Solubility Equilibrium
When a substance dissolves an equilibrium results
between the precipitate and the dissolved ions. The
solution becomes saturated. The particles dissolving
equals the particles precipitating.
1
:
2
Ba2+ (aq) + SO42- (aq)
BaSO4 (s)
When writing the equilibrium-constant expression for the dissolution of BaSO4, we
remember that the concentration of a solid is constant. The expression is therefore:
K = [Ba2+][SO42-]
K = Ksp, the solubility-product constant.
Ksp = [Ba2+][SO42-]
The relative solubilities can be deduced by comparing values of Ksp but only for salts
that have the same ion:ion ratio. Solubility changes with temperature. Some substances
become less soluble in cold while some become more soluble.
The molar solubility is the number of moles of the
salt that dissolves in a liter of water. The ice table for
the above reaction nis:
BaSO4 (s)
I
Ba2+ (aq) + SO42- (aq
0
0
3
C
E
+x
x
+x
x
X is the molar solubility of the salt in this case
Barium sulfate
Ksp = x2, x2 = 1.08 x 10-10, x = 1.03 x 10-5 M
Calculate the molar solubility for lead(II)iodide
Exercise 1
Copper(I) bromide has a measured molar solubility of 2.0 x 10-4 mol/L at 25
o
C. Calculate its Ksp value.
Exercise 2
Calculate the Ksp value for bismuth sulfide (Bi2S3) which has a molar
solubility of 1.0 x 10-15 mol/L at 25oC.
Exercise 3
The Ksp value for copper(II) iodate, Cu(IO3)2 is 1.4 x 10-7 at 25oC. Calculate
its solubility at 25oC.
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When determining relative solubility’s for salts the
equilibrium constants can be compared along with
the number of ions that are produced in solution.
Solubility and the common ion effect
A common ion is any ion in the solution that is common to
the ionic compound being dissolved.
The solubility of an ionic compound decreases in the
presence of a common ion because of LeChatelier’s
Principle.
For example, the chloride ion in a sodium chloride solution is common to the chloride in
lead(II) chloride. The presence of a common ion must be taken into account when
determining the solubility of an ionic compound. To do this, simply use the concentration
of the common ion as the initial concentration.
Example: Estimate the solubility of barium sulfate in a 0.020 M sodium sulfate solution.
The solubility product constant for barium sulfate is 1.1 x 10-10.
•
Write the equation and the equilibrium expression for the dissolving of barium
sulfate.
BaSO4(s) --> Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-]
•
Make an "ICE" chart.
5
Let "x" represent the barium sulfate that dissolves in the sodium sulfate solution
expressed in moles per liter.
Initial Concentration
Change in Concentration
Equilibrium Concentration
•
BaSO4(s)
Ba2+(aq)
SO42-(aq)
All solid
0
0.020 M (from Na2SO4)
- x dissolves
+x
+x
Less solid
x
0.020 M + x
Substitute into the equilibrium expression and solve for x. We will make the
assumption that since x is going to be very small (the solubility is reduced in the
presence of a common ion), the term "0.020 + x" is the same as "0.020." (You
can leave x in the term and use the quadratic equation or the method of successive
approximations to solve for x, but it will not improve the significance of your
answer.)
1.1 x 10-10 = [x][0.020 + x] = [x][0.020]
x = 5.5 x 10-9 M
Excersise 4
Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a .025 M NaF
solutions.
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Ksp and the reaction quotient, Q
The reaction quotient can decide whether a precipitate will
form
When :
Q < Ksp, the system is not at equilibrium
(unsaturated)
Q = Ksp, the system is at equilibrium
(unsaturated)
Q > Ksp, the system is not at equilibrium
(supersaturated)
Precipitates form when the solution is supersaturated
Example: 25.0 mL of 0.0020 M potassium chromate are mixed with 75.0 mL of
0.000125 M lead(II) nitrate. Will a precipitate of lead(II) chromate form. Ksp of lead(II)
chromate is 1.8 x 10-14.
•
First, determine the overall and the net-ionic equations for the reaction that occurs
when the two solutions are mixed.
K2CrO4(aq) + Pb(NO3)2(aq) --> 2 KNO3(aq) + PbCrO4(s)
Pb2+(aq) + CrO42-(aq) --> PbCrO4(s)
The latter reaction can be written in terms of Ksp as:
PbCrO4(s) --> Pb2+(aq) + CrO42-(aq)
Ksp = [Pb2+][CrO42-]
•
Using the dilution equation, C1V1 = C2V2, determine the initial concentration of
each species once mixed (before any reaction takes place).
(0.0020 M K2CrO4)(25.0 mL) = (C2)(100.0 mL)
C2 for K2CrO4 = 0.00050 M
Similar calculation for the lead(II) nitrate yields:
C2 for Pb(NO3)2 = 0.0000938 M
•
Using the initial concentrations, calculate the reaction quotient Q, and compare to
the value of the equilibrium constant, Ksp.
7
Q = (0.0000938 M Pb2+)(0.00050 M CrO42-) = 4.69 x 10-8
Q is greater than Ksp so a precipitate of lead(II) chromate will form.
Exercise 5
A solution is prepared by adding 750.0 ml of 4.00 x 10-3 M Ce(NO3)3 to
300.0 ml of 2.00 x 10-2 M KIO3. Will Ce(IO3)2 (Ksp = 1.0 x 10-10)
precipitate from this solution?
Exercise 6
A solution is prepared by mixing 150.0 ml of 1.00 x 10-2 M Mg(NO3)2 and
250 ml of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at
equilibrium with solid MgF2 (Ksp = 6.4 x 10-9)
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pH and solubility
The pH of a solution will affect a salt’s solubility. If
the ion is a strong acid. If an anion is a strong base
the solubility in acidic solution will be greater.
If HX is a weak acid then X- is a strong base and the
solubility of a salt of X will be higher.
Predicting whether a precipitate will form when two
solutions are mixed.
Ion product = Q
Q = [A+][B-]
The initial concentrations of A+ and B- are used
If Q is greater than Ksp, precipitation occurs.
If Q is less than Ksp, no precipitation occurs
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A solution is prepared by adding 750.0 ml of
4.00 x 10-3 M Ce(NO3)3 to 300.0 ml of 2.00 x 10-2 M KIO3, will Ce(IO3)3
(Ksp = 1.9 x 10-10) precipitate from this solution?
A solution contains 1.0 x 10-4 M Cu+ and 2.0 M Pb2+. If a source of I- is
added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3
x 10-12) precipitate first?
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Qualitative analysis
A qualitative analysis scheme introduces you to the basic
chemistry of various ions
Objective: Separate the following metal ions: silver, lead,
cadmium and nickel
1. Adding HCl causes silver and lead to precipitate
2. Separate nickel and cadmium by filitration
3. Add hot water to the precipitates and filter while
hot and the lead (II) chloride will redissolve
4. Separate the chloride precipitates by filtriation
5. Separating Cd and Ni is more subtle. Use their
sulfide Ksp values to determine which sulfide
precipitates first.
Exercise 7
A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of Iis added gradually to this solution will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp =
5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to
begin precipitation of each salt.
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