Answers included:
1. Solubility and Precipitation Reaction worksheet
2. Thermodynamics Worksheet
3. Problem Set 3b
4. Problem Set 3a
CHEM 1412 Solubility and Precipitation Reaction
1. Mg(OH)2
2. Ag3PO4
a. reaction for
dissolving
(dissolution)
Mg(OH)2(s) Mg2+ + 2OH-
b. Algebraic
expression of
Ksp and (S)
Ksp = S(2S)2=4S3= 1.5x10-11
S= molar solubility
Dr. Ya-Ping Huang
3. Fe4[Fe(CN)6]3
Ag3PO4(s)  3Ag+ + PO43-
Fe4[Fe(CN)6]3 4Fe3+
+ 3 [Fe(CN) 6] 4-
c. Molar
10 Ksp = [Ag+]3[PO43-] = (3S)3S =27S4
= 1.3x10-20
Ksp 3 1.5 x10 11

 1.55 x10  4 M
4
4
solubility
S= 3
d. Molar mass
58.33
e. Solubility,
1.55x10-4M)(58.33g/mol)=
9.04
x10-3
g/L
S= 4
Ksp 4 1.3x10 20

 4.68 x10 6 M
27
27
11 Ksp = (4S)4(3S)3 = 6912 S7
= 3.0 E-41
Ksp 7 3.0 x10 41

 4.59 x10 7 M
6912
6912
S= 7
418.58
859.24
1.96 E-3
3.95 E-4
0.0049 g
9.9 x10-4g
skip
g/L
f. g in 2.50 L
sat’d solution
M.V.MM =
g.
given
solution
0.10 M HCl
in 0.1 M AgNO3
Possible
reaction or
effect
WB + SA, 100 % neutralization
Common-ion (Ag+) effect on
Ag3PO4(s)  3Ag+ + PO43-
Molar
solubility
(1.55x10-4M)(2.5L)(58.33g/mol)= 0.0226g
Mg(OH)2 +
2H+Mg2++
2H2O
Neutralization between WB, Mg(OH)2 and SA
(HCl) is almost 100%, Solubility of Mg(OH)2 is
based on how many moles of Mg(OH)2 reacts
w/ HCl, and is limited by the amount H+
available
Ksp = (3S+0.1)3 S
= 1x 10-3 S
S= 1.3 E-17 M
skip
Solubility = ½ (.01MH+)=.005M
Given the combination of solutions,:
(25.0 mL 0.010M of 1st solution and 50.0 mL 0.020 M 2nd solution) After dilution,
[Solution1] = 0.010M(25.0mL)/75.0mL = 0.0033M, [Solution2] = 0.020M(50.0mL)/75.0mL = 0.0133M
Solution combination
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4. Cu(NO3)2 + NaOH
12
5. CaCl2 + Na3PO4
6. AgNO3 + K2CrO4
2/13/2016
Cu(NO3)2 + 2 HCl  Cu(OH)2 +
2 HNO3
Cu(OH)2
Cu(OH)2(s) Cu2+ + 2OH-
3CaCl2 +2 Na3PO4  Ca3(PO4)2(s) + 3
NaCl
Ca3(PO4)2
Ca3(PO4)2(s) 3Cu2+ + 2PO43-
2AgNO3 + K2CrO4  Ag2CrO4(s)
+2 KNO3
Ag2CrO4
Ag2CrO4(s)  2Ag+ + CrO42-
b. Ksp of the ppt
1.6x10-19
1.0x10-25
9.0x10-12
c. Qsp in solution
(0.0033)(0.0133)2 =5.83x10-7
(0.0033)3(0.0133)2 =6.36x10-12
(0.0033)2(0.0133) =1.45x10-7
d. Will ppt form?
Yes
yes
reaction
a. formula of ppt
Thermodynamics Worksheet
Reactions
Dr. Ya-Ping Huang
1. Decomposition of a fertilizer
2. Habor process
3. Ionization of water
2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)
N2(g) + 3H2(g)  2NH3(g)
H2O(ℓ)  H+ (aq) + OH-(aq)
a
H kJ/ mol rxn
-236
-92.22 /mol rxn
55.8 kJ/mol rxn
b
qp for 500 g (kJ)
NH4NO3(s) -737.5
NH3(g) -1356 kJ
H2O(ℓ): 1550 kJ
c
n
7
-2
0
d
E kJ/ mol rxn
-253.35
-87.26 kJ/mol rxn
55.8kJ/mol rxn
e
S
1.0406
-0.1987 kJ/mol.K
-0.0806 kJ/mol.K
f
Transition Temp
Does not exist
= 464.1 K = 190.9 C
Does not exist
g
G
-546.1
**a. -40.0 (by H & S)
b. -32.38 (Gf)
79.9 kJ
h
Keq
4.53 x 1095
a.1.0 x 107
b. 4.7 x 105
1.01 x 10-14
kJ/mol.K
25 C
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i
Spontaneous?
yes
yes
300 C
pH value for pure water: 7.00
90 C
j
G kJ/mol rxn
-832.47
21.64 kJ/mol rxn
85.06 kJ
k
Keq
7.08 x 1075
0.0107
5.78 x 10-13
l
Spontaneous?
yes
No
pH value for pure water; 6.12
** There is discrepancy in G calculated by Gf and by G = H -T S for this reaction (2) at 25 C
(1). 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)
always spontaneous
H = 2HfN2(g) + 4HfH2O(g) + HfO2(g) - 2HfNH4NO3(s) = 2(0) +4(-241.8) + 0 –2(-365.6) = -236 kJ/mol rxn
500 g
mole.of . lim iting .reagent
n

ceofficient.in.balanced .equation
80 g / mol
 3.125mole.reaction
2
qp = 3.125 mol rxn x (-236 kJ/mol rxn) = -737.5 kJ
H = E +nRT,
n=mole of gas on RHS – mole of gas on LHS = 7 -0 =7
E = H -nRT = -236 kJ – (7)(8.314E-3)(298.15) = -737.5 kJ –17.35 = -253.35 kJ
S = 2S.N2(g) + 4S.H2O(g) + S O2(g) – 2S.NH4NO3(s) = 2(191.5) +4(188.7) + 205 –2(151.1)
= 1040.6 J/mol.K = 1.0406kJ/mol.K
H o
 236kJ / mol.rexn
Ttransition 

 227 K ( Dose.not.exist , Temp.can.not.be.lower.than.0 K )
o
1.0406kJ / k .mole.rexn
S
(In this reaction, reaction is always spontaneous, favored by lower enthalpy, H < 0 and increased entropy, S> 0)
25 C :G = H -T S = -236kJ – 298(1.0406) = -546.1 kJ/mol rxn
(or G = 2GfN2(g) + 4GfH2O(g) + GfO2(g) - 2GfNH4NO3(s) = 2(0) +4(-228.6) + 0 –2(-184) = -546.4 kJ/mol rxn)
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G = -5.709logK
G o
 546.1
log K 

 95.66
 5.709  5.709
K  10 95.66  4.53x10 95 ,
300 C :G = H -T S = -236 – (300+273)(1.0406) = -832.47 kJ
G = -2.303RT(logK)
G o
 832.47
log K 

 75.85
 2.303RT  2.303(8.314 x10 3 )(573)
K  10 75.85  7.08 x10 75
(2) Habor Process
N2(g) + 3H2(g)  2NH3(g)
H = 2HfNH3(g) – (HfN2 (g) + 3HfH2(g) = 2(-46.11)-0 = -92.22 kJ/mol reaction
500 g
mole.of . lim iting .reagent
n

ceofficient.in.balanced .equation
17.03g / mol
 14.7mole.reaction
2
qp = 14.7 mol rxn x (-92.22 kJ/mol rxn) = -1356 kJ
(3) H2O(ℓ)  H+ (aq) + OH-(aq)
always nonspontaneous
H = HfH+ (aq) + HfOH-(aq) - HfH2O(l) = 0 + (-230.0) –(-285.8) = 55.8kJ/mol rxn
qp = [500g/(18)]mol x 55.8 = 1550 kJ
S = SH+ (aq) + SOH-(aq) - SH2O(l) = 0 + (-10.7) –(69.91) = -80.6J/K.mol rxn = -0.0806kJ/K.mol rexn
25 C
G = GfH+ (aq) + GfOH-(aq) - GfH2O(l) = 0 + (-157.3)-(-237.2) = 79.9 kJ/mol rexn
logK = G /-5.709 = 79.9/-5.709 = -13.995
K = Kw = 10-13.995 = 1.01 x 10-14
Kw = [H+][OH-] = [H+]2, [H+] = 1.01x10 14  1.005x10 7 ,
pH =-log(1.005x10-7) =6.998
(If use G = H -T S = 55.8kJ/mol - 298(-0.0806) = 79.8 kJ )
90 C: G = H -T S =55.8kJ/mol - 363(-0.0806) = 85.06 kJ
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G = -2.303RT(logK)
G o
85.06
log K 

 12.24
 2.303RT  2.303(8.314 x10 3 )(363)
Kw = [H+][OH-] = [H+]2, [H+] =
K  10 12.24  5.78 x10 13
5.78x10 13 = 7.5 x 10-7, pH = 6.12
Problem Set 3b:
(2)
qH2O = (4.184 J/goC)(100 mL)(1.02 g/mL)(26.65 – 23.35oC) = 1408 J
qCal = (24.0 J/oC)(26.65 – 23.35oC) = 79.2 J
(Total heat evolved = qH2O + qCal = 1487 J)
qtotal= qH2O+ qCal+ qrexn = 0
qrexn = -1487 J = nH
To find n, you have to find moles of each reagent and find which reactant is the limiting reagent.
mol of CuSO4 = (0.400M)(0.050L) = 0.020 mol x
1molrexn
 0.0200molrexn
1molCuSO4
mol of NaOH = (0.600M)(0.0500L) = 0.030 mol x
1molrexn
 0.0150molrexn
2molNaOH
So NaOH is limiting reagent, and n = 0.0150 mole reaction
qrexn = -1487 J = nH = (0.0150 mole reaction) H
H = -1487 J/(0.0150 mole reaction) = - 99133 J/mole rexn = -99.13 kJ/mole rexn
Complete thermochemical equation includes balanced chemical equation and H
CuSO4(aq) + 2NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)
H = -99.13 kJ/mol rxn
4. The thermite reaction, used for welding iron, is as following:
8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)
A. To calculate H, the best method is using Hf:
8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)
H = 9Hf, Fe(s) + 4 Hf,Al2O3(s) – 3 Hf,Fe3O4(s) –8 Hf, Al(s) = 4(-1676) – 3(-1118) = -3350 kJ/mol rxn
B. To find the energy released, given both reactants, it’s a limiting reagent type of problem. The approach is to calculate
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the energy released by the reaction of each reactant. The smaller value is the answer.
a. H based on 8.0 g Al(s):
8.0 gAlx
1molAl 1molrexn  3350kJ
x
x
 124kJ
27 gAl
8molAl 1molrexn
b. H based on 20.0 g Fe3O4
1molFe3O 4
1molrexn  3350kJ
20.0 gFe3O 4 x
x
x
 96.3kJ
232 gFe3O 4 3molFe3O 4 1molrexn
The answer is therefore 96.3 kJ released. And Fe3O4(s) is the limiting reagent.
c. Grams of Fe produced is based on the limiting reagent Fe3O4(s)
1molFe3O4
9molFe
55.8gFe
x
x
 14.4 gFe
232 gFe3O4 3molFe3O4 1molFe
Q7. Estimate G° and determine if the following reaction is spontaneous at 25, 1000 and 2000°C?
20.0 g Fe3O4 x
N2(g) + O2(g) 2NO(g)
[Ans]: G° = H° - TS°
H° = Hf°,RHS - Hf°,LHS = 2Hf°,NO(g) – (Hf°N2(g) + Hf°,O2(g))
= 2(90.25) – (0+0) = 180.5 kJ/mol rxn
S° = S°,RHS - S°,LHS = 2S°,NO(g) – (Hf°N2(g) + Hf°,O2(g))
= 2(210.7) – (191.5 + 205) = 24.9 J/K.mol rxn = 24.9x10-3 kJ/K.mol rxn
G° = H° - TS°
At 25°C (298.15 K) G° = 180.5 kJ/mol rexn –(298.15)( 24.9x10-3 kJ/K.mol rxn)
= 173.08 kJ/mol rxn
(non-spontaneous)
At 1000°C (1273 K) G° = 180.5 kJ/mol rexn –(1273)( 24.9x10-3 kJ/K.mol rxn)
= 148.80 kJ/mol rxn
(non-spontaneous)
At 2000°C (2273 K) G° = 180.5 kJ/mol rexn –(2273)( 24.9x10-3 kJ/K.mol rxn)
= 123.90 kJ/mol rxn
(non-spontaneous)
At 25°C (298.15 K) G° can be calculated by GHf°
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G° = Gf°,RHS - Gf°,LHS = 2Gf°,NO(g) – (Gf°N2(g) + Gf°,O2(g))
= 2(86.57) –(0+0) = 173.14 kJ/mol rxn, almost identical to the one
calculated by G° = H° - TS°
Transition temperature, Ttransition =
H 
180.5kJ / mol.rxn

 7249 K  6976C
S  24.9 x10 3 kJ / K .mol.rxn
CHEM 1412 Problem Set #3a
Dr. Ya-Ping Huang
Solubility Equilibrium
1. Calculate the molar solubility of CaF2 at 25oC in:
a. pure water
(2.1 x 10-4 M)
b. in 0.010M Ca(NO3)2 solution
(3.1 x 10-5 M)
c. in 0.010M KF solution
(3.9 x 10-7 M)
[Ans]: The 1st step is to write the equation for dissolution, and find the relationship between Ksp and molar solubility.
. a.
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

2+
S
-
Ksp = [Ca ]eq [F
S3
K sp
4
3
]eq2
Let S represent the molar solubility
2S
2
= (s)(2s) = 4S3 = 3.9x10-11
3.9 x1011 3
 9.75x1012  2.14 x10 4 M
4
b. In 0.010M Ca(NO3)2 solution, there is the common –ion effect. The extra Ca2+ from Ca(NO3)2 will
reduce the solubility of CaF2(s) (Le Chatelier’s principle) –the common-ion effect
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

S
Let S represent the molar solubility
2S
+0.010
2+
[F-]eq = 2S
[Ca ]eq = 0.010+ S
Ksp = [Ca2+]eq [F-]eq2 = (S+0.010)(2S)2  (0.010)(2S)2 = 0.04S2 = 3.9x10-11
(s is dropped from s+0.010, because when there is common-ion effect , the solubility will be
severely reduced such that the s term is negligible compared to the concentration of
commen-ion)
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S
Ksp
3.9 x1011

 3.12 x10 5 M
0.04
0.04
Ksp = [Ca2+]eq [F-]eq2 = (s)(2s)2 = 4S3 = 3.9x10-11
c. in 0.010M KF solution: there is the common –ion effect. The extra F- from KF will reduce the solubility
of CaF2(s) (Le Chatelier’s principle)
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

S
Let S represent the molar solubility
2S
+0.010
[Ca2+]eq = S
[F-]eq = 0.010+ 2S
Ksp = [Ca2+]eq [F-]eq2 = (S)(0.01+2S)2  (S)(0.01)2 = 1.0x10-4S = 3.9x10-11
S = Ksp/1.0x10-4 = 3.9x10-7 M
2. Calculate the molar solubility of BaSO4 at 25oC:
a. in pure water
b. in 0.10M Na2SO4 solution
c. Grams of BaSO4 dissolved in 2.0 Liters of water
e. LD50 of Ba2+ for rat is around 20 mg/Kg body weight of rats. Would the Ba2+ in part (d) dangerous if
ingested?
(1.05 x 10-5M, 1.1 x 10-9 M, 4.9 x 10-3 g, 2.16 mg)
[Ans]:
a. In pure water (no common-ion or other effects)
BaSO4(s)  Ba2+(aq) + SO42-(aq)
Let S represent the molar solubility, then [Ba2+] = S and [SO42-] = S
Ksp = [Ba2+] [SO42-] = (S)( S)= S2 = 1.1x10-10
S= 1.1x1010  1.05x105 M
b. in 0.10M Na2SO4 solution, the additional SO42- ions from Na2SO4 will reduce solubility of BaSO4
through common-ion effect
Ksp = [Ba2+] [SO42-] = (S)(0.10+S)= S(0.10) = 1.1x10-10
S=1.1x10-9 M
d. mg of Ba2+ in 1.5 L of saturated BaSO4 solution.
 1.05x105 mol 
137 g  1000mg 
mg of Ba2+ = (M)(V)(MM) = 
 x(1.5L)(
  2.16mg
)
L
1mol.Ba  1g 


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e. LD50 refers to lethal dose for 50% population exposed. Assume a typical adult weighs 50kg, then
LD50 = (20mg/Kg body weight)(50 kg) = 1000mg. So 2.16 mg is quite safe.
3. How would the solubility of the following substances be affected by  pH of a solution?
a. Ni(OH)2
b. CaCO3
c. BaSO4
d. AgCl
[Ans]: The key to see how does  pH of a solution affect the solubility is to see if additional H+ will affect the
dissolution process
a. Ni(OH)2 (s)  Ni2+(aq) + 2OH-(aq)
Dissolution reaction
Rexn w/ H+
dissolution
solubility
A
Ni(OH)2 (s)
Ni(OH)2 (s)  Ni2+(aq) + 2OH-(aq)
H + OH  H2O


B
CaCO3
CaCO3 Ca2+(aq) + CO32-
2H++CO3 2-
H2O + CO2


C
BaSO4
BaSO4(s)  Ba2+(aq) + SO42-(aq)
H++SO42HSO4-
Slightly
favored
barely
d
AgCl
AgCl Ag+(aq) + Cl-(aq)
No reaction
Not affected
same
+
-
4. A solution is 0.10 M in Mg(NO3)2 and the pH is adjusted to 8.0. Would Mg(OH)2 precipitate? (No)
5. A solution is 0.10M in Mg(NO3)2. What concentration of OH- is required to just start precipitation of
Mg(OH)2? If NH3-NH4+ buffer is used to control the pH, and [NH3] = 0.10 M. What concentration
of NH4+ is required to prevent the precipitation of Mg(OH)2?
(1.2 x 10-5M, >0.15 M)
[Ans]: To find out conditions for ppt, compare the requirements for different situation:
The easiest way is to calculate conditions for saturated solution and modify condition for different
situation
Qsp = [Mg2+][OH-]2
-
Saturated soln,
equilibrium
Unsaturated soln,
= Ksp = 1.5x10-11
< Ksp
no ppt
-5
> Ksp
[OH ]
1.2x10 M
< 1.2x10 M
> 1.2x10-5 M
[NH4+]
0.15M
> 0.15 M, need more CA
(NH4+] to keep solution
acidic
< 0.15M, need less
CA (NH4+] to keep
solution basic
For saturated solution, Qsp = [Mg2+] [OH-]2
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(0.10) [OH-]2 = 1.5x10-11
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1.5 x10 11
 1.2 x10 5 M
0.10
[OH-] =
If [OH-] is maintained by buffer: you can either use Handerson-Hasselbalch equation or Kb of NH3
Kb of NH3 will be more straight forward:
NH3 + H2O  NH4+ + OH-


Kb= [ NH 4 ][OH ]
[ NH 3 ]
1.8 x10 5 
[ NH 4 ][1.2 x10 5 ]
0.10
(1.8 x10 5 )(0.10)
 0.15M
1.2 x105
[ NH 4 ] 
6. 100mL 0.020 M NaF is mixed with 200 mL of 0.050 M Ca(NO3)2. What would be the possible reaction?
Would there be any precipitate? Will the ppt be visible with naked eyes?
(yes, yes)
7. Solid AgNO3 is slowly added to a solution that is 0.0010M in NaCl, NaBr and NaI. Calculate the [Ag+] required to initiate the ppt of
each silver salt. Assume the solution volume does not change in the process. (AgCl, requires [Ag +] = 1.8 x 10-7M, AgBr: [Ag+] = 3.3
x 10-10M, AgI: [Ag+] = 1.5 x 10-13M)
[Ans]; For each ppt, 1st calculate the [Ag+] required for saturation.
All 3 ppts (AgCl, AgBr, AgI) follow the same dissolution process:
AgX(s)  Ag+(aq) + X-(aq)
(X= Cl, Br or I)
In saturated solutions, Qsp = [Ag+]eq [X-]eq = Ksp
X
Ksp
[Ag+]eq = K sp
[X ]
[Ag+]eq= K sp  K sp

Order of ppt
[X ]
0.001
AgCl
Cl
1.8x10-10
1.8x10-7 M
3rd, requires most amt of Ag+
AgBr
Br
3.3x10-13
3.3x10-10 M
2nd, requires 2nd amt of Ag+
AgI
I
1.5x10-16
1.5x10-13 M
1st, requires least amt of Ag+
** Since all three ppts have the same ion ratio, the Ksp indicates the relative molar solubility
AgI has lowest Ksp, therefore the lowest molar solubility, will be the 1st to ppt
AgCl has highest Ksp, therefore the highest molar solubility, will be the last to ppt
8. Solid AgNO3 is slowly added to a solution that is 0.0010M in NaCl and Na2CrO4. Which one will ppt
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first, AgCl or Ag2CrO4? Calculate the [Ag+] required to initiate the ppt of each silver salt.
(AgCl 1st, requires [Ag+] = 1.8 x 10-7, Ag2CrO4 requires [Ag+] = 9.5 x 10-5)
[Ans]:
AgCl and Ag2CrO4 have different ion ratio, values of Ksp does not indicates the relative molar solubility
AgCl
Ion
ratio
Ksp
Ksp and conc
[Ag+]eq= K sp  K sp
1:1=1
1.8x10-10
Ksp= [Ag+]eq [Cl-]eq
9.0x10-12
Ksp=[Ag+]2eq [CrO42-]eq
[X  ]
Ag2CrO4 2:1 = 2
[Ag+]eq =
Order of ppt
0.001
K sp

[Cl  ]
1.8 x10 10
 1.8 x10 7
0.001
K sp
2
4
[CrO ]

9.0 x10 12
 9.5 x10 5 M
0.001
1st, requires less
Ag+
2nd, requires
more Ag+
9. Refer to problem 7, what is [Cl-] and [I-] when AgBr just starts to precipitate?
([Cl-] = 0.001, [I-] = 4.5 x 10-7 when AgBr starts to ppt.
[Ans]: When AgBr starts to ppt, [Ag+ ] = 3.3x10-10 M.
AgCl has not ppt’d yet. All Cl- remains in solution. [Cl-] = 0.001M
AgI ppt’d before AgBr starts to ppt. Once pptn starts, the Qsp of the compound that has ppt’d
always equals to its Ksp.
I   K[ Ag

sp, AgI

]

1.5 x1016
 4.5 x107 M
10
3.3x10
10. A 0.010 M solution of AgNO3 is made 0.50 M in NH3 and Ag(NH3)2+ complex forms.
Ag+(aq) + 2 NH3(aq) 
Ag(NH3)2+(aq)
Kf = 1.7 x 107.
a. What is the equilibrium concentration of Ag+ in the solution?
R
Ag+(aq)
+ 2 NH3(aq) 
Ag(NH3)2+(aq) Kf = 1.7 x 107
I
0.010
0.50 .
0
C
-x
-2x
+x
E
0.01-x
0.5-2x
(2.56 x 10-9 M)
x

K f  1.7 x10 7 
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[ Ag ( NH 3 ) 2 ]
x


2
[ Ag ][ NH 3 ]
(0.01  x)(0.5  2 x) 2
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This is a relatively complicated equation to solve.
But the large size of Kf, it is reasonably to assume that the reaction is almost 100%, and the limiting
reagent, Ag+, will be completely used up in the reaction. We can re-write the RICE as following
R
Ag+(aq)
+ 2 NH3(aq) 
Ag(NH3)2+(aq) Kf = 1.7 x 107
I
0.010
0.50 .
0
C
-0.01
-0.02
+0.01
0.48
0.01
E
0

K f  1.7 x10 7 
[ Ag ( NH 3 ) 2 ]
0.01

[ Ag  ][ NH 3 ]2 [ Ag  ](0.48) 2
[ Ag  ] 
0.01
 2.56 x109 M
(1.7 x107 )(0.48) 2
b. What % of the total silver is in the form of Ag+ (aq)?
(2.56 x 10-5 %)

9
% of silver as Ag+= [ Ag ] x100  2.56 x10 x100  2.56 x10 5%
0.01
0.01
Most of Ag exists in the form of complex ion, only very small % stays as Ag+
This in fact drives the dissolution of AgCl forward and greatly increase the solubility of AgCl
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