EoC
9-1
Briefly describe or define and given an example of…
(a) a weak
electrolyte
A weak electrolyte only partially ionizes when dissolved in water. H2CO3 is an example of a weak
electrolyte.
(b) a BrønstedLowry acid
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 The conjugate acid of a Brønsted-Lowry base is the potential proton donator formed when a Brønstedof a Brønsted-Lowry Lowry base accepts a proton. For example, the NH 4 + is a conjugate acid in the reaction, NH3 +
base
proton ⇌ NH4 +.
(d) neutralization, in
terms of the
Brønsted-Lowry
concept
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,
NH3 + H2 O ⇌ NH4 + + OHIn the example above, NH 3 acts as a base with NH4 + as its conjugate acid. H2 O acts as an acid
with OH– as its conjugate base.
(e) and amphiprotic An amphiprotic solvent can act either as an acid or a base depending on the solute. Water is an
solvent
example of an amphiprotic chemical species.
(f) a zwitterion
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. NH2 CH3 COOH ⇌ NH 3 +CH2 COO - .
(g) autoprotonolysis Autoprotolysis is the act of self-ionization to produce both a conjugate acid and a conjugate base.
EoC
9-2
(h) a strong acid
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.
(i) Le Châtelier
principal
The Le Châtelier principle states that the position of an equilibrium always shifts in such a direction
that it relieves the stress. A common ion like sulfate added to a solution containing sparingly
soluble BaSO 4 is an example
(j) the common-ion
effect
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. Chloride ion added to a AgCl solution decreases the solubility of Ag +
because of the common ion effect.
9-2 Briefly describe or define and given an example of…
An amphiprotic solute is a chemical species that possesses both acidic and basic properties. The
(a) an amphiprotic solute
dihydrogen phosphate ion, H2 PO 4 – , is an example of an amphiprotic solute.
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
(b) a differentiating solvent
acid.
(c) a leveling solvent
A leveling solvent shows no difference between strong acids. Perchloric acid and hydrochloric acid
ionize completely in water; thus, water is a leveling solvent.
A mass-action effect arises when a shift in the chemical equilibrium occurs due to the introduction of
(d) a mass-action ion effect
one of the participating chemical species (i.e., addition of one of the reactants or products.
EoC
9-3
9-3 Briefly explain why there is no term in an equilibrium constant expression for water or for a
pure solid, even though one (or both) appears in the balanced net ionic equation for the
equilibrium.
For dilute aqueous solutions, the concentration of water remains constant and is assumed to be independent of the
equilibrium. 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.
EoC
9-4
9-4 Identify the acid on the left and its conjugate base on the right in the following equations:
(a) HOCl + H2 O ⇌ H3 O + + OCl(b) HONH2 + H2 O ⇌ HONH3 + +
(c)OH
NH- 4 + + H2 O ⇌ NH3 + H3 O +
HOCl A
HONH2 B
NH4 + A
H2 O B
H2 O A
H2 O B
NH3 B
H3 O + A
HONH3 + A
OCl B
OH- B
H3 O + A
(d) 2 HCO 3 - ⇌ H2 PO 4 - + CO 3 2- (e) PO 4 3- + H2 PO 4 - ⇌ 2 HPO 4 22 HCO 3 - B and A
PO 4 3- B
H2 PO 4 A
H2 PO 4 - A
2CO 3 B
2HPO 4 2- B and A
EoC9
P5
9-5 Identify the base on the left and its conjugate acid on the right from equations the equations
for problem 9-4
(a) HOCl + H2 O ⇌ H3 O + + OCl(b) HONH2 + H2 O ⇌ HONH3 + +
(c)OH
NH- 4 + + H2 O ⇌ NH3 + H3 O +
HOCl BL acid
HONH2 BL base
NH4 + BL acid
H2 O BL base
H2 O BL acid
H2 O BL base
NH3 conj. Base
H3 O + conj. Acid
HONH3 + conj. Acid
OCl conj. Base
OH- conj. Base
H3 O + conj. Acid
(d) 2 HCO 3 - ⇌ H2 PO 4 - + CO 3 2- (e) PO 4 3- + H2 PO 4 - ⇌ 2 HPO 4 22 HCO 3 - amphiprotic species
PO 4 3- BL base
H2 PO 4 conj. Acid
H2 PO 4 - BL acid
2conj.
Base
CO 3
2HPO 4 2- amphiprotic species, both conj. acid and conj. base
EoC9
P6
9-6 Write the expressions for the autoprotonolysis of
(a) H2 O
2 H2 O ⇌ H3 O + + OH(b) CH3 COOH
2 CH3 COOH ⇌ CH 3 COOH 2 +
(c) CHNH2
2 CHNH2 ⇌ CHNH3 + +
(d) CH3 OH
2 CH3 OH ⇌ CH3 OH2 + +
EoC9
P7
9-7 Write the equilibrium-constant expressions and obtain numerical values for each constant in
(a) the basic dissociation of aniline,
C 6 H4 NH2 .
K eq = [C 6 H4 NH3 +][OH - ]
=
[C 6 H4 NH2 ]
(b) the acidic dissociation of hypochlorous
acid, HClO.
K eq =
[ClO - ][H3 O +]
[HClO]
=
(c) the acidic dissociaton of methyl
ammonium hydrochloride, CH3 NH3 Cl.
(d) the basic dissociation of NaNO3 .
EoC9
P8
K eq = [CH3 NH2 - ][H3 O +]
=
[CH3 NH3 +]
K eq =
[HNO 3 ][OH - ]
[NO 3 - ]
=
(e) the dissociation of H3 AsO 3 to H3 O +
and AsO3 3- .
K eq = [H3 O +] 3 [AsO 3 3- ]
[H3 AsO 3 ]
=
(f) the reaction C2 O 4 2- with H2 O to give
H2 CO 4 and OH
K eq =
[H2 CO 4 ][OH - ] 2
[C 2 O 4 2- ]
=
9-8 Generate the solubility-product expression for
(A) CuBr
(B) HgClI
(C) PbCl 2
(D) La(IO 3 )3
(E) Ag3 AsO 4
3+
+
3CuBr (s ) ⇌ Cu- + Br- HgClI (s ) ⇌ Hg2+ + ClPbCl
+ I-2 (s ) ⇌ Pb 2+ + 2Cl
La(IO
+ Ag
3IO
3 )3 (s ) ⇌ La
3 AsO
3
4 (s ) ⇌ Ag + AsO4
+
2+
2+
- 2
3+
- 3
+ 3
3K sp = [Cu ][Br ]
K sp = [Hg ][Cl ][I ] K sp = [Pb ][Cl ]
K sp = [La ][IO 3 ] K sp = [Ag ] [AsO 4 ]
EoC9
P9
9-9 Express the solubility-product constant for each substance in Problem 9-8 in terms of its
molar solubility S.
(A) CuBr
(B) HgClI
(C) PbCl 2
(D) La(IO 3 )3
(E) Ag3 AsO 4
S = [Cu+]=[Br - ]
S = [Hg2+]=[Cl - ]=[I - ]S = [Pb 2+]=½[Cl - ] S = [La 3+]=⅓[IO 3 - ] S = [AsO 4 3- ]=⅓[Ag +]
K sp = S × S = S 2
K sp = S × S × S = S 3 K sp = S × (2S )2 = 4S 3 K sp = S × (3S )3 = 27S K4 sp = 3S × S = 3S 2
EoC9
P10
9-10 Calculate the solubility-product constant for each of the following substances, given that the
molar concentrations of their saturated solutions as indicated
(A) CuBr
(B) HgClI
(C) PbCl 2
(D) La(IO 3 )3
(E) Ag3 AsO 4
2+
22+
3+
AgSeCN ⇌ Ag+ + SeCN
RaSO
⇌
Ra
+
SO
Pb(BrO
)
⇌
Pb
+
2BrO
Ce(IO
)
⇌
Ce
+
3IO
4
4
3 2
3 3 3
3
+
[Ag ] = [SeCN ]
EoC9
P11
9-11 Calculate the solubility of the solutes in Problem 9-10 for solutions in which the cation
concentration is 0.030 M.
(A) CuBr
(B) HgClI
(C) PbCl 2
(D) La(IO 3 )3
(E) Ag3 AsO 4
2+
22+
3+
AgSeCN ⇌ Ag+ + SeCN
RaSO
+ SO4Pb(BrO
+ 2BrO
Ce(IO
+ 3IO 3 4 ⇌ Ra
3 )2 ⇌ Pb
3 3 )3 ⇌ Ce
[Ag+] = [SeCN - ]
EoC9
P12
9-12 Calculate the solubility of the solutes in Problem 9-10 for solutions in which the anion
concentration is 0.030 M.
(A)
(B)
(C)
(D)
2+
22+
3+
AgSeCN ⇌ Ag+ + SeCN
RaSO
+ SO4Pb(BrO
+ 2BrO
Ce(IO
+ 3IO 3 4 ⇌ Ra
3 )2 ⇌ Pb
3 3 )3 ⇌ Ce
+
[Ag ] = [SeCN ]
EoC9
P13
9-13 What CrO 4 2- concentration is required to…
Ag2 CrO 4 (s ) ⇌ 2Ag+ + CrO 4 2K sp =
1.20E-12
[CrO 4 2- ] = K sp / [Ag+] 2
(a) initiate a precipitate
of Ag2 CrO 4 from a
solution that is
[Ag+] = 0.00413
[CrO 4 2- ] =7.04E-08
######
EoC9
P14
(b) lower the
concentration of Ag+
in a solution to
[Ag+] = 9E-07
[CrO 4 2- ] =1.48E+00
######
9-14 What hydroxide concentration is required to
(a) initiate
(a) lower
EoC9
P15
9-15 The solubility-product constant for Ce(IO 3 )3 is 3.2×10-10 . What is the Ce 3+ concentration
in a solution prepared by mixing 50.00 mL of 0.0450 M Ce 3+ with 50.00 mL of…
(a)
(b)
(c)
(d)
water
0.0450 M
0.250 M
0.0500 M
EoC9
P16
9-16 The solubility-product constant for K2 PdCl 6 is 6.0×10-6 (K2 PdCl 6 ⇌ 2 K+ + PdCl 6 2- ).
Wjhat is the K+ concentration of a solution prepared by mixing 50.0 mL of a 0.200 M KCl with
50.0 mL of…
(a)
(b)
(c)
0.160 M PdCl 6 2- ?
0.0800 M PdCl 6 2- ?
0.240 M PdCl 6 2- ?
EoC9
P17
9-17 The solubility product for a series of iodides are… list these four compounds in decreasing
order of molar solubility in…
CuI K sp
AgI K sp =
PbI 2 K sp
BiI 3 K sp
EoC9
P18
1 × 10-12
8.3 × 107.1 × 10-9
8.1 × 10-
(a) water
(b) 0.20 M
(c) a 0.020
9-18 The solubility products for a series of hydroxides are
+
Which
hydroxide…
BiOOH K sp = 4.0 × 10-10 = [BiO
][OH
]
-22
(a) has the lowest molar solubility in H 2 O
Be(OH) 2 K sp = 7.0 × 10
(b) the lowest molar solubility in a solution that is 0.30 M in NaOH?
Tm(OH)3 K sp = 3.0 × 10-24
Hf(OH)4 K sp = 4.0 × 10-26
EoC9
P19
9-19 Calculate the pH of water at 25°C and 75°C. The values for p K w at these temperatures
are 13.99 and 12.70, respectively.
EoC9
P20
9-20 At 25°C, what are the molar H3 O + and OH- concentrations in…
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
0.0300 M C6 H5 COOH
0.600 M HN3
0.100 M ethylamine
0.200 M trimethylamine
0.200 M C6 H5 COONa (sodium benzoate)
0.0860 M CH3 CH2COONa
0.250 M hydroxulamine hydrochloride?
0.0250 M ethyl ammonium chloride
EoC9
P21
9-21 at 25° C, what is the hydronium ion concentration?
(a)
(b)
(c)
(d)
(e)
(f)
0.200 M chloroacetic acid
0.200 M sodium chloroacetate
0.0200 M methylamine
0.0200 methylamine hydrochloride
2.00 × 10-3 M aniline hydrochloride
0.300 M HIO3
EoC9
P22
9-22 define buffer solution
a buffer solution resists change in pH with dilution or
with addition of acids or bases. A buffer is composed
EoC9
P23
9-23 define buffer capacity
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
EoC9
P24
9-24 Which has the greater buffer capacity?
(Kb NH3 = 1.75 × 10-5 )
pH = pKa + log([NaA]/[HA])
(b)
NH3
NH4 Cl
a mixture containing 0.100 mol NH
4.756962
0.026
mol NH
0.065
3 and 0.200
4 Cl
or
4.356962 0.13
0.325
a mixture containing 0.050 mol NH
4.356962
3 and 0.100 mol NH4 Cl
EoC9
P25
9-25 Consider solutions prepared by (a), (b), and (c). In what respects do these solutions
resemble one another? How do they differ?
(a)
(b)
(c)
dissolving 8.00 mmol of NaOAc in 200 mL of 0.100 M HOAc
adding 100 mL of 0.0500 M NaOH to 100 mL of 0.175 M HOAc
adding 40.0 mL of 0.1200 M HCl to 160.0 mL of 0.0420 M NaOAc
a)
EoC9
P26
(a)
(b)
(c)
(d)
<<< this one has more
because of higher concentration of HA (weak acid) and NaA (conjugate base
9-26 Consult Appendix 3, and pick out a suitable acid/base pair to prepare a buffer with a pH of
4.5
8.1
10.3
6.1
look at the K a of
each and then use
3.98E-10
9.09E-09
4.33E-04
9.52E-07
EoC9 9-27 What mass of sodium formalate should be added to 500.0 mL of 1.00 M formalic acid to
P27
produce a buffer solution with a pH of 3.50?
NaF = sodium formalate
pH = pKa + log [NaF]/[HF]
HF = formic acid
HF volume0.500
HF molarity
1.00
moles HF 0.5
HF Ka
0.00018
HF pKa 3.744727
pH =
3.50
pH - pKa =-0.24473
10(pH - pKa) =0.57
moles NaF0.284605
molar mass68.0069
NaF
1.00E-14
mass NF =19.3551 1.70E-01
9.400117
8.041436
3.363512
6.021363
2.51E-05
1.1E-06
2.31E-11
1.05E-08
4.599883
5.958564
10.63649
7.978637
anilium
hydroxyl
ethyl
hydrazine
5.88E-14
EoC
P9 28
9-28 What mass of sodium glycolate should be added to 400.0 mL of 1.00 M glycolic acid to
produce a buffer solution with a pH of 4.00?
pH = pKa + log [NaG]/[HG]
NaG = sodium glycolate
HG = glycolic acid
HG volume0.400
HG molarity
1.00
moles HG 0.4
HG Ka 0.000147
HG pKa 3.832683
pH =
4.00
pH - pKa =0.167317
10(pH - pKa) =1.47
moles NaG0.588
molar mass98.0327
NaG
mass NG =57.64323
EoC
P9 29
0.500
1.10
0.55
0.00018
3.744727
3.40
-0.34473
0.45
0.248677
69.0148
17.16238
9-29 What volume of 0.200 M HCl must be added to 500.0 mL of 0.300 M sodium mandelate to
produce a buffer solution with a pH of 3.37?
HCl molarity
0.2
vol NaA
0.5
NaA molarity
0.3
moles NaA
0.15
[HA (conjugate
0.0004
of NaA)] Ka
[HA]'s pKa 3.39794
pH
3.37
pH - pKa =-0.02794
10(pH-pKa) 0.937692
moles HCl 0.077412
volume HCl387.0585
=
EoC
P9 30
9-30 What volume of 2.00 M NaOH must be added to 150.0 mL of 1.40 M glycolic acid to
produce a buffer solution having a pH of 3.80?
NaOH molarity 2
vol HA
0.2
HA molarity
1
moles HA
0.2
Ka
0.000147
pKa
3.832683
pH =
4
pH - pKa =0.167317
[A-]/[HA] = 1.47
moles NaOH
0.119028
volume NaOH
59.51417
=
EoC
P9 31
(This
problem
requires
values in
your
textbook's
specific
appendic
es, which
you can
access
through
(This
problem
requires
values in
your
textbook's
specific
appendic
es, which
you can
access
through
9-31 Is the following statement true or false, or both? Define your answer with equations,
examples, or graphs. "a buffer maintains the pH of a solution constant."