Download Chp16,Solubility_Calc

Chang, 6th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2001
Predicting Solubility
Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a
slightly soluble salt and the equilibrium constant for the reaction is the solubility product
constant, Ksp. Note that since the reactant is a solid, its concentration does not appear in the Ksp
expression. For example, the solubility equilibrium and Ksp for the salt SrF2(s) are
SrF2(s)  Sr2+(aq) + 2 F – (aq)
Ksp = [Sr2+][F –]2 = 2.0 x 10-10
The molar solubility of a salt in water can be formed by setting up an equilibrium table and
solving for x. The solubility is the same quantity expressed in g/L (rather than M = mol/L).
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Example 1: Calculate (a) the molar solubility and (b) the solubility of SrF2(s) in water.
Solution: We first set up an equilibrium table:
Balanced Equation
Initial Concentrations (M)
Change (M)
Equilibrium Concentrations (M)
SrF2(s) 
co
-x
(co - x)
Sr2+(aq) +
0
x
x
2 F – (aq)
0
2x
2x
To determine the solubility we use the equilibrium concentrations from the table and the Ksp
value given above, and solve for x:
Ksp = [Sr2+][F –]2 = (x)( 2x)2 = 4x3 = 2.0 x 10-10
x3 = 5.0 x 10-11
x = 3.7 x 10-4 = molar solubility
To find the solubility we use the molar mass to convert the molar solubility to g/L:
3.7  10 4 mol SrF 2  108.62 g SrF2 
  4.0  10 2 SrF2 /L
? g/ L  
 

  1 mol SrF 2 
1L
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Predicting Precipitation
Away from equilibrium, the ion product Qsp may be defined. As always, Q has the same
form as K, but typically involves non-equilibrium concentrations. Precipitation can be predicted
by comparing Qsp with Ksp:
Relationship
Qsp < Ksp
Qsp = Ksp
Qsp > Ksp
Solution Type
Unsaturated
Saturated
Supersaturated
Result
More salt can dissolve without ppt forming
No more salt can dissolve
Salt will precipitate until Qsp = Ksp
Chemical Separations
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Chang, 6th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2001
If two precipitates are possible, the least soluble salt will precipitate first (the one with the
smaller Ksp). For example, if you add NaOH(aq) to a solution containing equal amounts of Ca2+
and Mg2+, Mg(OH)2(s) will precipitate before Ca(OH)2(s) since Ksp(Mg(OH)2) = 1.2 x 10-11 is
less than Ksp(Ca(OH)2) = 8.0 x 10-6. The reason is that for Mg(OH)2, Qsp > Ksp will occur at
lower [OH-] since the Ksp is smaller. If the Ksp values for two salts differ widely, the salts may
be separated by fractional precipitation.
Reduction in Solubility due to the Common Ion Effect
Addition of an ion “common” to a solubility equilibrium will reduce solubility. This can be
predicted in a qualitative way using Le Chatelier’s Principle. For example, adding fluoride ion,
F–(aq), to the SrF2(s) equilibrium above will shift it left. The shift will increase the amount of
SrF2(s) in solid form, and thus decrease solubility.
The new solubility can be calculated as illustrated in the following example.
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Example 2. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?
Again we set up an equilibrium table, but now we have an initial concentration of fluoride ion:
Balanced Equation
Initial Concentrations (M)
Change (M)
Equilibrium Concentrations (M)
SrF2(s) 
co
-x
(co - x)
Sr2+(aq) +
0
x
x
2 F – (aq)
0.10
2x
(0.10 + 2x)
We next use the equilibrium concentrations in the table and the Ksp value given above, and solve
for x. Note that since x is small, (0.10 + 2x)  0.10.
Ksp = [Sr2+][F –]2 = (x)( 0.10 + 2x)2  (x)( 0.10)2 = 0.010 x
x = 2.0 x 10-8 = molar solubility
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Note that as predicted by the common ion effect, this solubility is much lower than what we
calculated in Example 1 for pure water!
Increase in Solubility due to addition of a Species that Reacts with an Ion
The solubility of a salt will increase if a species is added which reacts with one of its ions.
Once again this is an example of shifts predicted by Le Chatelier’s Principle. For example if the
anion reacts with the added substance, the concentration of the anion will be reduced. Thus the
equilibrium will shift right to “undo, in part, the disturbance”. The shift right will reduce the
amount of salt in solid form, and thereby increase its solubility. There are two common
examples of this phenomena:
(1) Reaction of the basic anion of a salt with a strong acid. Salts such as Mg(OH)2(s), BaCO3(s),
and NiS(s) are much more soluble in a strong acid solution than in water. On the other hand,
the solubility of AgCl(s) does not increase when 6 M HNO3 is added.
(2) Reaction of the acidic cation (Lewis acid) of a salt with a Lewis base to form a soluble
complex ion. Salts such as AgCl(s) and Cu(OH)2(s) are much more soluble in 6 M NH3(aq)
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Chang, 6th Edition, Chapter 16, Worksheet #1
S. B. Piepho, Fall 2001
solution than in pure water due to the formation of Ag(NH3)2+(aq) and Cu(NH3)42+(aq)
complex ions respectively. Likewise, amphoteric hydroxides such as Al(OH)3(s) and
Zn(OH)2(s) are soluble in 6 M NaOH(aq) due to the formation of Al(OH)4-(aq) and
Zn(OH)42-(aq) ions.
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