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Overview
Equilibria of slightly soluble ionic compounds
Equilibria involving complex ions
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The ion-product expression Qsp and
the solubility-product constant Ksp
For the reaction
2−
+
−−
*
Ag2 CrO3 (s) )
−
− 2 Ag (aq) + CrO4 (aq)
We can write the ion-product expression:
2
Qsp = cAg
+ cCrO4
When the solution is saturated (in the presence of some solid, the
ion-product becomes the solubility-product constant Ksp
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Writing ion-product expressions
Problem: Write the ion-product expression at equilibrium for each
compound:
1.
2.
3.
4.
calcium sulfate
chromium(III) carbonate
magnesium hydroxide
arsenic(III) sulfide
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Writing ion-product expressions
Problem: Write the ion-product expression at equilibrium for each
compound:
1. calcium sulfate Ksp = cCa2+ cSO42−
2
3
2. chromium(III) carbonate Ksp = cCr
3+ c
CO32−
2
3. magnesium hydroxide Ksp = cMg 2+ cOH
−
2
3
3
4. arsenic(III) sulfide Ksp = cAs
3+ cHS − cOH −
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Calculations involving the solubility-product constant
Problem: When fluorite (CaF2 ) is pulverized and shaken in water at
18 ◦ C, 10.0 mL of the solution contain 1.5 ×10−4 g of solute. Find the
Ksp of fluorite.
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Calculations involving the solubility-product constant
1 mol
78.07 g
= 1.92 × 10−4 mol · L−1
1L
10.0 mL ×
1000 mL
1.5 × 10−4 g ×
s=
From the equilibrium
2+
−
*
CaF2 (s) −
)
−
− Ca (aq) + 2 F− (aq)
We can see that Ksp = cCa2+ cF2 − = s (2s)2
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Calculations involving the solubility-product constant
From the equilibrium
2+
−−
*
CaF2 (s) )
−
− Ca (aq) + 2 F− (aq)
We can see that Ksp = cCa2+ cF2 − = s (2s)2
therefore
Ksp = 2.8 × 10−11
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Determining the solubility from the value of Ksp
Problem: The Ksp for chromium(III) iodate is 5.0 ×10−6 . What is the
solubility ( mol·L−1 ) of this compound at 25 ◦ C?
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Determining the solubility from the value of Ksp
−
3+
3
−
*
Cr(IO3 )3 (s) −
)
−
− Cr (aq) + 3 IO3 (aq) Ksp = cCr 3+ cIO
−
3
Initial
Change
Equilibrium
Cr(IO3 )3
-
Cr3+
0
s
s
IO3–
0
3s
3s
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Determining the solubility from the value of Ksp
Ksp = s (3s)3
r
s=
4
Ksp
27
s = 0.021mol · L−1
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The effect of a common ion on solubility
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Calculating the effect of a common ion on solubility
Problem: To decrease the solubility of BaSO4 in oral preparations, a
dilute solution of Na2 SO4 is normally added. What is the solubility of
BaSO4 (Ksp = 1.1 × 10−10 in
1. water
2. 0.10 mol·L−1 Na2 SO4
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Calculating the effect of a common ion on solubility
Problem: To decrease the solubility of BaSO4 in oral preparations, a
dilute solution of Na2 SO4 is normally added. What is the solubility of
BaSO4 (Ksp = 1.1 × 10−10 in
1. water
p
In water s = Ksp = 1.0 × 10−5 mol · L−1
2. 0.10 mol·L−1 Na2 SO4
In 0.10 mol·L−1 Na2 SO4 s =
Ksp
= 1.1 × 10−9 mol · L−1
0.10
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Separation of ions by selective precipitation
If the ions in a mixture form insoluble compounds of different
solubility, a concentration can be found that will precipitate the most
insoluble almost completely before the more soluble starts to
precipitate.
Problem: A solution is 0.050 mol·L−1 in BaCl2 and 0.025 mol·L−1 in
CaCl2 . What concentration of SO42– will separate the ions?
Ksp BaSO4 = 1.1 × 10−10 ,
Ksp CaSO4 = 2.4 × 10−5 ,
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Separation of ions by selective precipitation
Problem: A solution is 0.050 mol·L−1 in BaCl2 and 0.025 mol·L−1 in
CaCl2 . What concentration of SO42– will separate the ions?
To make the solution is saturated in CaSO4 the concentration of
SO42– needs to be 9.6 ×10−4 mol·L−1 . At that point, the
concentration of Ba2+ left in solution is 1.2 ×10−7 mol·L−1
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[Cr(NH3 )63+ ], a typical complex ion
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The stepwise exchange of NH3 for H2 O in
[M(H2 O)4 ]2+
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Formation constants of complex ions
Complex ion
[Ag(CN)2 ]–
[Ag(NH3 )2 ]+
[Ag(S2 O3 )2 ]3–
[AlF6 ]3–
[Al(OH)4 ]–
[Be(OH)4 ]2–
[CdI4 ]2–
[Co(OH)4 ]2–
[Cr(OH)4 ]2–
[Cu(NH3 )4 ]2+
Kf
3.0×1020
1.7×107
4.7×1013
4×1019
3×1033
4×1018
1×106
5×109
8.0×1029
5.6×1011
[Fe(CN)6 ]4–
[Fe(CN)6 ]3–
[Hg(CN)4 ]2–
[Ni(NH3 )6 ]2+
[Pb(OH)3 ]–
[Sn(OH)3 ]–
[Zn(CN)4 ]2–
[Zn(NH3 )4 ]2+
[Zn(OH)4 ]2–
3.0×1035
4.0×1043
9.3×1038
2.0×108
8×1013
3×1025
4.2×1019
7.8×108
3×1015
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Calculating the concentration of a complex ion
An industrial chemist converts [Zn(H2 O)4 ]2+ to the more stable
[Zn(NH3 )4 ]2+ by mixing 50.0 L of 0.0020 mol·L−1 [Zn(H2 O)4 ]2+ and
25.0 L of 0.15 mol·L−1 NH3 . What is the final c[Zn(NH3 )4 ]2+ ?
Datum: Kf of [Zn(NH3 )4 ]2+ is 7.8×108
2+
−−
[Zn(H2 O)4 ]2+ (aq) + 4 NH3 (aq) )
−*
− [Zn(NH3 )4 ] (aq) + 4 H2 O(`)
Kf =
c[Zn(NH3 )4 ]2+
4
c[Zn(H2 O)4 ]2+ cNH
= 7.8 × 108
3
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Calculating the concentration of a complex ion
An industrial chemist converts [Zn(H2 O)4 ]2+ to the more stable
[Zn(NH3 )4 ]2+ by mixing 50.0 L of 0.0020 mol·L−1 [Zn(H2 O)4 ]2+ and
25.0 L of 0.15 mol·L−1 NH3 . What is the final c[Zn(NH3 )4 ]2+ ?
Datum: Kf of [Zn(NH3 )4 ]2+ is 7.8×108
2+
−−
[Zn(H2 O)4 ]2+ (aq) + 4 NH3 (aq) )
−*
− [Zn(NH3 )4 ] (aq) + 4 H2 O(`)
Kf =
c[Zn(NH3 )4 ]2+
4
c[Zn(H2 O)4 ]2+ cNH
= 7.8 × 108
3
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Calculating the concentration of a complex ion
An industrial chemist converts [Zn(H2 O)4 ]2+ to the more stable
[Zn(NH3 )4 ]2+ by mixing 50.0 L of 0.0020 mol·L−1 [Zn(H2 O)4 ]2+ and
25.0 L of 0.15 mol·L−1 NH3 . What is the final c[Zn(NH3 )4 ]2+ ?
Datum: Kf of [Zn(NH3 )4 ]2+ is 7.8×108
2+
−−
[Zn(H2 O)4 ]2+ (aq) + 4 NH3 (aq) )
−*
− [Zn(NH3 )4 ] (aq) + 4 H2 O(`)
Kf =
c[Zn(NH3 )4 ]2+
4
c[Zn(H2 O)4 ]2+ cNH
= 7.8 × 108
3
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Calculating the concentration of a complex ion
Calculating the initial concentrations
(50.0 L) (2.0 × 10−3 mol · L−1 )
= 1.3 × 10−3 mol · L−1
50.0 + 25.0 L
(25.0 L) (0.15 mol · L−1 )
=
= 5.0 × 10−2 mol · L−1
50.0 + 25.0 L
c[Zn(H2 O)4 ]2+ =
cNH3
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Calculating the concentration of a complex ion
Initial
Change
Equilibrium
[Zn(H2 O)4 ]2+
1.3×10−3
∼ (- 1.3×10−3 )
x
NH3
5.0×10−2
∼ (-4 × 1.3×10−3 )
4.5×10−2
[Zn(NH3 )4 ]2+
0
∼ (+ 1.3×10−3 )
1.3×10−3
x = 4.1×10−7 mol·L−1
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Calculating the concentration of a complex ion
Initial
Change
Equilibrium
[Zn(H2 O)4 ]2+
1.3×10−3
∼ (- 1.3×10−3 )
x
NH3
5.0×10−2
∼ (-4 × 1.3×10−3 )
4.5×10−2
[Zn(NH3 )4 ]2+
0
∼ (+ 1.3×10−3 )
1.3×10−3
x = 4.1×10−7 mol·L−1
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Effect of the formation of a complex ion on the
solubility
Calculate the solubility of lead(II) chloride in
1. water
2. 0.75 mol·L−1 NaOH
Data: Kf [Pb(OH)3 ]− = 8×1013 ; Ksp PbCl2 = 1.7×10−5
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Effect of the formation of a complex ion on the
solubility
Calculate the solubility of lead(II) chloride in
1. waterr
Ksp
= 1.6×10−2 mol·L−1
4
2. 0.75 mol·L−1 NaOH
S=
3
Data: Kf [Pb(OH)3 ]− = 8×1013 ; Ksp PbCl2 = 1.7×10−5
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Effect of the formation of a complex ion on the
solubility
Calculate the solubility of lead(II) chloride in
1. water
2. 0.75 mol·L−1 NaOH
4S 3
= 1.4×109 , S = 0.25 mol·L−1
(0.75 − 3S)3
Data: Kf [Pb(OH)3 ]− = 8×1013 ; Ksp PbCl2 = 1.7×10−5
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