Exam II
Chemical Equlibria
LeChatlier’s Principle
Acid Dissociation
Base Hydrolysis
Titration Curves
Solubility Product
pH Controlled Solubility
Amphoterism
Equilibria and LeChatliers Principle
1) For each of the following sets of compounds write the equilibrium reaction that would occur
when the compounds are mixed together.
a) HNO3 and NaBenz
b) NH4Cl and KOH
c) NaCN and NaOH
2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after
the system reaches equilibrium?
H2 + Cl2 ---> 2 HCl
K = 150
2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been
added to the container originally?
3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium
concentration of all ionic species. Ka = 6.76x10-4
4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5
5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M
KOH? Ka = 6.46x10-5
6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65
mL of HCl? Kb = 1.8x10-5
7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a
solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15
8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5
Ac- + H2O <---> HAc + OH-
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction,
2 NH3 ↔ N2 + 3 H2
Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total
pressure) be in the system?
10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic
acid? Ka = 1.8x10-5
11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M
NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will
not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5).
12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl
are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M.
13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and
0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M
14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M
NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10.
15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH
are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp2-
1.38x10-4
1.51x10-10
0 mL
200 mL
400 mL
700 mL
800 mL
900 mL
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of
H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a
new buffer of pH 4.5?
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the
following volumes of 1.20 M NaOH?
H3His ↔ H+ + H2HisH2His- ↔ H+ + HHis2HHis2- ↔ H+ + His3-
pKa1 = 2.40
pKa2 = 6.04
pKa3 = 9.33
0 mL
30 mL
50 mL
75 mL
100 mL
125 mL
135 mL
150 mL
175 mL
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH =
8.0? Kb = 1.8x10-5
18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make
a buffer of pH = 4.0? pKa = 4.74 for acetic acid.
19) Another way of making the buffer from 2 above would be to titrate the acid with a base.
What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer
of pH = 4.0? pKa = 4.74 for acetic acid.
20) How much 0.5 M NaOH must be added to 0.75 M H3PO4 to make a buffer of pH = 8? Ka1 =
2.12, Ka2 = 7.21, Ka3 = 12.67
Solubility Products
21) Are the following molecules acidic, basic or neutral in aqueous solution?
NaF
Cr(NO3)3
KCl
NH4CN
acid
acid
acid
acid
basic
basic
basic
basic
neutral
neutral
neutral
neutral
can't tell
can't tell
can't tell
can't tell
22) Are the following compounds soluble or insoluble in water?
NaIO3
Cr(OH)3
PbSO4
CaCl2
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
FeSO4
ScCl3
Na2O
PbSO4
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5
24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6
25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8
26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8
27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10
28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium?
Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20
29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl.
Assume a liter of solution and calculate the concentration of NH3(aq).
AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10
Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9
30) Copper(I) ions in aqueous solution react with NH3 according to,
Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010
Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium
concentration of NH3 is 0.185 M.
31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for
PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-).
PbS ↔ Pb2+ + S2- Ksp = 8x10-28
H2S ↔ 2 H+ + S2- Keq = 1x10-20
32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6.
Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-).
33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH
= 3?
CdS ↔ Cd2+ + S2- Ksp = 8x10-27
H2S ↔ 2H+ + S2Keq = 1x10-20
34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated
solution of H2S = 0.10 M at pH = 3?
Ag2S ↔ 2 Ag+ + S2- Ksp = 1.6x10-49
H2S ↔ 2H+ + S2Keq = 1x10-20
35) How much of each precipitate will form and what will the concentration of lead be if
0.75 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30 M
Mg(NO3)2 and 0.5 M Pb(NO3)2?
MgOx(s) ↔ Mg2+ + Ox2- Ksp = 8.6x10-5
PbOx(s) ↔ Pb2+ + Ox2- Ksp = 4.8x10-10
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid
(0.034M) and which is buffered at pH = 9?
Ag2CO3 ↔ 2 Ag+ + CO32H2CO3 ↔ 2 H+ + CO32-
Ksp = 8.1x10-12
Ka = 2.02x10-17
Equilibria and LeChatliers Principle
1) For each of the following sets of compounds write the equilibrium reaction that would occur
when the compounds are mixed together.
a) HNO3 and NaBenz
b) NH4Cl and KOH
c) NaCN and NaOH
HBenz ↔ H+ + BenzNH4OH ↔ NH4+ + OHCN- + H2O ↔ HCN + OH-
2a) If H2 and Cl2 are added to a container, both at 2 atm, what will the pressure of HCl be after
the system reaches equilibrium?
H2 + Cl2 ↔ 2 HCl K = 150
x
x
4 atm – 2x
Since the equilibrium constant is large (150) we must shift the ingredients to the right.
This means that 2 atm H2 and 2 atm Cl2 will produce 4 atm HCl. Using this value we can
now solve the equilibrium
[4 − 2 x]2
= 150
x2
You do not have to assume that 2x is small. You can solve this exactly.
Take the square root of both sides and then solve for x. So, x = 0.2808
atm so the pressure of HCl = 4 atm – 2(0.2808 atm) = 3.44 atm
2b) If the equilibrium pressure of HCl is 2 atm what pressures of H2 and Cl2 must have been
added to the container originally?
[2atm]2
= 150
x2
x = 0.1633 atm H2 and Cl2 left after equilibrium. To make 2 atm of HCl, 1
atm H2 and 1 atm Cl2 must have been consumed. Therefore there must
have been 1 atm + 0.1633 = 1.1633 atm H2 and Cl2 to start.
3) If 100 grams of NaF and 70 grams of KOH are added to 250 mL water what is the equilibrium
concentration of all ionic species. Ka = 6.76x10-4 Note: HF is not ionic but was included.
100g NaF/42 g/mol = 2.381 mol NaF and 70g KOH/56.1 g/mol = 1.248 mol KOH
[ HF ][OH − ] [ HF ][4.99 + x]
1x10 −14
F + H2O ↔ HF + OH
=
=
⇒ [ HF ] = 2.822 x10 −11
−
−4
[F ]
[9.52 − x]
6.76 x10
-14
-14
-15
+
[H+] = 1x10 /[OH ] = 1x10 /[4.99 M] = 2x10 M H
Ion
Moles Before Rxn
Moles After Rxn
Conc (250 mL total)
Na+
FK+
OHHF
H+
2.381
2.381
1.248
1.248
0
0
2.381
2.381
1.248
1.248
9.52 M
9.52 M
4.99 M
4.99 M
2.822x10-11
2x10-15
From equilibrium
From equilibrium
4) What is the pH of 100 mL of 1.2 M HAc after 20 mL of 2 M NaOH is added? Ka = 1.8x10-5
(1.2M HAc)(0.100 L) = 0.12 mol HAc and (2M)(0.020L) = 0.040 mol NaOH
HAc + NaOH ↔ H2O + NaAc
0.12
0.040
0
-0.04
-0.040
+0.040
0.08
0
0.040
HAc ↔ H+ + Ac0.08-x x 0.04+x
[ H + ][ Ac − ] [ H + ][0.040 + x]
=
= 1.8 x10 −5 ⇒ [ H ] = 3.6 x10 −5 pH = 4.44
[ HAc]
[0.080 − x]
5) What is the pH of a solution that is made from 0.10 M HBenz, 0.25 M NaBenz, and 0.25 M
KOH? Ka = 6.46x10-5
Benz- + H2O ↔ OH- +
0.25
0.25
+0.10
-0.10
0.35-x
0.15+x
HBenz
0.10
-0.10
0+x
If you ignore x, [OH-] = 0.15 M and [H+] = 1x10-14/0.15 = 6.66x10-14 so pH = 13.18
6) What is the pH of a solution where 100 mL of 0.50 M NH4OH is completely neutralized by 65
mL of HCl? Kb = 1.8x10-5
If the NH4OH is completely neutralized, then the only thing left in solution is NH4Cl (an acid).
The only question is, what is the concentration of the NH4Cl? Total volume = 100 mL + 65 mL
(0.50 M)(0.100 L) = 0.050 mol NH4OH = 0.050 mol NH4Cl
0.050 mol NH 4 Cl
= 0.3030 M NH 4 Cl
0.165 L
NH4+ ↔ H+ + NH3
0.3030-x x
x
[H + ][NH3 ]
x2
=
= 5.55x10−10 ⇒ x = 1.297 x10−5 M H + pH = 4.89
[NH +4 ]
0.3030 − x
7) Ca(IO3)2 is a slightly soluble precipitate. What would the concentration of IO3- be in a
solution saturated with Ca(IO3)2? Ksp = 1.5 x 10-15
[Ca2+] [IO3-]2 = x(2x)2 = 4x3 = 1.5x10-15
x = 7.21x10-6
8) What is the pH of a 0.10 M solution of NaAc? Ka = 1.8x10-5
Ac- + H2O ↔ HAc + OH-
[OH − ][HAc]
x2
=
= 5.55x10−10 ⇒ x = 7.45x10 −6 M OH − pH = 8.87
[Ac ]
0.1 − x
9) When ammonia is heated it decomposes to N2 and H2 according to the following reaction,
2 NH3 ↔ N2 + 3 H2
Given 3 atm of NH3 and an equilibrium constant of 3x10-3, what will the final pressure (total
pressure) be in the system?
2 NH3 ↔ N2 + 3 H2
3 – 2x
x
3x
x(3 x)3
27 x 4
=
= 3 x10 −3
2
2
(3 − 2 x)
(3 − 2 x)
ignore the 2x on the bottom and solve. x = 0.1778 atm
Total pressure = 3-2x+x+3x = 3+2x = 3+2(0.1778) = 3.3556 atm
10) What is the pH of a solution made by adding 0.2 mole of NaAc to 250 mL of 1 M acetic
acid? Ka = 1.8x10-5
(1 M)(0.250 L) = 0.250 mol HAc
Use the HH equation pH = 4.74 + log 0.20/0.25
pH = 4.64
11) A 0.10 M solution of calcium ions can be precipitated with 0.10 M NaOH but not in 0.10 M
NH4OH. The solubility product for Ca(OH)2 is 7.88x10-6. Please explain why 0.10 M Ca2+ will
not precipitate in 0.10 M NH4OH (Kb = 1.8x10-5).
A 0.10 M Ca2+ solution will require.
[Ca2+] [OH-]2 = [0.10 M] [OH-]2 = 7.88x10-6 [OH-] = 0.00888 M OH- to precipitate
0.10 M NH4OH can only produce,
[OH − ][NH +4 ]
x2
=
= 1.8x10 −5 ⇒ x = 0.001342 M OH − this is not enough to ppt. Ca2+
[NH 4OH]
0.1 − x
12) What is the pH of the resulting solution when 0.25 mole of NaCH3CO2 and 0.15 mole HCl
are added to 200 mL water? The Ka for CH3COOH is 1.8x10-5M.
Ac- + H+ → HAc
[H + ][Ac− ] [H + ][0.10 + x]
0.25 0.15
=
= 1.8x10 −5 ⇒ [H + ] = 2.7 x10 −5 pH = 4.57
[HAc]
[0.15
−
x]
-0.15 -0.15 +0.15
0.10
0
0.15
13) What is the pH of a solution made by adding 0.30 mole NH3(aq) to 0.50 mole NH4Cl and
0.25 mole KOH? Kb for NH3(aq) = 1.8x10-5M
NH4OH ↔ NH4+ + OH0.30
0.50
0.25
Shift
+0.25
-0.25
-0.25
0.55
0.25
0
→ trickle
0.55-x
0.25+x
0+x
[OH − ][NH 4+ ] x(0.25 + x)
=
= 1.8x10−5 ⇒ x = 3.96 x10 −5 M OH −
[NH 4OH]
0.55 − x
pH = 9.60
14) What is the concentration of all ionic and molecular species when you add 30 mL of 0.5 M
NaOH to 120 mL of 0.75M HCN? Ka for HCN is 4.8x10-10.
(0.5 M)(0.030 L) = 0.015 mol NaOH and (0.75 M)(0.120 L) = 0.090 mol HCN
HCN + OH- ↔ H2O + CN0.09
0.015
-0.015 -0.015
+0.015
0.075
0
0.015
[H+] = 2.4x10-9
[ H + ][CN − ] [ H + ][0.015 + x]
=
= 4.8 x10 −10
[ HCN ]
[0.075 − x]
[OH-] = 1x10-14/[2.4x10-9] = 4.17x10-6 M OH-
Ion
Moles Before Rxn
Moles After Rxn
Conc (150 mL total)
Na+
OHHCN
H+
CN-
0.015
0.015
0.090
0
0
0.015
0.10 M
4.17x10-6
0.05 M
2.4x10-9
0.05 M
From equilibrium
0.075
From equilibrium
0.015
15a) What is the pH of the following solutions when the following amounts of 1.20 M NaOH
are added to 160 mL of 3 M dl Aspartic acid?
H2Asp ↔ H+ + HAspHAsp- ↔ H+ + Asp20 mL
1.38x10-4
1.51x10-10
pKA1 = 3.86
pKA2 = 9.82
x2
= 1.38x10 −4 ⇒ x = 0.002035 M H + pH = 1.69
3− x
M1V1 = M2V2
(3 M)(0.160 L) = (1.20 M)(V2)
V2 = 400 mL
Also, (3 M)(0.160 L) = 0.48 mol H2Asp
200 mL
400 mL
700 mL
pH = 3.86
pH =
pK A1 + pK A2 3.86 + 9.82
=
2
2
300 mL into 2nd Region (1.2 M)(0.300 L) = 0.36 mol OHpH = 9.82 + log
800 mL
[Asp2- ] =
0.36
0.48 − 0.36
pH = 10.30
0.48 mol
1x10-14
= 0.50 M and Kb =
= 6.62 x10 −5
-10
0.960 L
1.51x10
x2
= 6.62 x10 −5
0.50 − x
900 mL
pH = 6.84
x = 5.75x10-3 M OH-
(0.100L)(1.20M) = 0.12 mol OH0.12 mol OH −
= 0.1132 M OH −
1.060 L
pH = 11.76
pH = 13.05
15b) An Aspartic acid buffer of pH 5 was made by adding some NaHAsp to 0.5 moles of
H2Asp. How much 1.2 M NaOH or 1.2M HCl must you add to this buffer in order to make a
new buffer of pH 4.5?
First, calculate the composition of the existing buffer, pH = pKa + log
B
A
[ HAsp − ]
[HAsp-] = 6.90 mol
0 .5
Now find out how much acid must be added to make new buffer (pH is going down so add acid).
5 = 3.86 + log
4.5 = 3.86 + log
6.90 − x
0 .5 + x
x = 0.8792 mol H+ = (1.2 M)(V)
V = 732.7 mL of 1.2 M HCl
16) What is the pH of the solution when 75 mL of 0.8 M dl-Histidine is titrated with the
following volumes of 1.20 M NaOH?
H3His <---> H+ + H2HisH2His- <---> H+ + HHis2HHis2- <---> H+ + His3-
M1V1 = M2V2
(0.8 M)(0.075 L) = (1.20 M)(V2)
V2 = 50 mL
pKa1 = 2.40
pKa2 = 6.04
pKa3 = 9.33
Also, (0.8 M)(0.075 L) = 0.060 mol H3His
0 mL
x2
= 3.98x10 −3 ⇒ x = 0.0564 M H +
0 .8 − x
30 mL
(1.2 M)(0.030 L) = 0.036 mol OHpH = 2.40 + log
0.036
0.060 − 0.036
pK A1 + pK A2 2.40 + 6.04
=
2
2
pH = 1.248
pH = 2.58
50 mL
pH =
75 mL
pH = 6.04
100 mL
pH =
125 mL
pH = 9.33
135 mL
35 mL into 3rd Region (1.2 M)(0.035 L) = 0.042 mol OH-
pK A2 + pK A3 6.04 + 9.33
=
2
2
pH = 9.33 + log
150 mL
[His3- ] =
0.042
0.060 − 0.042
pH = 7.69
pH = 9.70
0.060 mol
1x10-14
= 0.2667 M and Kb =
= 2.14 x10 −5
0.225 L
4.68x10-10
x2
= 2.14 x10 −5
0.2667 − x
175 mL
pH = 4.22
x = 2,39x10-3 M OH-
(0.025 L)(1.20M) = 0.030 mol OH0.030 mol OH −
= 0.120 M OH −
pH = 13.08
0.250 L
pH = 11.38
17) How much NH4Cl and NH4OH must you add to 250 mL of water to make a buffer of pH =
8.0? Kb = 1.8x10-5
Using the base version of the HH equation;
6 = 4.74 + log
[NH 4 Cl]
[NH 4 Cl] 18.19 0.1819
⇒
=
=
[NH 4 OH]
[NH 4 OH]
1
0.01
Adding 18.19 moles to 250 mL of water will make a solution that is 72.76M, so divide
top and bottom by 100. Add 0.1819 mol NH4Cl and 0.010 mole NH4OH to 250 mL of
water to make a buffer of pH = 8
18) How many moles of sodium acetate must be added to 100 mL of 0.25 M acetic acid to make
a buffer of pH = 4.0? pKa = 4.74 for acetic acid.
(0.25 M HAc)(0.100 L) = 0.025 mol HAc
(x)
4 = 4.74 + log
⇒ x = 0.00455 mol NaAc
(0.025 mol HAc)
19) Another way of making the buffer from 18 above would be to titrate the acid with a base.
What volume of 0.40 M NaOH must be added to 100 mL of 0.25 M acetic acid to make a buffer
of pH = 4.0? pKa = 4.74 for acetic acid.
4 = 4.74 + log
(x)
⇒ x = 0.003849 mol NaOH
(0.025 - x)
(0.40 M)(V) = 0.003849 mol NaOH
V = 9.62 mL
20) How much 0.5 M NaOH must be added to 80 mL of 0.75 M H3PO4 to make a buffer of pH =
8? Ka1 = 2.12, Ka2 = 7.21, Ka3 = 12.67
The pH is nearest the 2nd pKa so the buffer is in Region II. You must add,
(0.75 M)(0.080 L) = (0.50 M)(V) V = 120 mL so add 120 mL to get to the starting point.
8 = 7.21 + log
(x)
⇒ x = 0.05163 mol NaOH = (0.50 M)(V)
(0.060 - x)
Total volume = 103.25 mL + 120 mL = 223.35 mL NaOH
V = 103.25 mL
Solubility Products
21) Are the following molecules acidic, basic or neutral in aqueous solution?
NaF
Cr(NO3)3
KCl
NH4CN
acid
acid
acid
acid
basic
basic
basic
basic
neutral
neutral
neutral
neutral
can't tell
can't tell
can't tell
can't tell
22) Are the following compounds soluble or insoluble in water?
NaIO3
Cr(OH)3
PbSO4
CaCl2
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
FeSO4
ScCl3
Na2O
Hg2SO4
Soluble
Soluble
Soluble
Soluble
Insoluble
Insoluble
Insoluble
Insoluble
23) What is the solubility of PbBr2 in pure water? Ksp = 4x10-5
[Pb2+] [Br-]2 = x(2x)2 = 4x3 = 4x10-5
x = 2.15x10-2 M
24) What is the solubility of Ca(OH)2 in 0.05 M NaOH? Ksp = 5.5x10-6
[Ca2+] [0.50 M]2 = 5.5x10-6
x = 2.20x10-5
25) What is the solubility of Ca3(PO4)2 in water? Ksp = 5.87x10-8
[Ca2+]3 [PO43-]2 = (3x)3(2x)2 = 108x5 = 5.87x10-8
x = 1.40x10-2 M
26) What is the solubility of Ca3(PO4)2 in 0.5 M Na3PO4? Ksp = 5.87x10-8
[Ca2+]3 [PO43-]2 = (3x)3(0.50)2 = 0.75x3 = 5.87x10-8
x = 4.278x10-3 M
27) What is the solubility of AgCl in a solution of 1 M HCl? Ksp = 1.8x10-10
[Ag+] [1 M] = 1.8x10-10
[Ag+] = 1.8x10-10
28) At what pH will the concentration of Cu2+ exceed 0.02 M given the following equilibrium?
Cu(OH)2 ↔ Cu2+ + 2 OH- Ksp = 2.2x10-20
[Cu2+] [OH-]2 = [0.02 M] [OH-]2 = 2.2x10-20 [OH-] = 1.05x10-9 pH = 6.02
29) How much NH3(aq) must you add to 100 grams of AgCl in order to dissolve all of the AgCl.
Assume a liter of solution and calculate the concentration of NH3(aq).
AgCl ↔ Ag+ + Cl- Ksp = 1.8x10-10
Ag(NH3)2+ ↔ Ag+ + 2 NH3(aq) Keq = 1.6x10-9
If you reverse the bottom reaction and add it to the top reaction you get the following (see
below).
AgCl ↔ Ag+ + ClKsp = 1.8x10-10
Ag+ + 2 NH3(aq) ↔ Ag(NH3)2+
Keq = 6.25x10-8
+
AgCl + 2 NH3(aq) ↔ Ag(NH3)2 + Cl
K = 0.1125
This means that all the silver in solution will end up being Ag(NH3)2+. If you dissolve all of the
AgCl in 1 liter of solution, the concentration of Ag(NH3)2+ must be,
100 g AgCl/143.35 g/mol = 0.6976 mole AgCl =>
0.6976 mole Ag+ becomes,
0.6976 mole Ag(NH3)2+ and 0.6976 mole Cl- in 1 liter of solution.
So, [Ag(NH3)2+] = 0.6976 M and [Cl-] = 0.6976 M
Solving for the concentration of NH3(aq),
[Ag(NH3 ) 2+ ] [Cl− ] [0.6976 M] [0.6976 M]
=
= 0.1125 [NH3] = 2.08 M
[NH 3 ]2
[NH 3 ]2
30) Copper(I) ions in aqueous solution react with NH3 according to,
Cu+ + 2 NH3 ↔ Cu(NH3)2+ Keq = 6.3x1010
Calculate the solubility of CuBr (Ksp = 5.3x10-9) in a solution in which the equilibrium
concentration of NH3 is 0.185 M.
CuBr ↔ Cu+ + BrCu+ + 2 NH3 ↔ Cu(NH3)2+
CuBr + 2 NH3 ↔ Cu(NH3)2+ + Br-
Ksp = 5.3x10-9
Keq = 6.3x1010
K = 333.9
For every Cu(NH3)2+ there must be one Br-, so [Cu(NH3)2+] = [Br-] = x
x2
x2
=
= 333.9 x = 3.38 M CuBr
[NH 3 ]2 [0.185]2
31) What is the solubility of PbCl2 in a solution of 0.10 M H2S at a pH=0? Ksp =1.6x10-5 for
PbCl2, Ksp = 8x10-28 for PbS, Keq = 1x10-20 for H2S (to 2 H+ + S2-).
PbS <---> Pb2+ + S2- Ksp = 8x10-28
H2S <---> 2 H+ + S2- Keq = 1x10-20
Note: the Pb2+ concentration can only be as high as the least soluble compound so the
solubility of PbS will determine the concentration of Pb2+
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1]2 = 1x10-21 M S2Substitute this result into the following equation,
[Pb2+] [S2-] = 8x10-28
[Pb2+] = 8x10-28/[S2-] = 8x10-28/[1x10-21] = 8x10-7 M
Now, all equilibriums must be satisfied simultaneously. Since this is the maximum
concentration of Pb2+ that can be in solution, we can use this value to calculate the Clconcentration from the PbCl2.
PbCl2(s) ↔ Pb2+ + 2 Cl[Pb2+] [Cl-]2 = 1.6x10-5 => [8x10-7 M] [Cl-]2 = 1.6x10-5 [Cl-] = 4.47 M
This concentration of Cl- is fairly high. It can only be this high if all of the PbCl2
dissolved. So, if you put H2S in a solution of PbCl2, all of the PbCl2 would dissolve and
become PbS.
32) Calculate the maximum concentration of Fe3+ in an 0.10 M H2S solution buffered at pH=6.
Ksp = 1.6x10-21 for Fe2S3 and Keq = 1x10-20 for H2S (to 2 H+ + S2-).
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-6]2 = 1x10-9 M S2Substitute this result into the following equation,
Fe2S3(s) ↔ 2 Fe3+ + 3 S2[Fe3+]2 [S2-]3 = 1.6x10-21
[Fe3+] = (1.6x10-21/[S2-]3)½ = (1.6x10-21/[1x10-9]3)½ = 1264.9 M Fe3+ => 632.45 M Fe2S3
33) What is the solubility cadmium sulfide (CdS) in a saturated solution of H2S = 0.10 M at pH
= 3?
CdS <--> Cd2+ + S2- Ksp = 8x10-27
H2S <--> 2H+ + S2Keq = 1x10-20
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation,
[Cd2+] [S2-] = 8x10-27
[Cd2+] = 8x10-27/[S2-] = 8x10-27/[1x10-17] = 8x10-12 M
34) What is the solubility (silver ion concentration) of silver sulfide (Ag2S) in a saturated
solution of H2S = 0.10 M at pH = 3?
Ag2S <--> 2 Ag+ + S2- Ksp = 1.6x10-49
H2S <--> 2H+ + S2Keq = 1.1x10-20
[H+]2 [S2-]/[H2S] = 1x10-20
[S2-] = 1x10-20 [H2S]/[H+]2 = 1x10-20 [0.1]/[1x10-3]2 = 1x10-15 M S2Substitute this result into the following equation,
[Ag+]2 [S2-] = 1.6x10-49
[Ag+] = 1.6x10-49/[S2-] = 1.6x10-49/[1x10-17] = 1.6x10-32 M Ag+
35) How much of each precipitate will form and what will the concentration of lead be if
0.075 mole sodium oxalate (Na2C2O4 = Na2Ox) is added to 100 mL of solution containing 0.30
M Mg(NO3)2 and 0.5 M Pb(NO3)2?
MgOx(s) ↔ Mg2+ + Ox2PbOx(s) ↔ Pb2+ + Ox2-
Ksp = 8.6x10-5
Ksp = 4.8x10-10
Neither compound is very soluble so you can begin by assuming that 100% of the Pb2+ and
Mg2+ will react to form solids.
(0.30 M)(0.100 L) = 0.03 mole Mg(NO3)2
(0.50 M)(0.100 L) = 0.05 mole Pb(NO3)2
Start with the least soluble compound (PbOx),
You have 0.075 mole Ox2- and 0.50 mole of Pb2+, so all of the Pb2+ will react to make 0,50
mole of PbOx and leave 0.025 mole of Ox2- in solution.
The remaining 0.025 mole of Ox2- will react with the 0.03 mole Mg(NO3)2 leaving 0.005
mole of Mg2+ in solution and no Ox2-. The [Mg2+] = (0.005 mol)(0.100 L) = 0.05 M Mg2+.
[Mg2+] [Ox2-] = [0.05 M] [Ox2-] = 8.6x10-5
[Ox2-] = 1.72x10-3 M
[Pb2+] [Ox2-] = [Pb2+] [1.72x10-3 M] = 4.8x10-10
[Pb2+] = 2.79x10-7 M
36) What is the solubility of silver carbonate (Ag2CO3) in a solution saturated with carbonic acid
(0.034M) and which is buffered at pH = 9?
Ag2CO3 <---> 2 Ag+ + CO32- Ksp = 8.1x10-12
H2CO3 <--> 2 H+ + CO32Ka = 2.02x10-17
[H + ]2 [CO32− ] (1x10−9 M) 2 (CO32− )
=
= 2.02x10−17 ⇒ [CO32− ] = 0.6868 M
[H 2 CO 3 ]
0.034M
[Ag+]2 [CO32-] = [Ag+]2 [0.6868 M] = 8.1x10-12
[Ag+] = 3.43x10-6 M
Solubility of Ag2CO3 = (3.43x10-6 M)/2 = 1.72x10-6 M Ag2CO3
Chemistry 121
Second Exam
Name_____________
April 18, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Circle one. Show your
work.
BaSO4
Ksp = 1.6x10-10
PbI2
Ksp = 1.2x10-13
Ag3PO4
Ksp = 1x10-22
2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of
concentration. I wanted to know the concentration so I measured the pH and found it to be pH =
2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts
of 0.8 M NaOH
H3AsO4 <---> H2AsO4- + H+
H2AsO4- <---> HAsO42- + H+
HAsO42- <---> AsO43- + H+
Ka1= 5.62x10-5
Ka2= 1.7x10-8
Ka3= 3.95x10-12
a) 0 mL
b) 150 mL
c) 225 mL
d) 400 mL
e) 450 mL
f) 600 mL
f) If you wanted to titrate to the first endpoint which indicator would you use? Circle one.
Indicator
Methyl Red
Bromothymol Blue
Cresol Red
Phenolphthalein
Range
4.8 - 6.0
6.2 - 7.6
7.0 - 8.8
8.2 - 10
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of
pH = 11.6?
4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic
acid (0.034M) and which is buffered at pH = 8?
MgCO3 <---> Mg2+ + CO32- Ksp = 8.1x10-12
H2CO3 <--> 2 H+ + CO32- Ka = 2.02x10-17
5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example.
Chemistry 121
Second Exam
Name_____________
April 18, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Circle one. Show your
work.
BaSO4
Ksp = 1.6x10-10
(x)(x) = x2 = 1.6x10-10
x = 1.26x10-5
PbI2
Ksp = 1.2x10-13
(x)(2x)2 = 4x3 = 1.2x10-13
x = 3.11x10-5
Ag3PO4
Ksp = 1x10-22
(3x)3(x) = 27x4 = 1x10-22
x = 1.39x10-6
2) I found a bottle marked “Acetic Acid” in our back room, but there was no indication of
concentration. I wanted to know the concentration so I measured the pH and found it to be pH =
2.87. What was the concentration of the acetic acid in the bottle? Ka = 1.8x10-5
-log[H+] = 2.87
[H+] = 1.35x10-3 M also, when HAc dissociates, [H+] = [Ac-]
HAc ↔ H+
+
Acx
1.35x10-3 1.35x10-3
(1.35x10 )
-3 2
x
= 1.6 x10 −5
x = 0.1013 M HAc
3) Calculate the pH of 100 mL of 1.2 M Arsenous acid upon addition of the following amounts
of 0.8 M NaOH
H3AsO4 <---> H2AsO4- + H+
H2AsO4- <---> HAsO42- + H+
HAsO42- <---> AsO43- + H+
a) 0 mL
x2
= 5.62 x10 −5
1.2 − x
Ka1= 5.62x10-5 pKA1 = 4.25
Ka2= 1.7x10-8 pKA2 = 7.74
Ka3= 3.95x10-12 pKA3 = 11.40
x = 8.21x10-3 pH = 2.09
M1V1 = M2V2
(1.2M)((0.100L) = (0.08 M)(V2)
V2 = 150 mL for each region
Also (1.2M)(0.100L) = 0.12 mol H3AsO4
pK A1 + pK A2 4.25 + 7.74
=
2
2
b) 150 mL
pH =
c) 225 mL
pH = 7.74
d) 400 mL
100 mL into 3rd Region (0.8 M)(0.100 L) = 0.08 mol OHpH = 11.40 + log
e) 450 mL
[AsO34− ] =
0.08
0.12 − 0.08
pH = 11.70
0.12mol
1x10-14
= 0.2182M and Kb =
= 2.53 x10 −3
-12
0.550L
3.95x10
x2
= 2.54 x10 −3
0.2182 − x
f) 600 mL
pH = 5.995
x = 7.43x10-2 M OH- pH = 12.87
(0.150L)((0.80M) = 0.12 mol OH0.12 mol OH −
= 0.1714 M OH −
0.700 L
pH = 13.23
g) If you wanted to titrate to the first endpoint which indicator would you use? Circle one.
Indicator
Methyl Red
Bromothymol Blue
Cresol Red
Phenolphthalein
Range
4.8 - 6.0
6.2 - 7.6
7.0 - 8.8
8.2 – 10
since pH = 5.995 @ first endpoint
choose an indicator that changes
just after 5.995
g) How much 0.8 M NaOH must you add to 100 mL of 1.2 M Arsenous acid to make a buffer of
pH = 11.6?
The pH is in the 3rd region so you will have to add 300 mL of NaOH to your answer,
11.6 = 11.4 + log
x
0.12 − x
x = 0.7357 mol OH-
0.7357 mol OH- = (0.8 M)(V)
V = 91.97 mL + 300 mL = 391.97 mL total
4) What is the solubility of magnesium carbonate (MgCO3) in a solution saturated with carbonic
acid (0.034M) and which is buffered at pH = 8?
MgCO3 <---> Mg2+ + CO32- Ksp = 8.1x10-12
H2CO3 <--> 2 H+ + CO32- Ka = 2.02x10-17
[H + ][CO32− ] (1x10−8 M)(CO 32− )
=
= 2.02x10−17 ⇒ [CO32− ] = 6.868x10−3 M
[H 2CO 3 ]
0.034M
[Mg2+] [CO32-] = [Mg2+] [6.868x10-3 M] = 8.1x10-12
[Mg2+] = 1.18x10-9 M
5) Define amphoterism. Draw a graph of pH vs. concentration as part of your example.
Concentration of Metal Ion
Amphoterism = the ability of a metal ion to be soluble in acidic and basic solutions
0
7
pH
14
Chemistry 121
Second Exam
Name ________________
April 10, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Show your work.
BaF2
Ksp = 1.84x10-9
Mg3(PO4)2
Ksp = 1.04×10-15
Ag3PO4
Ksp = 1.2x10-15
2) Please write the equilibrium reaction that will occur when the following compounds are
mixed. Write all reactions as dissociations where appropriate.
NaOH + HAc
Na3PO4 + HCl
HCl + KOH
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of
0.6 M NaOH
H2CO3 <---> H+ + HCO3HCO3- <---> H+ + CO32-
Ka1= 4.3x10-7
Ka2= 4.7x10-11
a) 0 mL
b) 100 mL
c) 150 mL
d) 200 mL
e) 300 mL
f) 365 mL
g) 400 mL
h) 450 mL
i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH =
10?
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3?
Ag2S <---> 2 Ag+ + S2- Ksp = 6.0x10-50
H2S <---> 2 H+ + S2- Keq = 6.84x10-23
5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble
in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and
FeS.
5b) Define amphoterism and give an example.
Chemistry 121
Second Exam
Name
April 10, 2008
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1(10)
2(15)
3(45)
4(20)
5(10)
Total
1) Which of the following compounds is the most soluble in water? Show your work.
BaF2
Ksp = 1.84x10-9
(x)(2x)2 = 4x3 = 1.84x10-9
x = 7.72x10-4
Mg3(PO4)2
Ksp = 1.04×10-15
(3x)3(2x)2 = 108x5 = 1.04x10-15 x = 3.95x10-4 most
soluble
Ag3PO4
Ksp = 1.2x10-15
(3x)3 (x) = 27x4 = 1.2x10-15 x = 8.16x10-5
2) Please write the equilibrium reaction that will occur when the following compounds are
mixed. Write all reactions as dissociations where appropriate.
NaOH + HAc
H2O + Ac- –> HAc + OHNa3PO4 + HCl
HPO42- –> H+ + PO43HCl + KOH
H2O –> H+ + OH-
3) Calculate the pH of 80 mL of 1.5 M Carbonic acid upon addition of the following amounts of
0.6 M NaOH
H2CO3 <---> H+ + HCO3HCO3- <---> H+ + CO32-
Ka1= 4.3x10-7
Ka2= 4.7x10-11
a) 0 mL
x2/(1.5-x) = 4.3x10-7
x = 8.03x10-4M H+
pH = 3.095
b) 100 mL
½ way point so pH = 6.37
c) 150 mL - use H-H equation
(0.6M)(0.150L) = .09mole NaOH and (1.5M)(0.080L) = 0.12mole H2CO3
pH = 6.37 + log(0.09/(0.12-0.09))
pH= 6.847
d) 200 mL
first end point so pH = (pka1 + pka2)/2 –> pH = (6.37 + 10.33)/2 = 8.35
e) 300 mL
2nd midpoint so pH = pKa2 = 10.33
f) 365 mL - you are now 165 mL into the second region - use H-H equation
(0.06M)(0.165L) = 0.099 mole NaOH and you still have 0.12 mole of acid (HCO3-)
pH = 10.33 + log(0.099/(0.12-0.099)) pH = 11.00
g) 400 mL - at final endpoint pKb = Kw/Ka2 = 1x10-14/4.7x10-11 = 2.13x10-4
also [CO32-] = 0.12 mole/ (0.480L total) = 0.25M CO32x2/(0.25 - x) = 2.13x10-4 –> x = 0.007297 M OH- pH = 11.86
h) 450 mL - this is 50 mL past the final endpoint
(0.050L)(0.60M) = 0.030 mole excess OH- in a total of 530 mL of solution.
Therefore, [OH-] = 0.030 mole / 0.530 L = 0.0566 M OH- –> pH = 12.75
i) How much 0.8 M NaOH must be added to 80 mL of 1.5 M H2CO3 to make a buffer of pH =
10?
The pH is in the 2nd region so you must add 200 mL to start this titration. Use H-H,
10 = 10.33 + log(x/(0.12-x)) –> x = 0.0382 mole NaOH
0.0382 mole NaOH = MV = (0.80M)(V) –> V = 0.0478 L or 47.8 mL, therefore,
200 mL NaOH to get to first endpoint + 47.8 mL = 247.8 mL of 0.80 M NaOH
4) What is the solubility of Ag2S in a saturated solution of H2S = 0.01 M at pH = 3?
Ag2S <---> 2 Ag+ + S2- Ksp = 6.0x10-50
H2S <---> 2 H+ + S2- Keq = 6.84x10-23
[H+]2 [S2-]/[H2S] = 6.84x10-23
[S2-] = 6.84x10-23 [H2S]/[H+]2 = 6.84x10-23 [0.01]/[1x10-3]2 = 6.84x10-19 M S2Substitute this result into the following equation,
[Ag+]2 [S2-] = 6x10-50
[Ag+]2 = 6x10-50/[S2-] = 6x10-50/[6.84x10-19] = 8.77x10-32
taking the square root of this answer to find [Ag+],
[Ag+] = 2.96x10-16 M Ag+
5a) Iron can be separated from copper using pH controlled solubility in H2S. CuS is not soluble
in acid solutions and FeS is soluble. Draw and approximate ion separation curve for CuS and
FeS.
[Ions]
CuS FeS
0
At this pH FeS is highly soluble
and CuS is not very soluble.
7
pH
14
5b) Define amphoterism and give an example.
Amphoterism is the ability of an ion to be soluble in either an acid or a base but not in
between. This property is commonly exhibited with transition metals like Zinc,
Zn(OH)2(s) –> Zn2+ + 2 OHZn(OH)2(s) + 2 OH- –> Zn(OH)42-
soluble in acid
soluble in base