Transcript Chemistry 1011 - Marine Institute of Memorial University

Chemistry 1011
TOPIC
Solubility Equilibrium
TEXT REFERENCE
Masterton and Hurley Chapter 16.1
Chemistry 1011 Slot 5
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16.1 Solubility Equilibrium
YOU ARE EXPECTED TO BE ABLE TO:
• Write an expression for the solubility product constant, Ksp, for a
substance
• Calculate the concentration of ions at equilibrium, given Ksp
• Calculate the solubility product constant for a substance given its
solubility and formula
• Predict whether a combination of ions will form a precipitate, given
Ksp and ion concentrations
• Calculate the solubility of a substance in water, given Ksp
• Use Le Chatelier’s Principle to determine the effect of adding a
common ion to a solution.
• Calculate the solubility of a substance in the presence of a common
ion
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Formation of Precipitates
• A precipitate is formed when
– Two solutions are mixed, and
– The cation from one solution combines with the anion
from the other solution to form an insoluble solid
NaCl(aq) + AgNO3(aq)
NaNO3(aq) + AgCl(s)
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)
Ag+(aq) + Cl-(aq)
Na+(aq) + NO3-aq) + AgCl(s)
AgCl(s)
• An equilibrium is established between the solid and the
corresponding ions in solution
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Solubility Equilibrium
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AgCl(s)
Ag+(aq) + Cl-(aq)
An expression can be written for the equilibrium
constant:
Ksp = [Ag+]x[Cl-]
[solid] does not appear in the equilibrium
constant expression
Ksp is known as the solubility product constant
Solubility product data are normally measured at
25oC
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Determining Ion Concentrations
Ag3PO4(s)
3Ag+(aq) + PO43-(aq)
Ksp = [Ag+]3x[PO43-] = 1 x 10-16
PbCl2(s)
Pb2+(aq) + 2Cl-(aq)
Ksp = [Pb2+]x[Cl-]2 = 1.7 x 10-5
Q: Calculate [Pb2+] and [Cl-] in a solution of PbCl2 at 25oC
[Cl-] = 2 x [Pb2+]
Ksp = [Pb2+] x [2Pb2+]2 = 1.7 x 10-5
Ksp = 4 [Pb2+]3 = 1.7 x 10-5
[Pb2+] = 1.6 x 10-2 mol/L
[Cl-] = 3.2 x 10-2 mol/L
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Calculating Ksp
• The solubility of a salt can be determined by experiment
• Ksp for the salt can be determined from these results
• Q: The solubility of magnesium hydroxide is found to be 8.4 x 10-4
g/100cm3 at 25oC. Find Ksp
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
Ksp = [Mg2+]x[OH-]2 = ??
Solubility = 8.4 x 10-4 g/100cm3 at 18oC
Solubility = (8.4 x 10-4)g x 1000cm3/L = 1.44 x 10-4 mol/L
58.3 g/mol 100cm3
[Mg2+] = 1.44 x 10-4 mol/L; [OH-] = 2.88 x 10-4 mol/L
Ksp = [Mg2+]x[OH-]2 = 1.2 x 10-11
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Determining Precipitate Formation
•
To determine whether a precipitate will form
when two solutions are mixed:
1. Determine the concentrations of the reacting ions in
the mixture
2. Calculate the ion product, P
3. Compare the ion product, P, with Ksp
4. If P > Ksp then precipitate will form
5. If P < Ksp then no precipitate
6. If P = then no precipitate – solution is saturated
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Determining Precipitate Formation
Q: Will a precipitate form when 5.0mL of 1.0 x 10-3 mol/L silver
nitrate is added to 5.0mL of 1.0 x 10-5 mol/L potassium chromate?
Ksp Ag2CrO4 = 1.0 x 10-12
2AgNO3(aq) + K2CrO4(aq)
2Ag+(aq) + CrO42-(aq)
2KNO3(aq) + Ag2CrO4(s)
Ag2CrO4(s)
Ksp = [Ag+]2 x [CrO42-] = 1.0 x 10-12
[Ag+] = 5.0 x 10-4 mol/L
[CrO42-] = 5.0 x 10-6 mol/L
Ion Product, P = [Ag+]2 x [CrO42-] = (5.0 x 10-4 )2 x (5.0 x 10-6 )
P = 1.25 x 10-12
P > Ksp A precipitate will form
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Determining Precipitate Formation
Q: 1.0 mL of 1.0 mol/L barium chloride is added
to 10.0 mL of a solution containing a small
amount of magnesium sulfate. Determine the
minimum concentration of magnesium sulfate
that will cause a precipitate to form.
Ksp BaSO4 = 1.1 x 10-10
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Determining Precipitate Formation
•
1.0 mL of 1.0 mol/L barium chloride is added to 10.0 mL of a solution containing a small
amount of magnesium sulfate. Determine the minimum concentration of magnesium sulfate
that will cause a precipitate to form.
Ksp BaSO4 = 1.1 x 10-10
BaCl2(aq) + MgSO4(aq)
Ba2+(aq) + SO42-(aq)
MgCl2(aq) + BaSO4(s)
BaSO4(s)
Ksp = [Ba2+]x[SO42-] = 1.1 x 10-10
[Ba2+] = 1.0 x 10-3L x 1.0 mol/L  1.00 x 10-2L = 1.0 x 10-1 mol/L
[SO42-] = x mol/L
Ksp = [Ba2+]x[SO42-] = (1.0 x 10-1 ) x (x) = 1.1 x 10-10
x = [SO42-] = minimum [MgSO4] =1.1 x 10-10 mol/L
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Selective Precipitation
• Suppose that a solution contains two different cations,
for example Ba2+ and Ca2+
• Each forms an insoluble sulfate, BaSO4 and CaSO4
Ksp BaSO4 = 1.1 x 10-10
Ksp CaSO4 = 7.1 x 10-5
• If sulfate ions are added to a solution containing equal
amounts of Ba2+ and Ca2+, then the BaSO4 will
precipitate first
• Only when the Ba2+ ion concentration becomes very
small will the SO42- ion concentration rise to the point
that CaSO4 will be precipitated
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Determining Solubility
• The solubility, s, of a salt can be determined from Ksp data
Q: Determine the solubility of lead chloride in water at 25oC.
1.7 x 10-5
Let solubility of lead chloride = s mol/L
• For every mole of PbCl2 that dissolves,
1 mole of Pb2+(aq) and 2 moles of Cl-(aq) are formed
PbCl2(s)
Ksp =
Pb2+(aq) + 2Cl-(aq)
[Pb2+] = s mol/L
[Cl-] = 2 x [Pb2+] = 2s mol/L
Ksp = [Pb2+]x[Cl-]2 = (s) x(2s)2 = 1.7 x 10-5
4s3 = 1.7 x 10-5
s = 1.6 x 10-2 mol/L (Can also be expressed in grams/Litre)
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The Common Ion Effect
• The presence of a common ion will reduce the solubility
of an ionic salt (Le Chatelier)
• If a common ion is added to a saturated solution of a salt,
then the salt will be precipitated (Le Chatelier)
For example, CaCO3 is less soluble in a solution
containing CO32- ions than in pure water
CaCO3(s)
Ca2+(aq) + CO32-(aq)
Ksp = [Ca2+] x [CO32-] = 4.9 x 10-9
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The Common Ion Effect
CaCO3(s)
Ca2+(aq) + CO32-(aq)
Ksp = [Ca2+] x [CO32-] = 4.9 x 10-9
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Solubility of CaCO3 = [Ca2+]
In pure water
[Ca2+] = [CO32-] = (4.9 x 10-9) = 7.0 x 10-5 mol/L
Solubility of CaCO3 in 1.0 x 10-1 sodium carbonate solution = ???
• [CO32-] = 1.0 x 10-1 mol/L (ignore CO32- from CaCO3 )
• [Ca2+] =
Ksp
= 4.9 x 10-9 = 4.9 x 10-8 mol/L
[CO32-]
1.0 x 10-1
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One More Example
• Ksp for manganese II hydroxide is 1.2 x 10-11
• Solid sodium hydroxide is added slowly to a 0.10 mol/L
solution of manganese II chloride. What will be the pH when
a precipitate forms?
Mn(OH)2(s)
Mn2+(aq) + 2OH-(aq)
Ksp = [Mn2+] x [OH-]2 = 1.2 x 10-11
• [Mn2+] = 0.10 mol/L
• [OH-] =  Ksp  [Mn2+] =  1.2 x 10-11  0.10
= 1.1 x 10-5 mol/L
• pH = 9.0
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