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Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
Aqueous equilibria II Solubility Products
BaSO4(s) ↔ Ba2+(aq) + SO42­(aq)
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Solubility Products
The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility product.
Which Ksp expression is correct for AgCl?
A
[Ag+]/[Cl­]
B
[Ag+][Cl­]
C
[Ag2+]2[Cl2­]2
D
[Ag+]2[Cl­]2
E
None of the above.
There is never any denominator in Ksp expressions because pure solids are not included in any equilibrium expressions.
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Given the reaction at equilibrium: Zn(OH)2 (s) <­­> Zn2+ (aq) + 2OH­ (aq)
what is the expression for the solubility product constant, Ksp, for this reaction?
A Ksp= [Zn2+][OH­]2 / [Zn(OH)2]
B
Ksp= [Zn(OH)2] / [Zn2+][2OH­]
C
Ksp= [Zn2+][2OH­]
Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
D Ksp= [Zn2+][OH­]2
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Solubility
Solubility
The term "solubility" represents the maximum amount of solute that can be dissolved in a certain volume before any precipitate is observed.
The solubility of a substance can be given in terms of grams per liter
g/L
or in terms of moles per liter
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mol/L
The latter is sometimes referred to as "molar solubility."
Example #1
Consider the slightly soluble compound barium oxalate, BaC2O4. The solubility of BaC2O4 is 1.3 x 10­3 mol/L.
The ratio of cations to anions is 1:1.
• This means that 1.3 x 10­3 moles of Ba2+ can dissolve in one liter.
• Also, 1.3 x 10­3 moles of C2O42­ can dissolve in one liter. • What is the maximum amount (in grams) of BaC2O4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)?
Solubility
Solubility
What is the maximum amount (in grams) of BaC2O4 that could dissolve in 2.5 L (before a precipitate occurs)?
The solubility of BaC2O4 is 1.3 x 10­3 mol/L.
BaC2O4 ­­> Ba2+ + C2O42­
1 mol BaC2O4 225.3g
­­­­­­­­­­­­­­­­­­­­ = ­­­­­­­­­­­
1.3 x 10­3 mol BaC2O4 x g
= 0.293g BaC2O4 x 2.5L = 0.73g
1L
Example #2
Consider the slightly soluble compound lead (II) chloride, PbCl2. The solubility of PbCl2 is 0.016 mol/L.
The ratio of cations to anions is 1:2.
• This means that 0.016 moles of Pb2+ can dissolve in one liter. Molar solubility always refers to the ion with the lower molar ratio.
• Twice as much, or 2(0.016) = 0.032 moles of Cl­ can dissolve in one liter. This is the maximum amount that could dissolve in 2.5 L before a precipitate occurs.
2.5 L 1 1.3 x 10­3 mol BaC2O4 225.3 g BaC2O4
x 1 mol BaC2O4
1 L
x Solubility
Example #3
Consider the slightly soluble compound silver sulfate, Ag2SO4. Solubility
Remember that molar solubility refers to the ion with the lower mole ratio. It does not always refer to the cation, although in most cases it does.
The solubility of Ag2SO4 is 0.015 mol/L.
The ratio of cations to anions is 2:1.
•
in one liter.
This means that 0.015 moles of SO42­ can dissolve • Twice as much, or 2(0.015) = 0.030 moles of Ag+ can dissolve in one liter. Compound Molar [Cation] Solubility of Compound 1.3 x 10­3 mol [Anion] BaC2O4 1.3 x 10­3 mol 1.3 x 10­3 mol PbCl2 0.016 mol/L 0.016 mol/L 0.032 mol/L
Ag2SO4 0.015 mol/L 0.030 mol/L 0.015 mol/L 2
solubility­presentation­2012­03­31.notebook
If the solubility of barium carbonate, BaCO3 is 7.1 x 10­5 M, this means that a maximum of _______barium ions, Ba2+ ions can be dissolved per liter of solution.
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7.1 x 10 ­5 moles
B half of that
B
half of that
C twice as much
C
twice as much
D one­third as much
D
one­third as much
E one­fourth as much
E
one­fourth as much
7.1 x 10­5 moles
If the solubility of Ag2CrO4 is 6.5 x 10­5 M, this means that a maximum of _______chromate ions, CrO42­, can be dissolved per liter of solution.
A
If the solubility of barium carbonate, BaCO3 is 7.1 x 10­5 M, this means that a maximum of _______carbonate ions, CO32­ ions can be dissolved per liter of solution.
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A
A
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If the solubility of Ag2CrO4 is 6.5 x 10­5 M, this means that a maximum of _______ Ag+ ions can be dissolved per liter of solution.
6.5 x 10­5 moles ­5
A
6.5 x 10­5 moles B
twice 6.5 x 10 moles B
twice 6.5 x 10­5 moles C
half 6.5 x 10­5 moles C
half 6.5 x 10­5 moles D
one­fourth 6.5 x 10­5 moles D
one­fourth 6.5 x 10­5 moles E
four times 6.5 x 10­5 moles E
four times 6.5 x 10­5 moles 7
Calculating Ksp from the Solubility
Sample Problem
The molar solubility of lead (II) bromide, PbBr2 is 1.0 x 10­2 at 25oC. Calculate the solubility product, Ksp, for this compound.
For the slightly soluble compound, AB, the molar solubility is 3 x 10­8 moles per liter. Calculate the Ksp for this compound. No calculator.
AB <­­> A + + B­
A 3 x 10­8
B
1/2 (3 x 10­8)
C (3 x 10­8)^1/2
[Pb2+] = 1.0 x 10­2 mol/L
[Br­] = 2.0 x 10­2 mol/L
Substitute the molar concentrations into the Ksp expression and solve.
2+
­ 2
Ksp = [Pb ][Br ]
­2
­2 2
= (1.0 x 10 )(2.0 x 10 ) D
2 (3 x 10­8)
E
(3 x 10­8)^2
­6
= 4.0 x 10
AgCl <−−> Ag+ + Cl­
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For the slightly soluble compound, XY, the molar solubility is 5 x 10­5 M. Calculate the Ksp for this compound. No calculator.
XY <­­> X + + Y­
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For the slightly soluble compound, MN, the molar solubility is 4 x 10­6 M. Calculate the Ksp for this compound. No calculator.
MN <­­> M+ + N­
A 4 x 10­6
A 5 x 10­5
B 10 x 10­5
B
16 x 10­6
C
25 x 10­5
C
16 x 10­12
D
5 x 10­10
D
16 x 10­36
E
25 x 10­10
BaCO3 <−−> Ba2+ + CO32­
10 For the slightly soluble compound, AB2, the molar solubility is 3 x 10­4 M. Calculate the solubility­
product constant for this compound. No calculator.
AB2 <­­> A2+ + 2B­
11 For the slightly soluble compound, X3Y, the molar solubility is 1 x 10­4 M. Calculate the solubility product for this compound. No calculator.
X3Y <­> 3X+ + Y3­
A 9 x 10­4
A
3 x 10­4
9 x 10­8
B
3 x 10­8
C 18 x 10­8
C
27 x 10­12
D 36 x 10­8
D
27 x 10­16
B
E
108 x 10­12
Fe(OH)3 <−−> Fe3+ + 3(OH)­
Na3P <­­> 3Na+ + P3­
PbCl2 <−−> Pb2+ + 2Cl­
Calculating Solubility from the Ksp
Calculating Solubility from the Ksp
a) pure water
CaF2 <−−> Ca2+ + 2F­ If we assume x as the dissociation then, Ca2+ ions = x
and [F­] = 2x Sample Problem
Sample Problem
Calculate the solubility of CaF
2 in grams per liter in The molar solubility of lead (II) bromide, PbBr
2 a) pure water
is 1.0 x 10­2 at 25oC. Calculate the solubility b) a 0.15 M KF solution
product, Ksp, for this compound.
c) a 0.080 M Ca(NO3)2 solution
Ksp = [Ca2+] [F­]2
The solubility product for calcium fluoride, CaF2 is 3.9 x 10­11
So x = 2.13 x 10­4 mol/L x (78 g/mol CaF2)
= (x)(2x)2 Ksp = 3.9 x 10­11 = 4x3 Solubility is 0.0167 g/L
2.13 x10­4 Ca 2+ mol/L x 1mol/L CaF2 78g/L
­­­­­­­­­­­­­­­­­­­­­ x ­­­­­­­­­­­­­
1mol/L Ca2+ ions 1 mol CaF2
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Calculating Solubility from the Ksp
Calculating Solubility from the Ksp
Calculate the solubility of CaF2 in grams per liter in c) a 0.080 M Ca(NO3)2 solution
[Ca2+ ] = 0.08M
b) a 0.15 M KF solution
remember KF is a strong electrolyte, is completely ionized. the major source of F­ ions, then [F­] =0.15M
The solubility product for calcium fluoride,CaF2 is 3.9 x 10­11
The solubility product for calcium fluoride, CaF2 is 3.9 x 10­11
CaF2 (s) <­­­>
[ F­] = 0.15M Ksp = [Ca2+] [F­]2
Ca2+ (aq)
+
2 F­ (aq)
Ksp = [Ca2+] [F­]2 = (0.080)(x)2 = (x)(0.15)2 Ksp = 3.9 x 10­11 = 0.0225x Ksp = 3.9 x 10­11 = 0.080x2 So x = ______ mol/L So x = 2.2 x 10­5 mol/L * (78 g/mol CaF2)/ 2
Solubility is = ______ x (78 g/mol CaF2) = ______ g/L
Solubility is 0.000858 g/L
1.73 x10­9 Ca 2+ mol/L x 1mol/L CaF2 78g/L
­­­­­­­­­­­­­­­­­­­­­ x ­­­­­­­­­­­­­
1mol/L Ca2+ ions 1 mol CaF2
Calculating Solubility from the Ksp
12 Calculate the concentration of silver ion when the solubility product constant of AgI is 10­16.
Recall from the Common­Ion Effect that adding a strong electrolyte to a weakly soluble solution with a common ion will decrease the solubility of the weak electrolyte.
A 0.5 (1 x 10­16)
Compare the solubilities from the previous Sample Problem
CaF2 (s) <­­­> Ca2+ (aq) + 2 F­ (aq)
CaF2 dissolved with: Solubility of CaF2 pure water 0.016 g/L B
2 (1 x 10­16)
C
(1 x 10­16)2
D
(1 x 10­16)
0.015 M KF 0.080 M Ca(NO3)2 0.0017 g/L These results support Le Chatelier's Principle that increasing a product concentration will shift equilibrium to the left.
13 The Ksp of a compound of formula AB3 is 1.8 x 10­18. The molar solubility of the compound is ­­­­
­
The Ksp of a compound of formula AB3 is 1.8 x 10­18. The solubility of the compound is ­­­­
The molar mass is 210g/mol
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­
The concentration of hydroxide ions in a saturated solution of Al(OH)3 is 1.58 x10­15. The concentration of Al 3+ is ­­­­­­­
Factors Affecting Solubility
• Recall The Common­Ion Effect
• If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
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­
The concentration of hydroxide ions in a saturated solution of Al(OH)3 is 1.58 x10­15. The ksp of Al (OH)3 is­­­­­
Factors Affecting Solubility
• The Common­Ion Effect applies when a basic anion is involved, such as conjugate base of a weak acid.
• Consider the dissociation of the salt calcium fluoride:
CaF2 <­­> ______ + ______
BaSO4(s) ↔ Ba2+(aq) + SO42­(aq)
• So adding any soluble salt containing either Ba2+ or SO42­ ions will decrease the solubility of barium sulfate.
Factors Affecting Solubility
Write the dissociation equation for magnesium hydroxide:
Suppose you have a saturated solution of magnesium hydroxide in a flask. What will happen if you add a small amount of strong acid to it? (Think Le Châtelier’s Principle.) What do you expect will happen to equilibrium point if the pH of this system is lowered by adding a strong acid? Factors Affecting Solubility
Main idea: Adding an acid, or otherwise lowering the pH of a solution will _______________________ the solubility of a salt containing a ___________________________________.
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Factors Affecting Solubility
• pH
• If a substance has a basic anion, it will be more soluble in an acidic solution.
• Substances with acidic cations are more soluble in basic solutions.
Factors Affecting Solubility
• Complex Ions
• Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
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14 Given the system at equilibrium
AgCl (s) <­­> Ag+ (aq) + Cl­ (aq)
When 0.01M HCl is added to the sytem, the point of equilibrium will shift to the ________.
A
right and the concentration of Ag+ will decrese
B
right and the concentration of Ag+ will increase
C
left and the concentration of Ag+ will decrease
D
left and the concentration of Ag+ will increase
Factors Affecting Solubility
• Complex Ions
• The formation of these complex ions increases the solubility of these salts.
Factors Affecting Solubility
Precipitation Problems
• Amphoterism
• Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
• Examples of such cations are Al3+, Zn2+, and Sn2+.
Solubility rules to memmorize by this time. Will help you to identify which will precipitate as you mix two solutions.
• Any salt made with a Group I metal is soluble.
• All salts containing nitrate ion are soluble.
• All salts containing ammonium ion are soluble. 7
solubility­presentation­2012­03­31.notebook
15 What is the name of the solid precipitate that is formed when a solution of sodium chloride is mixed with a solution of silver nitrate?
16 What is the name of the solid precipitate that is formed when a solution of potassium carbonate is mixed with a solution of calcium bromide?
A potassium bromide
B calcium carbonate
C
A sodium silver
B
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potassium calcium
D carbonate bromide
sodium nitrate
E Not enough information
C chloride nitrate
D silver chloride
E
Not enough information
17 What is the name of the solid precipitate that is formed when a solution of lead (IV) nitrate is mixed with a solution of magnesium sulfate?
A
PbSO4
Precipitation Problems
Most of the problems in this section ask you whether a precipitate will form after mixing certain solutions together.
General Problem­Solving Strategy
Step 1 ­ Determine which of the products is the precipitate. Write the Ksp expression for this compound.
B Pb(SO4)2
C Pb2SO4
Step 2 ­ Calculate the cation concentration of this slightly soluble compound. D Mg(NO3)2
E Not enough information
Step 3 ­ Calculate the anion concentration of this slightly soluble compound. Step 4 ­ Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K.
Step 5 ­ Compare Q to K to determine whether a precipitate will form.
Precipitation Problems
Analyzing quantitative results:
If Q = Ksp
If Q > Ksp
then you have an exactly perfect saturated solution with not one speck of undissolved solid.
then YES you will then NO precipitate observe a precipitate; will form; there are so the number of cations few cations and and anions exceeds anions that they all the solubility
remain dissolved
18 The Ksp for Zn(OH)2 is 5.0 x10­17. Will a precipitate form in a solution whose solubility is 8.0x10­2 mol/L Zn(OH)2?
If Q < Ksp
A yes, because Qsp < Ksp
B yes, because Qsp > Ksp
C no, because Qsp = Ksp
D no, because Qsp < Ksp
E no, because Qsp > Ksp
In a solution,
• If Q = Ksp, the system is at equilibrium and the solution is saturated.
• If Q > Ksp, the salt will precipitate until Q = Ksp.
• If Q < Ksp, more solid can dissolve until Q = Ksp.
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Precipitation Problems
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Sample Problem ­ Answers
Will a precipitate form if you mix 50.0 mL of 0.20 M barium chloride, BaCl2, and 50.0 mL of 0.30 M sodium sulfate, Na2SO4?
Step 1 ­ Determine which of the products is the precipitate. Write the Ksp expression for this compound.
Sample Problem
Will a precipitate form if you mix 50.0 mL of 0.20 M barium chloride, BaCl2, and 50.0 mL of 0.30 M sodium sulfate, Na2SO4?
BaSO4 (s) <­­> Ba2+ (aq) + SO42­ (aq)
Step 2 ­ Calculate the cation concentration of this slightly soluble compound. M1V1 =M2V2
• Qualitatively, you know that ____________ will be the solid that would theoretically form. • Quantitatively, however, the question is, are there enough _________________ ions and ____________________ ions to produce a solid precipitate, or will there be so few of them that they will all dissolve?
M2 = (M1V1) / V2 M2= (0.20M*50.0mL) / 100 mL M2 = 0.10 M BaCl2 [Ba2+] = 0.10 M Step 3 ­ Calculate the anion concentration of this slightly soluble compound. M1V1 =M2V2
M2 = (M1V1) / V2 M2= (0.30M*50.0mL) / 100 mL M2 = 0.15 M Na2SO4 [SO42­] = 0.15 M 19 The Ksp for zinc carbonate is 1 x 10­10.
If equivalent amounts of 0.1M sodium carbonate and 0.1M zinc nitrate are mixed, what happens?
Sample Problem ­ Answers (con't)
Will a precipitate form if you mix 50.0 mL of 0.20 M barium chloride, BaCl2, and 50.0 mL of 0.30 M sodium sulfate, Na2SO4?
A A zinc carbonate precipitate forms, since Q>K.
B
A zinc carbonate precipitate forms, since Q<K. C A sodium nitrate precipitate forms, since Q>K.
D No precipitate forms, since Q=K. Step 4 ­ Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K.
Ksp = [Ba2+] [SO42­] = (0.10) (0.15) = 0.015
Step 5 ­ Compare Q to K to determine whether a precipitate will form.
The Ksp for barium sulfate is 1 x 10­10.
Therefore, since Q > K, there will be a precipitate formed when you mix equal amounts of 0.20 M BaCl2, and 0.30 M Na2SO4.
20 Which of the following factors affect solubility?
A
pH
B
Concentration
C
Common­Ion Effect D
A and C
E
A, B, and C
Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
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solubility­presentation­2012­03­31.notebook
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HW question
http://www.ktf­split.hr/periodni/en/abc/kpt.html
To a solution containing 2.0 x 10­5 M Barium ions and 1.8 x 10­4 M lead ions, Na2CrO4 is added. Which would precipitate first from this solution?
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