#9 Homework holcombe (51160)
1
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Multiple-choice questions may continue on
19
4. 5.2 10
before answering.
5. 1.69 10
the next column or page find all choices
001
10.0 points
A student titrated a sample containing 102
mL of 0.3 M NaOH with 0.67 M HCl. What
is the pH of the solution after 20 mL of the
hydrochloric acid has been added?
14
13
6. 7.744 10
17
7. 8.281 10
correct Explanation:
9
S = 9.1 10
mol/L
The solubility equilibrium is
1. 7
AgI(s)
2. 14.16
+
Ag+(aq) + I(aq)
9
[Ag ] = [I ] = S = 9.1 10
3. 0.84
+
Ksp = [Ag ][I ]
4. 13.16 correct
= 9.1 109
5. 0.16
2
17
Explanation:
VNaOH = 102 mL
= 8.281 10
[NaOH] = 0.3 M
VHCl = 20 mL
[HCl] = 0.67 M
Initially
nNaOH : (102 mL)(0.3 M) = 30.6 mmol
nHCl : (20 mL)(0.67 M) = 13.4 mmol
NaOH
+
ini 30.6 mmol
13.4 mmol
fin 17.2 mmol
HCl
13.4 mmol
13.4 mmol
0 mmol
NaCl
+H2O
Vnew = 102 mL + 20 mL = 122 mL
[NaOH] =
mol/L
17.2 mmol
122 mL
= 0.141 M = [OH]
pOH = log[OH ] = log(0.141) = 0.84
pH = 14 pOH = 14 0.84 = 13.16
002
10.0 points
What is Ksp for AgI, if its molar
9
solubility is 9.1 10
mol/L?
003
10.0 points
What is Ksp for Ca(OH)2, if its
molar solu-bility is 0.011 mol/L?
25
1. 5.2 10
21
2. 1.4 10
6
3. 5.3 10
correct
14
4. 1.7 10
11
5. 6.9 10
30
6. 4.7 10
8
7. 8.3 10
Explanation:
S = 0.011 mol/L
The solubility equilibrium is
9
1. 6.9 10
21
2. 1.4 10
3. 0.0053
Ca(OH)2(s)
2+
2+
Ca (aq) + 2 OH (aq)
[Ca ] = S = 0.011 mol/L
[OH ] = 2 S = 0.022 mol/L
Homework #9 holcombe (51160)
Explanation:
All of the salts are composed of a
single cation and anion, so the solubility
of each is the square root of its Ksp.
2+
2
Ksp = [Ca ][OH ]
2
= (0.011) (0.022)
6
= 5.324 10
004
10.0 points
Rank following salts from least to most solu-
ble: BiI
19
Cd3(AsO4)2
AlPO4 CaSO4
Ksp = 7.7 10
33
Ksp = 2.2 10
21
Ksp = 9.8 10
5
Ksp = 4.9 10
1.
AlPO4 < BiI < Cd3(AsO4)2 < CaSO4
correct
2. CaSO4 < AlPO4 < BiI < Cd3(AsO4)2
3. Cd3(AsO4)2 < CaSO4 < AlPO4 < BiI
4. BiI < Cd3(AsO4)2 < CaSO4 < AlPO4
Explanation:
Molar solubility can be approximated by
th
taking the n root of the Ksp where n is the
number of ions in the salt. Doing so results
10
in approximate molar solubilities of 10
,
7
11
3
10 , 10
and 10
for bismuth iodide,
cadmium arsenate, aluminum phosphate
and calcium sulfate, respectively. Arranging
these from least to greatest produces:
AlPO4 < BiI < Cd3(AsO4)2 < CaSO4.
005
10.0 points
Rank the following salts from greatest to
least molar solubility:
10
A) BaCrO4
Ksp = 1.17 10
21
B) AlPO4
Ksp = 9.84 10
C) CaSO4
D) FeS
2
006
10.0 points
The solubility product constant of PbCl2 is
5
1.7 10 . What is the maximum
2+
concentra-tion of Pb
that can be in
ocean water that contains 0.0500 M NaCl?
1. 3.4 10
2. 8.5 10
3. 1.7 10
4. 4.2 10
5. 6.8 10
3
7
3
8
3
M
M
M
M
M correct
Explanation:
5
Ksp of PbCl2 = 1.7 10
PbCl2 Pb
2+
+ 2FCl
2+
2
Ksp = [Pb ] [Cl ]
5
2+
2
1.7 10
= [Pb ] (0.05)
2+
[Pb ] = 0.0068
007
10.0 points
AgCl would be least soluble in
1. 0.1 M NH3.
2. 0.1 M CaCl2. correct
3. pure water.
5
Ksp = 4.93 10
Ksp 8 = 10
19
1. C > A > B > D
4. 0.1 M HCl.
5. 0.1 M HNO3.
Explanation:
2. A > C > D > B
3. C > A > D > B correct
4. A > C > B > D
008
10.0 points
Does a precipitate form when 100 mL of
0.00250 M AgNO3 and 100 mL of 0.00200
M NaBr solutions are mixed?
Homework #9 holcombe (51160)
3
Explanation:
1. AgBr does not precipitate.
010 10.0 points
Determine the molar solubility of some salt
2. AgBr precipitates. correct
2
Explanation:
VAgNO3 = 0.1 L
VNaBr = 0.1 L
(AB2) if Ksp = 2.56 10 .
[AgNO3] = 0.0025 M
AgNO3 + NaBr AgBr + NaNO3
+
2. 0.1 M
13
Ksp = [Ag ] [Br ] = 5.0 10
+
(0.1 L) (0.0025 M Ag ) = 0.00025 mol
AgBr(s) Ag
1. 10 M
[NaBr] = 0.002 M
(aq) + Br (aq)
+
3. 0.005 M
4. 1 M
5. 4 M correct
(0.1 L) (0.002 M Br ) = 0.0002 mol
Explanation:
total volume = 0.1 L + 0.1 L = 0.2 L
2
3
The equation for Ksp = x(2x) = 4x
0.000250. 2molL
Ag+
1
1/3
x = 4 Ksp . Plugging in numbers gives
you 3.97 M, or about 4 M.
= 0.00125 mol/L Ag+
0.0002 mol Br
= 0.001 mol/L Br
0.2 L
011
10.0 points
A saturated solution of MgF 2(s) is found to
6
= 1.25 10
have a Mg
concentration of 2.0
>K
M.
9
1. 8.0 10
8
2. 3.2 10
correct
9
3. 4.0 10
sp
10.0 points
Fe(OH)3(s) is very insoluble in water (Ksp =
39
10
What is the value of Ksp for MgF2?
AgBr precipitates.
009
3
2+
+
[Ag ] [Br ] = (0.00125 mol/L)
(0.001 mol/L)
3+
1.6 10
; however, Fe
forms a strong
complex with EDTA (FeEDTA,
Kf =
25
1.3 10 ). For a solution which is at equilibrium and contains Fe(OH)3(s) precipitate,
which of the following occurs if EDTA is
added to the solution?
2. No more Fe(OH)3(s) dissolves, but Fe
complexes with EDTA.
5. 1.0 10
12
9
Explanation:
2+
3
[Mg ] = 2.010
M
MgF2 Mg
1. Nothing happens because Ksp is much
smaller than Kf .
4. 4.0 10
3+
3. More Fe(OH)3(s) dissolves. correct
4. More Fe(OH)3(s) precipitates because
3+
[Fe ] decreases.
Ksp = [Mg
=4x
3
2+
2+
+2F
2
2
] [F ] = (x) (2x)
3 3
8
= 4 2 10
= 3.2 10
012 10.0 points
The molar solubility of PbSO4 is 1.34
Homework #9 holcombe (51160)
4
10 mol/L. What is the value for the Ksp
for PbSO4?
1. 3.45 10
2. 1.16 10
3. 8.8 10
4. 1.8 10
4
=
2.37 104
(1) 3
11
correct
1. 0.1 M KOH
2. 0.1 M HCl correct
PbSO4 Pb
2
+ SO4
4 2
8
= (1.34 10 ) = 1.8 10 .
3. 0.1 M Ca(OH)2
4. 0.1 M FeCl2
013 10.0 points
Arrange the compounds
36
Ksp = 1.3 10
I) CuS
5
Ksp = 1.6 10
II) PbCl2
18
Ksp = 6.3 10
III) FeS
Ksp = 2.6 1018
IV) Hg2Cl2
Ksp = 2.0 1047
V) Cu2S
in increasing order of molar solubility.
1. II, III, IV, I, V
2. V, I, IV, III, II
3. I, V, III, IV, II correct
4. II, IV, III, V, I
5. I, II, III, IV, V
Explanation:
014
10.0 points
What would be the molar solubility of Li3PO4
(Ksp = 2.37 10
1. 2.37 10
2. 1.54 10
3. 5.44 10
4. 1.24 10
4
4
2
2
1
=
= 2.37 104
015
10.0 points
Ferrous hydroxide is a slightly soluble
base. In which of the following would
Fe(OH)2 be most soluble? It is not
necessary to know Ksp for Fe(OH)2.
2+
Ksp = x
solubility
K
+3
[Li ]
99
5. 3.0 10
Explanation:
2
Explanation:
molar
sp
7
8
4
) in a 1 M LiCl solution?
correct
Explanation: