Chemistry 2410: Problem Sheet 3 Solutions
1.
+
Calculate the activity of H3O in pure water at 0 °C.
2 H2O (l) <==> H3O+ (aq) + OH- (aq)
Kw = AH3O+ • AOH- = 1.15 x 10-15 at 0˚C (from CRC tables)
AH3O+ = √Kw = √1.15 x 10-15 = 3.40 x 10-8
2.
Calculate the activity coefficient of Hg2+ in a solution that has an ionic strength of 0.085.
Use 5 Angstroms for the effective diameter of the ion.
−0.51z 2 µ
where z = +2, µ = 0.085, and α = 5 Angstroms = 500 pm
α µ 
1+ 

 305 
2
−0.51(+2) 0.085
log γ Hg 2+ =
=− 0.402
 500 0.085 
1+ 

305


-0.402
γ Hg2+ = 10
= 0.396
log γ Hg 2+ =
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3.
Calculate the activity coefficient of;
log γ Hg 2+ =
−0.51z 2 µ
where z = charge, µ = ionic strength, and α = ion size in pm
α µ 
1+ 

 305 
(a) Fe3+ at I=0.075
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€
€
2
−0.51(+3) 0.075
log γ Fe 3+ =
=− 0.695
 900 0.075 
1+ 

305


γ Fe3+ = 10-0.695 = 0.202
(b) Pb2+ at I=0.012
2
−0.51(+2) 0.012
log γ Pb 2+ =
=− 0.192
 450 0.012 
1+ 

305


3+
-0.192
γ Fe = 10
= 0.642
(c) Ce4+ at I=0.080
3.
2
−0.51(+4 ) 0.080
log γ Ce4 + =
=−1.142
1100 0.080 
1+ 

305


3+
-1.142
γ Fe = 10
= 0.072
(d) Sn4+ at I=0.060
€
2
−0.51(+4 ) 0.060
log γ Sn 4 + =
=−1.061
 1100 0.060 
1+ 

305


3+
-1.061
γ Fe = 10
= 0.089
€ 4.
Use activities to calculate the acid dissociation constant of HNO2 in a 0.05 M solution of
NaCl.
HNO2 (aq) + H2O (l) <==> H3O+ (aq) + NO2- (aq)
[H O ]γH O [NO ]γNO
=
+
Ka
3
€
1
1
2
2
2
−
2
Na + ] zNa
=
0.050 • (+1) + 0.50 • (−1)
+ + [Cl ] z
[
Cl −
2
2
) {(
(
µ = 0.050 M
for µ = 0.050 M: γH3O+ = 0.86 and γNO2- = 0.805
(from Harris, Quantitative Chemical Analysis 7th edition, p. 144)
also: γHNO2 = 1 (by definition)
[H O ][NO ] = K
=
−
+
K
'
a
3
[HNO 2 ]
Ka’ = 1.03 x 10-3
€
−
2
[HNO 2 ]γHNO 2
IonicStrength= µ =
€
−
2
+
3
Ka = 7.1 x 10-4 (The question
asks to solve for Ka’, not Ka)
2
−4
• γHNO 2 ( 7.1x10 )(1)
=
γH 3O +γNO 2− (0.86 x0.805)
a
)}
5.
Using activities calculate the hydronium ion concentration in a 0.120 M solution of HNO2
that is also 0.050 M in NaCl at 25.0˚C.
HNO2 (aq) + H2O (l) <==> H3O+ (aq) + NO2- (aq)
Ka = 7.1 x 10-4
−
−
H 3O + ]γH 3O + NO 2 γNO 2
[
Ka =
[HNO 2 ]γHNO 2
1
1
2
2
2
−
2
IonicStrength= µ = [Na + ] zNa
=
0.050 • (+1) + 0.50 • (−1)
+ + [Cl ] z
Cl −
2
2
µ = 0.050 M
for µ = 0.050 M: γH3O+ = 0.86 and γNO2- = 0.805
(from Harris, Quantitative Chemical Analysis 7th edition, p. 144)
also: γHNO2 = 1 (by definition)
[
]
) {(
(
€
€
[H O ][NO ] = K
=
−
+
K
€
'
a
3
−4
• γHNO 2 ( 7.1x10 )(1)
=
=1.03x10−3
γH 3O +γNO 2− (0.86 x0.805)
2
[HNO 2 ]
[Initial] (M)
[Change] (M)
[Equil.] (M)
a
HNO2 (aq) + H2O (l) <==>
0.120
-x
0.120 – x
[H O ][NO ] =
=
−
+
'
a
K =1.03x10
−3
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€
€
3
2
[HNO 2 ]
€
H3O+ (aq) +
0
+x
x
NO2- (aq)
0
+x
x
x2
0.120− x
Using the quadratic equation:
x2
1.03x10−3 =
==> x 2 =1.03x10−3 (0.120− x )
0.120− x
x 2 =1.231x10−4 −1.03x10−3 x
x 2 +1.03x10−3 x −1.231x10−4 = 0
−3
−b+ b 2 − 4ac −1.03x10 +
x=
=
2a
x = 1.06 x 10-2 M
−3 2
(1.03x10 )
2(1)
OR: Assume that x << 0.120 M:
x2
1.03x10−3 =
==> x 2 =1.03x10−3 (0.120)
0.120− x
x = 1.231x10−4 =1.11x10−2 M
x = 1.11 x 10-2 M
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)}
− 4 (1)(-1.231x10−4 )
6.
Use activities to calculate the molar solubility of amorphous Mg(OH)2 in;
Mg(OH)2(s) <==> Mg2+(aq) + 2 OH-(aq)
2
Ksp = 6 x 10-10
2
K sp = [Mg 2+ ]γMg 2+ [OH− ] (γOH− ) =6 x10−10
(a) 0.0100 M NaNO3
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NaNO3 is a 1:1 electrolyte, so µ = 0.0100 M
γMg2+ = 0.69; γOH- = 0.899
K sp
6 x10−10
'
2+
− 2
−9
K sp = [Mg ][OH ] =
=
2 =1.08 x10
2+
− 2
γMg (γOH ) (0.69)(0.899)
Mg(OH)2 is a 1:2 compound, so
K sp' 3 1.08 x10−9
2+
3
[Mg ] = 4 =
4
2+
-4
[Mg ] = 6.46 x 10 M
(b) 0.0167 K2SO4
1
2
2
 1
2−
IonicStrength= µ =  [K + ] zK2 + + SO 4 z 2 2 −  =
2 • 0.0167) • (+1) + 0.0167 • (−2)
(
SO
 2
4
2
µ = 0.0501 M ≈ 0.050 M
γMg2+ = 0.52; γOH- = 0.81
K sp
2
6 x10−10
−9
K'sp = [Mg 2+ ][OH− ] =
=
2 =1.76 x10
2+
− 2
γMg (γOH ) (0.52)(0.81)
Mg(OH)2 is a 1:2 compound, so
K sp' 3 1.76 x10−9
2+
3
[Mg ] = 4 =
4
2+
-4
[Mg ] = 7.60 x 10 M
[
€
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]
{(
)}
7.
Use activities to calculate the solubilities of the following compounds in a 0.0167 M
solution of Mg(NO3)2.
+
1
2
IonicStrength= µ =  Mg 2 zMg
2+ + NO 3

2
µ = 0.0501 M ≈ 0.050 M
[
€
[
]
−
]z
2
NO 3 −
 1
=
 2
{((0.0167) • (+2) + (2 • 0.0167) • (−1) )}
2
(a) AgIO3
γAg+ = 0.80; γIO3- = 0.82; Ksp = 3.1 x 10-8
K sp
3.1x10−8
−
K'sp = [ Ag + ] IO 3 =
=
= 4.73x10−8
−
+
(0.80)(0.82)
γAg γIO 3
[
]
(
)
AgIO3 is a 1:1 compound, so:
[Ag ] =
+
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€
Ksp' = 4.73x10−8
[Ag+] = 2.17 x 10-4 M
(b) Mg(OH)2
γMg2+ = 0.52; γOH- = 0.81; Ksp = 6 x 10-10
K sp
2
6 x10−10
−9
K'sp = [Mg 2+ ][OH− ] =
=
2 =1.76 x10
2+
− 2
γMg (γOH ) (0.52)(0.81)
€
The solution already contains some Mg2+ ions from the Mg(NO3)2, so we must
account for that amount:
Mg(OH)2 (s) <==> Mg2+ (aq) + 2 OH- (aq)
[Initial] (M)
0.0167
0
[Change] (M)
+x
+2x
[Equil.] (M)
0.0167 + x
2x
2
K'sp =1.76 x10−9 = [Mg 2+ ][OH− ] = (0.0167+ x )(2x )
2
Assume that x << 0.0167 M:
€
1.76 x10−9 = (0.0167)( 4 x 2 ) ==> x=
1.76 x10−9
( 4)(0.0167)
x = 1.62 x 10-4 M
€
Check assumption: 1.62 x 10-4 M << 0.0167 M => assumption is valid.
2
7.
(c) PbSO4
γPb2+ = 0.455; γSO42- = 0.445; Ksp = 6.3 x 10-7
K sp
6.3x10−7
2−
K'sp = [Pb 2+ ] SO 4 =
=
= 3.11x10−6
2−
2+
(0.455)(0.445)
γPb γSO 4
[
]
(
)
PbSO4 is a 1:1 compound, so:
[Pb ] =
2+
Ksp' = 3.11x10−6
€
[Pb2+] = 1.76 x 10-3 M
€
(d) La(IO3)3
γLa3+ = 0.245; γIO3- = 0.82; Ksp = 1.0 x 10-11
K sp
1.0 x10−11
− 3
'
3+
−11
K sp = [La ] IO 3 =
=
3 = 7.40 x10
− 3
3+
(0.245)(0.82)
γLa γIO 3
[
€
€
]
(
)
La(IO3)3 is a 1:3 compound, so:
K sp' 4 7.40 x10−11
3+
4
[La ] = 27 =
27
3+
-3
[La ] = 1.29 x 10 M
8.
Use activities to calculate the equilibrium concentration of Mg2+ in a solution that is
0.0167 M in Mg(ClO4)2 and is saturated with MgNH4PO4.
(Ksp(MgNH4PO4) = 3 x 10-13)
Mg(ClO4)2 is highly soluble.
+
1
1
2
2
−
2
2
IonicStrength= µ = Mg 2 zMg
zClO
(0.0167) • (+2) + (2 • 0.0167) • (−1)
2+ + ClO 4
− =
4
2
2
µ = 0.0501 M ≈ 0.050 M
γMg2+ = 0.52; γNH4+ = 0.80; γPO43- = 0.16; Ksp = 3 x 10-13
K sp
3x10−13
+
3−
K'sp = [Mg 2+ ] NH 4 PO 4 =
=
= 4.51x10−12
+
3−
2+
0.52
0.80
0.16
(
)
(
)
(
)
γ
Mg
γ
NH
γ
PO
(
) 4
4
([
€
[
][
]
[
]
2+
€
(
)(
MgNH4PO4 (s) <==> Mg (aq) +
[Initial] (M)
0.0167
[Change] (M)
+x
[Equil.] (M)
0.0167 + x
[
K'sp = 4.51x10−12 = [Mg 2+ ] NH 4
+
) {(
]
+
)
NH4 (aq) + PO43- (aq)Ksp = 3 x 10-13
0
0
+x
+x
x
x
][PO ] = (0.0167+ x )( x)( x )
3−
4
Assume that x << 0.0167 M:
€
4.51x10−12 = (0.0167)( x 2 ) ==> x=
4.51x10−12
(0.0167)
x = 1.6 x 10-5 M
€
Check assumption: 1.6 x 10-5 M << 0.0167 M => assumption is valid.
)}
9.
Calculate the molar solubility of calcium oxalate in a solution that has a constant pH of 4.00
at a temperature of 25.0˚C. (Use the systematic method)
Ksp(CaC2O4) = 2.3 x 10-9; Ka1 (oxalic acid) = 5.36 x 10-2; Ka2 (oxalic acid) = 5.42 x 10-5
Step 1) Give reaction equations for all pertinent equilibrium processes in the system:
CaC2O4 (s) <==> Ca2+ (aq) + C2O42- (aq)
Ksp = 2.3 x 10-9
C2O42- + H2O (l) <==> HC2O4- (aq) + OH- (aq)
Kb1 = Kw/Ka2 = 1.0 x 10-14/5.42 x 10-5
Kb1 = 1.85 x 10-10
HC2O4- + H2O (l) <==> H2C2O4 (aq) + OH- (aq)
Kb2 = Kw/Ka1 = 1.0 x 10-14/5.36 x 10-2
Kb2 = 1.87 x 10-13
2 H2O (l) <==> H3O+ (aq) + OH- (aq)
Kw = 1.0 x 10-14 (at 25˚C)
Step 2) Define an overall charge balance equation:
2[Ca2+] + [H3O+] = 2[C2O42-] + [HC2O4-] + [OH-]
Step 3) Define a mass balance equation(s):
[Ca2+] = [C2O42-] + [HC2O4-] + [H2C2O4]
Step 4) Define all equilibrium constants using expressions:
Ksp = [Ca2+][C2O42-]
= 2.3 x 10-9
2Kb1 = [HC2O4 ][OH ]/[C2O4 ] = 1.85 x 10-10
Kb2 = [H2C2O4][OH-]/[HC2O4-] = 1.87 x 10-13
Kw = [H3O+][OH-]
= 1.0 x 10-14 (at 25.0˚C)
Step 5) Check:
6 equations, 6 unknowns
Step 6) Solve:
pH = 4.00, so [H3O+] = 1.0 x 10-4 M
[OH-] = Kw/[H3O+] = 1.0 x 10-14/1 x 10-4 = 1.0 x 10-10 M
Use Kb expressions to convert [HC2O4-] and [H2C2O4] to expressions involving [C2O42-]
and known values, to substitute into mass balance equation.
[HC2O4-] = Kb1[C2O42-]/[OH-] = (1.85 x 10-10)[C2O42-]/1.0 x 10-10
=> [HC2O4-] = 1.85[C2O42-]
[H2C2O4] = Kb2[HC2O4-]/[OH-] = Kb2Kb1[C2O42-]/[OH-]2
= (1.85 x 10-10)(1.87 x 10-13)[C2O42-]/(1.0 x 10-10)2
=> [H2C2O4] = 0.00346[C2O42-]
[Ca2+] = [C2O42-] + 1.85[C2O42-] + 0.00345[C2O42-] = 2.85[C2O42-]
=> [C2O42-] = [Ca2+]/2.85
Substitute this expression for [C2O42-] into Ksp expression:
Ksp = [Ca2+][Ca2+]/2.85 => [Ca2+] = √(Ksp•2.85)
[Ca2+] = √(2.3 x 10-9 x 2.85) => [Ca2+] = 8.10 x 10-5 M
10. Calculate the solubility of silver sulphide in pure water at 25.0˚C using the systematic
method.
Ksp(Ag2S) = 6 x 10-50; Ka1(H2S) = 5.7 x 10-8; Ka2(H2S) = 1.2 x 10-15
Ag2S (s) <==> 2 Ag+ (aq) + S2- (aq)
S2- + H2O (l) <==> HS- (aq) + OH- (aq)
Kb1 = Kw/Ka2 = 1.0 x 10-14/1.2 x 10-15
HS- + H2O (l) <==> H2S (aq) + OH- (aq)
Kb2 = Kw/Ka1 = 1.0 x 10-14/5.7 x 10-8
2 H2O (l) <==> H3O+ (aq) + OH- (aq)
Ksp = 2.3 x 10-9
Kb1 = 8.33
Kb2 = 1.75 x 10-7
Kw = 1.0 x 10-14 (at 25˚C)
charge balance equation:
[Ag+] + [H3O+] = 2[S2-] + [HS-] + [OH-]
mass balance equation:
½ [Ag+] = [S2-] + [HS-] + [H2S]
equilibrium constant expressions:
Ksp = [Ag+]2[S2-]
= 6 x 10-50
Kb1 = [HS-][OH-]/[S2-] = 8.33
Kb2 = [H2S][OH-]/[HS-] = 1.75 x 10-7
Kw = [H3O+][OH-]
= 1.0 x 10-14 (at 25.0˚C)
6 equations, 6 unknowns
Solve:
Pure water & Ksp for Ag2S is very small, so pH ≈ 7.00 and [H3O+] ≈ [OH-] ≈ 1.0 x 10-7 M
[HS-] = Kb1[S2-]/[OH-] = (8.33)[S2-]/1.0 x 10-7
=> [HS-] = 8.33 x 107[S2-]
[H2S] = Kb2[HS-]/[ OH-] = Ka2[S2-]/[ OH-]
= (1.75 x 10-7)(8.33 x 107[S2-])/(1.0 x 10-7)
=> [H2S] = 1.46 x 108[S2-]
½ [Ag+] = [S2-] + 8.33 x 107[S2-] + 1.46 x 108[S2-] = 2.29 x 108[S2-] [S2-]
=> [S2-] = [Ag+]/(2 x 2.29 x 108)
Substitute this expression for [C2O42-] into Ksp expression:
Ksp = [Ag+]2[Ag+]/4.58 x 108=> [Ag+] = √(Ksp•4.58 x 108)
[Ca2+] = 3√(6 x 10-50 x 4.58 x 108) = 3√(2.75 x 10-41) => [Ag+] = 3.02 x 10-14 M
Check assumption: 0.0111 M << 0.120 M => assumption is valid.