Solubility Equilibria
Reading: from chapter 19 of Petrucci, Harwood and Herring
(8th edition):
Required: Sections 19-1 through 19-8.
Examples: 19-1 through 19-8, 19-11, 19-13.
Problem Set:
Chapter 19 questions: 4, 5, 7a-c, 10, 11a,b, 14, 47, 53, 58,
62, 64.
Additional problems from Chapter 19: 65, 71.
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Solubility
C Some liquids can dissolve in each other in any amount.
Example: ethanol and water.
C Many liquids can only dissolve in each other to only a
limited degree. Example: benzene and water.
C Solids will dissolve in liquids, but there is always a limit.
C Solubility is the amount of a substance that can dissolve in
a given amount of another substance (usually a liquid).
C Units should be M but other units are often used (such as
grams per 100 mL).
C Important for separations and quantitative analysis.
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Non-Dissociating Solute
We can treat dissolution as a chemical reaction.
Example:
sucrose(s) W sucrose(aq)
At 25 °C, KC = 1.97 and KC = [sucrose(aq)].
Y The solubility of sucrose in water is 1.97 M.
C If [sucrose(aq)] = 1.97 M, the solution is saturated;
additional sucrose can not dissolve.
C If [sucrose(aq)] < 1.97 M, the solution is unsaturated;
additional sucrose can dissolve.
C If [sucrose(aq)] > 1.97 M, the solution is supersaturated;
if solid sucrose is present, sucrose will precipitate.
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The Solubility Product Constant
When salts dissolve in water, they usually dissociate. The
equilibrium constant for this process is called the solubility
product constant, Ksp.
Example:
AgCl(s) W Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = 1.8×10-10 at 25 °C.
Q: How much AgCl can dissolve in water at 25 °C?
From stoichiometry, [Ag+] = [Cl-] = x.
At equilibrium Ksp = x2 = 1.8×10-10 Y
x = 1.3×10-5 M
The solubility of AgCl is 1.3×10-5 M.
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Calculating Solubility from Ksp
For Mg3(AsO 4)2, Ksp = 2.0×10-20. Find the solubility.
Solution:
Mg3(AsO 4)2(s) W 3Mg2+(aq) + 2AsO 43-(aq)
Ksp = [Mg2+]3[AsO43-]2
Let S = solubility (moles of Mg3(AsO 4)2 dissolved per liter).
[Mg2+] = 3S
[AsO43-] = 2S
Y Ksp = (3S)3(2S)2 = 27S3 × 4S2 = 108S5 = 2.0×10-20
Y S = 4.5×10-5 M
The solubility of Mg3(AsO 4)2 is 4.5×10-5 M.
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Calculating Ksp from Solubility
The solubility of MgNH4PO3 is 5.8×10-5 M. Determine Ksp.
Solution:
MgNH4PO3 W Mg2+(aq) + NH4+(aq) + PO33-(aq)
KSP = [Mg2+][NH4+][PO33-]
Let S = solubility. [Mg2+] = [NH4+] = [PO33-] = S
Y Ksp = S×S×S = S3 = (5.8×10-5)3 = 2.0×10-13
For MgNH4PO3, Ksp = 2.0×10-13.
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The Common Ion Effect
Adding a common ion to a solution of a slightly soluble salt
suppresses the solubility of the salt.
Examples:
AgCl(s) W Ag+ + ClAdd:
Ksp = 1.8×10-10
NaCl(s) W Na+ + Cl-
C Cl- is a common ion. By LeChâtelier's principle, adding it
shifts the AgCl equilibrium to the left.
Add:
AgNO3(s) W Ag+ + NO3-
C The common ion, Ag+, suppresses the solubility of AgCl.
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The Common Ion Effect - Example
For PbI2, Ksp = 7.1×10-9. By how much does 0.10 M
Pb(NO3)2 reduce the solubility of PbI2?
Let C = [Pb(NO3)2(aq)] = [Pb2+] if no PbI2 dissolves.
Reaction
Initial
Change
Equil.
PbI2(s)
-
W
Pb2+ +
C
S
C+S
2I0
2S
2S
M
M
M
In water (C=0), Ksp = [Pb2+][I-]2 = 4S3 Y S = 1.2×10-3 M
In 0.10 M Pb(NO3)2, C=0.10 o S Y Ksp . 4CS 2
Y S = 1.3×10-4 M
Nearly a factor of 10 smaller.
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Other Influences on Solubility
C Salt effect - Ions influence each other electrostatically. This
causes activities to be lower than concentrations and
increases solubility.
C Ion pairs - At high concentrations, ions tend to "clump".
This usually increases activities and decreases solubility.
C Simultaneous reactions can change concentrations of ions.
Example:
CaCO3(s) W Ca2+ + CO32CO32- + H2O(l) W OH- + HCO3Reaction of CO32- with H2O increases solubility of CaCO3.
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Criteria for Precipitation
The ion product, Qsp, is used to decide if a salt may precipitate.
Example:
PbI2(s) W Pb2+ + 2I-
Qsp = [Pb2+]init[I-]2init
Ksp = [Pb2+]eq[I-]2eq
C If Qsp < Ksp, the solution is unsaturated. A precipitate will
not form and PbI2(s), if present, will dissolve.
C If Qsp = Ksp, the solution is saturated. It is in equilibrium
with PbI2(s).
C If Qsp > Ksp, the solution is supersaturated. A precipitate
may form. PbI2(s) will precipitate until solution becomes
saturated. Precipitation may initially be slow.
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Applying the Criteria for Precipitation
10.0 mL of 0.015 M Pb(NO3)2 are mixed with 20.0 mL of
0.030 M KI. Will a precipitate form?
Solution: Pb(NO3)2, KI, and KNO3 are all very soluble.
PbI2(s) W Pb2+ + 2I-
Ksp = 7.1×10-9
[Pb2+] = (0.015 M) (10.0 mL) /(10.0+20.0 mL) = 0.0050 M
[I-] = (0.030 M) (20.0 mL) /(10.0+20.0 mL) = 0.020 M
Qsp = [Pb2+]init[I-]2init = (0.0050)(0.020)2 = 2.0×10-6
Qsp > Ksp Y A precipitate of PbI2 will form.
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Completeness of Precipitation
The concentration of an ion in solution can be determined by
precipitating a salt of the ion and weighing the precipitate.
Example: SO42- can be determined by adding BaCl2:
BaCl2 6 Ba2+ + 2ClBaSO4(s) W Ba2+ + SO42-
Ksp = 1.1×10-10
It is impossible to precipitate all the SO42-. The analysis will
be valid if the fraction of SO42- left in solution is negligible.
If this is so, the precipitation is said to be complete.
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Completeness of Precipitation - Example
60 mL of 0.040 M BaCl2(aq) is added to 140 mL of solution
containing about 2.2 mmol SO42-. What fraction of the SO42is left in solution?
Solution:
BaSO4(s) W Ba2+ + SO42-
Ksp = 1.1×10-10
Is Ba2+ in excess? nBa = (60 mL)(0.040 M) = 2.4 mmol
Y 2.2 mmol of BaSO4(s), 0.2 mmol Ba2+(aq)
[Ba2+]f = (0.2 mmol) / (60+140 mL) = 1×10-3 M
[SO42-]f = Ksp/[Ba2+]f = (1.1×10-10)/(1×10-3) = 1.1×10-7 M
amount in solution = (1.1×10-7 M) (200 ml) = 2.2×10-5 mmol
Y 0.001% is left in solution. Precipitation is complete.
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Fractional Precipitation
Differing solubilities can be used to precipitate different
ions one at a time. Can be used for
C separation and/or purification
C precipitation titrations
Concept: To separate two ions, find a counterion that forms
salts of differing solubility. Gradually add the counterion to
the solution.
Example: Separate Br- and CrO 42- using Ag+ (counterion).
Ag2CrO4(s) W 2Ag+ + CrO42-
Ksp = 1.1×10-12
AgBr(s) W Ag+ + Br-
Ksp = 5.0×10-13
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Fractional Precipitation - Example
Slowly add AgNO3 to 0.010 M [Br-] and 0.010 M [CrO 42-].
Which precipitates first?
For AgBr:
Ksp = [Ag+][Br-]
5.0×10-13 = [Ag+](0.010)
[Ag+] = 5.0×10-11 M
For Ag2CrO4:
Ksp = [Ag+]2[CrO42-] Y [Ag+] = (Ksp/[CrO42-])½ = 1.0×10-5 M
Y AgBr precipitates first.
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Fractional Precipitation - continued
How much Br- is left in solution when Ag2CrO4 starts to
precipitate?
[Ag+] = 1.0×10-5 M (from previous slide)
Using Ksp for AgBr:
[Br-] = Ksp / [Ag+] = 5.0×10-8 M
Y 0.0005% of Br- is left in solution
Conclusion: Precipitation of AgBr is complete before
precipitation of Ag2CrO4 begins.
Also, red-brown Ag2CrO4 can be used as an indicator in
titrating Cl- or Br- with Ag+.
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Influence of pH on Solubility
If a salt contains the anion of a weak acid or the cation of
a weak base, then its solubility will depend on pH.
Example:
CaCO3(s) W Ca2+ + CO32-
Hydrolysis: CO32- + H2O(l) W OH- + HCO3-
Ksp = 2.8×10-9
pKb = 3.67
Consequences:
C Unless pH is high,
solubility is greater than
implied by Ksp.
C Solubility increases as
pH decreases.
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Change of Solubility with pH - Example
Given Ksp(CaF2) = 5.3×10-9 and pKa(HF) = 3.18, how does
the solubility of CaF2 vary as pH changes from 5.0 to 1.0?
Assume that the pH is controlled by using buffers.
CaF2(s) W Ca2+ + 2FHF + H2O(l) W H3O+ + FSignificant species: Ca2+, F-, HF (H3O+ is fixed)
Electroneutrality: Not helpful in buffer solution.
Material balance: 2[Ca2+] = [F-] + [HF]
Equilibria: Ksp = [Ca2+] [F-]2
Ka [HF] = [H3O+] [F-]
3 equations, 3 unknowns
Solution:
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Change of Solubility with pH - continued
From equilibria:
[F-] = (Ksp/[Ca2+])½
[HF] = [H3O+][F-]/Ka = ([H 3O+]/Ka) (Ksp/[Ca2+])½
Substitute into material balance:
2[Ca2+] = (Ksp/[Ca2+])½ (1 + [H 3O+]/Ka)
Solve for [Ca2+]:
[Ca2+] = (Ksp/4)a (1 + [H 3O+]/Ka)b
At pH = 5.0, [H3O+] n Ka Y [Ca2+] = (Ksp/4)a = 1.1×10-3 M
At pH = 3.18, [H3O+] = Ka Y [Ca2+] = Kspa = 1.7×10-3 M
At pH = 1.0, [H3O+] o Ka Y [Ca2+] = (Ksp[H3O+]2/(4Ka2))a
Y [Ca2+] = 3.1×10-2 M
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Solubility at Low pH - Another Approach
If the pH is low enough that [HF] o [F-], the dissolution
reaction is
CaF2(s) + 2H3O+ W Ca2+ + 2HF + 2H2O(l)
K=?
This is equivalent to:
CaF2(s) W Ca2+ + 2F2H3O+ + 2F- W 2HF + 2H2O(l)
K = Ksp
K = (1/Ka)2
CaF2(s) + 2H3O+ W Ca2+ + 2HF + 2H2O(l) K = Ksp/(Ka2)
Let S / [Ca2+] Y K[H3O+]2 = [Ca2+][HF]2 = S(2S)2 = 4S3
Y S = (K[H3O+]2/4)a = (Ksp[H3O+]2/(4Ka2))a
Same as result on previous slide for pH n pKa.
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Complex Ions
A complex ion consists of:
C a central positive ion (usually a transition metal)
C a number of ligands
The ligands are anions or neutral molecules.
The complex ion may be positively or negatively charged.
Examples:
Ag+ + 2NH3 W [Ag(NH3)2]+
Cu+ + 4CN- W [Cu(CN)4]3Fe3+ + 2SCN- W [Fe(SCN)2]+
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The Formation Constant
The equilibrium constant for the reaction producing the
complex ion from its central ion and ligands is called the
formation constant, Kf.
Co3+ + 6NH3 W [Co(NH3)6]3+
Pb2+ + 3Cl- W [PbCl3]Fe3+ + 6CN- W [Fe(CN)6]3Formation constants can be very large.
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The Formation Constant - Example
A solution is prepared containing 0.010 M NaCN,
0.010 M HCN, and 1.0×10-4 M AgNO3. Find 'free' [Ag+].
Solution:
Ag+ + 2CN- W [Ag(CN)2]-
Kf = 5.6×1018
Find the relative amounts of Ag+ and [Ag(CN)2]-:
Y
[CN-] = 0.010 M Y [Ag+] = 1.8×10-15[[Ag(CN)2]-]
Y [[Ag(CN)2]-] = 1.0×10-4 M, [Ag+] = 1.8×10-19 M
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Complex Ions and Solubility
Complex ions can substantially increase solubility.
Example: Find the solubility of AgCl in 0.10 M HCl(aq).
AgCl(s) W Ag+ + Cl-
Ksp = 1.8×10-10
[Ag+] = Ksp / [Cl-] = (1.8×10-10)/(0.10) = 1.8×10-9 M
But:
Ag+ + 2Cl- W AgCl2-
Kf = 1.1×105
[AgCl2-] = Kf[Ag+][Cl-]2 = (1.1×105)(1.8×10-9)(0.10)2
Y [AgCl2-] = 2.0×10-6 M
1000 times as large as [Ag+]
Y solubility = [Ag+] + [AgCl2-] = 2.0×10-6 M
Could also be done by the method of example 19-13.
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Solubility Equilibria - Summary
C Solids dissolve in liquids to a finite extent.
C The solubility of salts in water is governed by the
solubility product constant.
C Solubility is reduced by the presence of common ions.
C Solubility can be enhanced by hydrolysis, so it may
depend on pH.
C Solubility can be enhanced by the formation of complex
ions, so it may depend on the presence of ligands.
C Precipitation of an ion is "complete" if only a negligible
amount remains in solution.
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