 N + H 2 O →
(CH 3 ) 3 NH + + OH −
←
Kb =
[(CH 3 ) 3 NH + ][OH − ] K w
1.0 × 10 −14
=
=
= 6.33 × 10 −5
K a 1.58 × 10 −10
[(CH 3 ) 3 N]
[OH − ] = [(CH 3 ) 3 NH + ] [(CH 3 ) 3 N] = 0.200 M − [OH − ]
[OH − ]2
= 6.33 × 10 −5
(0.200 M − [OH − ])
[OH − ]2 = 6.33 × 10 −5 (0.200 M − [OH − ])
[OH − ]2 + 6.33 × 10 −5 [OH − ] − 1.27 × 10 −5 = 0
6.33 × 10 −5 + (6.33 × 10 −5 ) 2 + 4(1.27 × 10 −5 )
[OH ] = −
= 3.53 × 10 −3 M
2
−14
1.0 × 10
[H 3O + ] =
= 2.83 × 10 −12 M
3.53 × 10 −3 M
−
Chapter 9
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(e)
−
→
OCl + H 2 O ← HOCl + OH
−
[HOCl][OH − ] K w 1.0 × 10 −14
Kb =
=
=
= 3.3 × 10 −7
−
−8
[OCl ]
Ka
3.0 × 10
[OH − ] = [HOCl] [OCl − ] = 0.200 M − [OH − ]
[OH − ]2
= 3.3 × 10 −7
−
(0.200 M − [OH ])
[OH − ]2 = 3.3 × 10 −7 (0.200 M − [OH − ])
[OH − ]2 + 3.3 × 10 −7 [OH − ] − 6.7 × 10 −8 = 0
3.3 × 10 −7 + (3.3 × 10 −7 ) 2 + 4(6.7 × 10 −8 )
= 2.6 × 10 −4 M
2
−14
1.0 × 10
[H 3O + ] =
= 3.9 × 10 −11 M
−4
2.6 × 10 M
[OH − ] = −
(f)
C 2 H 5COO − + H 2 O →
C 2 H 5COOH + OH −
←
[C 2 H 5COOH][OH − ] K w 1.0 × 10 −14
Kb =
=
=
= 7.42 × 10 −10
−
−5
K a 1.34 × 10
[C 2 H 5COO ]
[OH − ] = [C 2 H 5COOH] [C 2 H 5COO − ] = 0.0860 M − [OH − ]
[OH − ]2
= 7.42 × 10 −10
−
(0.0860 M − [OH ])
[OH − ]2 = 7.42 × 10 −10 (0.0860 M − [OH − ])
[OH − ]2 + 7.42 × 10 −10 [OH − ] − 6.42 × 10 −11 = 0
7.42 × 10 −10 + (7.42 × 10 −10 ) 2 + 4(6.42 × 10 −11 )
= 8.01 × 10 −6 M
2
−14
1.0 × 10
[H 3O + ] =
= 1.25 × 10 −9 M
8.01 × 10 −6 M
[OH − ] = −
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(g)
+
→
HONH 3 + H 2 O ← HONH 2 + H 3O
+
[HONH 2 ][H 3O + ]
Ka =
= 1.10 × 10 −6
+
[HONH 3 ]
+
[H 3O + ] = [HONH 2 ] [HONH 3 ] = 0.250 M − [H 3O + ]
[H 3O + ]2
= 1.10 × 10 −6
(0.250 M − [H 3O + ])
[H 3O + ]2 = 1.10 × 10 −6 (0.250 M − [H 3O + ])
[H 3O + ]2 + 1.10 × 10 −6 [H 3O + ] − 2.75 × 10 −7 = 0
1.10 × 10 −6 + (1.10 × 10 −6 ) 2 + 4(2.75 × 10 −7 )
= 5.24 × 10 −4 M
2
1.0 × 10 −14
[OH − ] =
= 1.91 × 10 −11 M
−4
5.24 × 10 M
[H 3O + ] = −
(h)
+
HOC 2 H 4 NH 3 + H 2 O →
HOC 2 H 4 NH 2 + H 3O +
←
Ka =
[OHC 2 H 4 NH 2 ][H 3O + ]
= 3.18 × 10 −10
+
[HOC 2 H 4 NH 3 ]
+
[H 3O + ] = [HOC 2 H 4 NH 2 ] [HOC 2 H 4 NH 3 ] = 0.0250 M − [H 3O + ]
[H 3O + ]2
= 3.18 × 10 −10
+
(0.0250 M − [H 3O ])
[H 3O + ]2 = 3.18 × 10 −10 (0.0250 M − [H 3O + ])
[H 3O + ]2 + 3.18 × 10 −10 [H 3O + ] − 7.95 × 10 −12 = 0
3.18 × 10 −10 + (3.18 × 10 −10 ) 2 + 4(7.95 × 10 −12 )
= 2.82 × 10 −6 M
2
−14
1.0 × 10
[OH − ] =
= 3.55 × 10 −9 M
−6
2.82 × 10 M
[H 3O + ] = −
Fundamentals of Analytical Chemistry: 8th ed.
9-21
Chapter 9
(a)
→
−
ClCH 2 COOH + H 2 O ← ClCH 2 COO + H 3O
+
[ClCH 2 COO − ][H 3O + ]
= 1.36 × 10 −3
Ka =
[ClCH 2 COOH]
[H 3O + ] = [ClCH 2 COO − ] [ClCH 2 COOH] = 0.100 M − [H 3O + ]
[H 3O + ]2
= 1.36 × 10 −3
+
(0.100 M − [H 3O ])
[H 3O + ]2 = 1.36 × 10 −3 (0.100 M − [H 3O + ])
[H 3O + ]2 + 1.36 × 10 −3 [H 3O + ] − 1.36 × 10 −4 = 0
[H 3O + ] = −
1.36 × 10 −3 + (1.36 × 10 −3 ) 2 + 4(1.36 × 10 −4 )
= 0.0110 M
2
(b)
ClCH 2 COO − + H 2 O →
ClCH 2 COOH + OH −
←
Kb =
[ClCH 2 COOH][OH − ] K w 1.0 × 10 −14
=
=
= 7.35 × 10 −12
−
−3
[ClCH 2 COO ]
K a 1.36 × 10
[OH − ] = [ClCH 2 COOH] [ClCH 2 COO − ] = 0.100 M − [OH − ]
[OH − ]2
= 7.35 × 10 −12
−
(0.100 M − [OH ])
[OH − ]2 = 7.35 × 10 −12 (0.100 M − [OH − ])
[OH − ]2 + 7.35 × 10 −12 [OH − ] − 7.35 × 10 −13 = 0
7.35 × 10 −12 + (7.35 × 10 −12 ) 2 + 4(7.35 × 10 −13 )
= 8.57 × 10 −7 M
2
1.0 × 10 −14
+
= 1.17 × 10 −8 M
[H 3O ] =
−7
8.57 × 10 M
[OH − ] = −
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(c)
+
CH 3 NH 2 + H 2 O →
CH 3 NH 3 + OH −
←
+
[CH 3 NH 3 ][OH − ] K w 1.0 × 10 −14
Kb =
=
=
= 4.35 × 10 −4
[CH 3 NH 2 ]
K a 2.3 × 10 −11
+
[OH − ] = [CH 3 NH 3 ] [CH 3 NH 2 ] = 0.0100 M − [OH − ]
[OH − ]2
= 4.35 × 10 − 4
(0.0100 M − [OH − ])
[OH − ]2 = 4.35 × 10 −4 (0.0100 M − [OH − ])
[OH − ]2 + 4.35 × 10 − 4 [OH − ] − 4.35 × 10 −6 = 0
4.35 × 10 − 4 + (4.35 × 10 −4 ) 2 + 4(4.35 × 10 −6 )
= 1.88 × 10 −3 M
2
1.0 × 10 −14
[H 3O + ] =
= 5.32 × 10 −12 M
−3
1.88 × 10 M
[OH − ] = −
(d)
+
→
CH 3 NH 3 + H 2 O ← CH 3 NH 2 + H 3O
[CH 3 NH 2 ][H 3O + ]
Ka =
= 2.3 × 10 −11
+
[CH 3 NH 3 ]
+
+
[H 3O + ] = [CH 3 NH 2 ] [CH 3 NH 3 ] = 0.0100 M − [H 3O + ]
[H 3O + ]2
= 2.3 × 10 −11 [H 3O + ]2 = 2.3 × 10 −11 (0.0100 M − [H 3O + ])
+
(0.0100 M − [H 3O ])
[H 3O + ]2 + 2.3 × 10 −11 [H 3O + ] − 2.3 × 10 −13 = 0
[H 3O + ] = −
2.3 × 10 −11 + (2.3 × 10 −11 ) 2 + 4(2.3 × 10 −13 )
= 4.8 × 10 −7 M
2
(e)
+
→
C 6 H 5 NH 3 + H 2 O ← C 6 H 5 NH 2 + H 3O
+
[C 6 H 5 NH 2 ][H 3O + ]
Ka =
= 2.51 × 10 −5
+
[C 6 H 5 NH 3 ]
+
[H 3O + ] = [C 6 H 5 NH 2 ] [C 6 H 5 NH 3 ] = 0.0010 M − [H 3O + ]
[H 3O + ]2
= 2.51 × 10 −5
(0.0010 M − [H 3O + ])
[H 3O + ]2 = 2.51 × 10 −5 (0.0010 M − [H 3O + ])
[H 3O + ]2 + 2.51 × 10 −5 [H 3O + ] − 2.51 × 10 −8 = 0
[H 3O + ] = −
2.51 × 10 −5 + (2.51 × 10 −5 ) 2 + 4(2.51 × 10 -8 )
= 1.46 × 10 − 4 M
2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
(f)
−
HIO 3 + H 2 O
→
←
−
IO 3 + H 3O
+
[IO 3 ][H 3O + ]
Ka =
= 1.7 × 10 −1
[HIO 3 ]
−
[ H 3O + ] = [IO 3 ] [HIO 3 ] = 0.200 M − [H 3O + ]
[H 3O + ]2
= 1.7 × 10 −1 [H 3O + ]2 = 1.7 × 10 −1 (0.200 M − [H 3O + ])
(0.200 M − [H 3O + ])
[H 3O + ]2 + 1.7 × 10 −1 [H 3O + ] − 0.034 = 0
[H 3O + ] = −
9-22
1.7 × 10 −1 + (1.7 × 10 −1 ) 2 + 4(0.034)
= 0.12M
2
A buffer solution resists changes in pH with dilution or with addition of acids or bases. A
buffer is composed of a mixture of a weak acid and its conjugate base.
9-23
Buffer capacity of a solution is defined as the number of moles of a strong acid (or a
strong base) that causes 1.00 L of a buffer to undergo a 1.00-unit change in pH.
9-24
(a) pH = pK a + log
[NH 3 ]
+
[NH 4 ]
= − log(5.7 × 10 −10 ) + log
(0.100M)
= 8.943
(0.200M)
(b) pH = pK a + log
[NH 3 ]
+
[NH 4 ]
= − log(5.7 × 10 −10 ) + log
(0.050M)
= 8.943
(0.100M)
The solutions have identical pH values, but the solution in part (a) has the greater buffer
capacity because it has the higher concentration of weak acid and conjugate base.
Fundamentals of Analytical Chemistry: 8th ed.
9-25
(a)
HOAc + H 2 O →
H 3O + + OAc −
←
OAc − + H 2 O →
HOAc + OH −
←
0.100 M HOAc ≡
0.100 mole
L
×
× 200 mL = 0.020 mole
L
1000 mL
pH = − log(1.75 × 10 −5 ) + log
0.008
= 4.359
0.020
(b)
0.175 M HOAc ≡
0.175 mole
L
×
× 100 mL = 0.0175 mole
L
1000 mL
0.0500 M NaOH ≡
0.0500 mole
L
×
× 100 mL = 0.005 mole
L
1000 mL
[HOAc] =
(0.0175 − 0.0050)mole 1000 mL
×
= 6.25 × 10 −2 M
200 mL
L
[OAc − ] =
0.0050 mole 1000 mL
×
= 2.50 × 10 − 2 M
200 mL
L
pH = −log(1.75 × 10 −5 ) + log
2.50 × 10 − 2
= 4.359
6.25 × 10 −2
(c)
0.0420 M OAc − ≡
0.1200 M HCl ≡
0.042 mole
L
×
× 160 mL = 0.00672 mole
L
1000 mL
0.1200 mole
L
×
× 40.0 mL = 0.0048 mole
L
1000 mL
[OAc − ] =
(0.00672 − 0.0048)mole 1000 mL
×
= 9.6 × 10 −3 M
200 mL
L
[HOAc] =
0.0048 mole 1000 mL
×
= 2.4 × 10 −2 M
200 mL
L
9.6 × 10 −3
pH = −log(1.75 × 10 ) + log
= 4.359
2.4 × 10 − 2
−5
Chapter 9
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
Since the ratios of the amounts of weak acid to conjugate base are identical the three
solutions will have the same pH. They will differ in buffer capacity, however, with (a)
having the greatest and (c) the least.
9-26
(a) Malic acid / sodium hydrogen malate
(b) HOCl / OCl(c) NH4Cl / NH3
(d) pyridine / pyridinium
9-27
pH = 3.50 = pK a + log
log
[HCOO − ]
[HCOO − ]
[HCOO − ]
= − log(1.8 × 10 − 4 ) + log
= 3.74 + log
[HCOOH]
[HCOOH]
[HCOOH]
[HCOO − ]
[HCOO − ]
= −0.245
= 10 −0.245 = 0.569
[HCOOH]
[HCOOH]
1.00 M HCOOH ≡
mole HCOO −
= 0.569
mole HCOOH
1.00 mole
L
×
× 400.0 mL = 0.400 mole HCOOH
L
1000 mL
mole HCOO − = (0.569) × (0.400 mole HCOOH) = 0.228 mole
weight HCOONa = 0.228 mole HCOO − ×
1 mole HCOONa 67.997 g
×
= 15.5 g
1 mole HCOO −
1 mole
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
9-28
[HOCH 2 COO − ]
pH = 4.00 = pK a + log
[HOCH 2 COOH]
= − log(1.47 × 10 −4 ) + log
log
[HOCH 2 COO − ]
[HOCH 2 COO − ]
= 3.83 + log
[HOCH 2 COOH]
[HOCH 2 COOH]
[HOCH 2 COO − ]
= 0.167
[HOCH 2 COOH]
[HOCH 2 COO − ]
= 10 0.167 = 1.47
[HOCH 2 COOH]
mole HOCH 2 COO −
= 1.47
mole HOCH 2 COOH
1.00 M HOCH 2 COOH ≡
1.00 mole
L
×
× 300.0 mL = 0.300 mole HOCH 2 COOH
L
1000 mL
mole HOCH 2 COO − = (1.47) × (0.300 mole HOCH 2 COOH) = 0.441 mole
weight HOCH 2 COONa = 0.441 mole HCOO − ×
1 mole HOCH 2 COONa 98.01g
×
= 43.2 g
1 mole HOCH 2 COO −
1 mole
Fundamentals of Analytical Chemistry: 8th ed.
9-29
0.300 M C 6 H 5CHOHCOONa ≡
0.300 mole
L
×
× 250.0 mL
L
1000 mL
= 0.0750 mole C 6 H 5CHOHCOONa
[C 6 H 5CHOHCOO − ]
pH = 3.37 = pK a + log
[C 6 H 5CHOHCOOH]
= − log(4.0 × 10 −4 ) + log
log
[C 6 H 5CHOHCOO − ]
[C H CHOHCOO − ]
= 3.40 + log 6 5
[C 6 H 5CHOHCOOH]
[C 6 H 5CHOHCOOH]
[C 6 H 5CHOHCOO − ]
[C 6 H 5CHOHCOO − ]
= −0.03
= 10 −0.03 = 0.933
[C 6 H 5CHOHCOOH]
[C 6 H 5CHOHCOOH]
mole C 6 H 5CHOHCOO −
= 0.933
mole C 6 H 5CHOHCOOH
(mole C 6 H 5CHOHCOO − − x mole HCl)
= 0.933
x mole HCl
mole C 6 H 5CHOHCOO − − x mole HCl = 0.933 ( x mole HCl)
mole C 6 H 5CHOHCOO − 0.075 mole
=
= 0.0388
1.933
1.933
L
1000 mL
volume HCl = 0.0388 mole ×
×
= 194 mL HCl
0.200 mole
L
x mole HCl =
Chapter 9
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 9
9-30
1.00 M HOCH 2 COOH ≡
pH = 4.00 = pK a + log
[ HOCH 2 COO − ]
[ HOCH 2 COOH ]
= − log(1.47 × 10 −4 ) + log
log
1.00 mole
L
×
× 300.0 mL = 0.300 mole HOCH 2 COOH
L
1000 mL
[ HOCH 2 COO − ]
[ HOCH 2 COO − ]
= 3.83 + log
[ HOCH 2 COOH ]
[ HOCH 2 COOH ]
[ HOCH 2 COO − ]
= 0.167
[ HOCH 2 COOH ]
−1
[ HOCH 2 COO − ]
= 101.67×10 = 1.47
[ HOCH 2 COOH ]
mole HOCH 2 COO −
= 1.47
mole HOCH 2 COOH
x mole NaOH
= 1.47
( mole HOCH 2 COOH − x mole NaOH )
x mole NaOH = 1.47 × ( mole HOCH 2 COOH − x mole NaOH )
1.47 × ( mole HOCH 2 COO − ) 1.47 × (0.300 mole)
=
= 0.179 mole
2.47
2.47
L
1000 mL
volume NaOH = 0.179 mole ×
×
= 89.3 mL NaOH
2.00 mole
L
x mole NaOH =
9-31
The statement “A buffer maintains the pH of a solution constant” is false. The change in
pH of a buffered solution is relatively small with the addition of a small volume of acid
or base as shown in the example below.
pH = pK a + log
[ NaA ]
[ HA ]
mL of 0.050M NaOH
[ NaA ]
[ HA ]
∆pH
1.48
0.170
100
1.59
0.200
200
1.70
0.230
300
1.83
0.262