Chem 30S Review…Solubility Rules
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Solubility Equilibria
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Salts are generally more soluble in HOT water(Gases
are more soluble in COLD water)
Alkali Metal salts are very soluble in water.
NaCl, KOH, Li3PO4, Na2SO4 etc...
Ammonium salts are very soluble in water.
NH4Br, (NH4)2CO3 etc…
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Salts containing the nitrate ion, NO3-, are very soluble
in water.
Most salts of Cl-, Br- and I- are very soluble in water exceptions are salts containing Ag+ and Pb2+.
soluble salts: FeCl2, AlBr3, MgI2 etc...
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“insoluble” salts: AgCl, PbBr2 etc...
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Dissolving a salt...
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A salt is an ionic compound usually a metal cation bonded
to a non-metal anion.
The dissolving of a salt is an
example of equilibrium.
The cations and anions are
attracted to each other in the
salt.
They are also attracted to the
water molecules.
The water molecules will start
to pull out some of the ions
from the salt crystal.
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At first, the only process
occurring is the dissolving of
the salt - the dissociation of
the salt into its ions.
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However, soon the ions floating
in the water begin to collide with
the salt crystal and are “pulled
back in” to the salt.
(crystalization)
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Eventually the rate of
dissociation is equal to the rate
of crystalization.
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The solution is now
“saturated”. It has reached
equilibrium.
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Solubility Equilibrium:
Dissociation = Crystalization
Na+ and Cl - ions
surrounded by
water molecules
Dissolving silver sulfate, Ag2SO4, in H2O
✦ When silver sulfate dissolves it dissociates into ions.
In a saturated solution,
there is no change in
amount of solid
precipitate at the
bottom of the beaker.
When the solution is saturated, the following
equilibrium exists:
Ag2SO4 (s) ⇔ 2 Ag+ (aq) + SO42- (aq)
✦ Since this is an equilibrium, we can write an
Concentration of the
solution is constant.
NaCl Crystal
Dissolving NaCl in water
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equilibrium expression for the reaction:
The rate at which the
salt is dissolving into
solution equals the rate
of crystalization.
Ksp = [Ag+]2 [SO42-]
✦ Ksp (solubility product constant)
Notice that the Ag2SO4 is left out of the expression! Why?
Ag2SO4 is a solid and not included in the expression.
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Some Ksp Values
Writing solubility product expressions...
For each salt below, write a balanced equation
showing its dissociation in water.
Then write the Ksp expression for the salt.
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Iron (III) hydroxide, Fe(OH)3
Nickel sulfide, NiS
Silver chromate, Ag2CrO4
Zinc carbonate, ZnCO3
Note:
These are experimentally determined, and may
be slightly different on a different Ksp table.
Calcium fluoride, CaF2
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Calculate Solubility from Ksp
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Calculate Solubility from Ksp
Ksp for AgCl = 1.8 x 10-10 , what is the
molar solubility in pure water?
✦ AgCl (s) ⇔ Ag+ (aq) + Cl- (aq)
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1.8 x 10-10 = x2
✦ Square root both sides
✦ 1.3 x 10-5 = x = [Ag+] and [Cl-]
✦ Applying the concentration values to
the balanced solubility equation (1:1:1
ratio) indicates the molar solubility is
1.3 x 10-5 mol/L
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Ksp = [Ag+] [Cl-]
✦ At equilibrium the concentrations of Ag+
and Cl- are equal
✦ let x = [Ag+] and also [Cl-]
✦ 1.8 x 10-10 = (x) (x)
✦ 1.8 x 10-10 = x2
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Calculate Solubility from Ksp 2
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Calculate Solubility from Ksp 2
Lead iodide (PbI2) has a Ksp = 7.9 x 10-9
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What is the molar solubility of PbI2 in water?
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PbI2 (s) ⇔
Pb2+
[Pb2+]
(aq)
+2
I-
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(aq)
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Ksp =
let x = [Pb2+] and let 2x = [I-]
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7.9 x 10- 9 = (x) (2x)2
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7.9 x 10- 9 = (x) (4x2)
7.9 x 10-9 = 4x3
1.3 x 10- 3 = x = [Pb2+]
[I-]2
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Applying the concentration values to the
balanced solubility equation (1:1:2 ratio)
indicates the molar solubility is
1.3 x 10-3 mol/L
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Algebraic form of Ksp
Calculate Ksp given Solubility
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For “salts” the algebraic form of the Ksp
are …
AB: (x) (x) = x2 = Ksp
e.g. AgCl
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AB2: (x) (2x) = 4x = Ksp
e.g. PbI2
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AB3: (x) (3x)3 = 27x4 = Ksp
e.g. AlBr3
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A2B3:
e.g. Au2O3
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The solubility of Ag2CrO4 is 6.7 x 10-5 mol/L
What is the Ksp?
Ag2CrO4 (s) ⇌ 2 Ag+ (aq) + CrO42− (aq)
Ksp = [Ag+]2 [CrO42−]
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The equilibrium concentrations of the ions are
determined from the solubility of Ag2CrO4
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The ratios of the coefficients from the balanced
solubility equation are used.
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Calculate Ksp given Solubility
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Calculate Ksp given Solubility
e.g. 1 Ag2CrO4 : 2 Ag+
✦ there is a 1:2 ratio,
✦ if 6.7 x 10-5 mol/L is the Ag2CrO4 solubility
the [Ag+]eq = 2 ( 6.7 x 10-5) = 1.3 x 10-4 M
✦ e.g. 1 Ag2CrO4 : 1 CrO42−
✦ there is a 1:1 ratio, if 6.7 x 10-5 mol/L is the
Ag2CrO4 solubility
the [CrO42-]eq= 1( 6.7 x 10-5) = 6.7 x10-5 M
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Apply the equilibrium values to the Ksp
Ksp = [Ag+]2 [CrO42−]
= (1.3 x 10-4)2 (6.7 x 10-5)
= 1.1 x 10-12
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Calculate Ksp given Solubility
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Ag2CrO4
(s)
More Calculate Ksp given Solubility
⇔ 2 Ag+ (aq) + CrO42− (aq)
s
solubility
2s
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s
Solubility 3.42 x 10-7
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Cu(OH)2
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Answer: Ksp = 1.6 x 10-19
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Fe(OH)3
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Answer: Ksp = 2.6 x 10-39
(s)
equilm conc.
Ksp = [Ag+]2 [CrO42−] = (2s)2(s) = (4s2)(s) = 4s3
(s)
Solubility 9.9 x 10-11
s = 6.7 x 10-5 mol/L
Ksp = 4(6.7 x 10-5 )3 = 1.1 x 10-12
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Ksp and Solubility
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Generally, salts with very small solubility product
constants (Ksp) are only slightly soluble in water.
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When comparing the solubilities of two salts, you can
sometimes simply compare the relative sizes of their
Ksp values.
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This works if the salts have the same ratio of ions!
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For example… CuI has Ksp = 5.0 x 10-12 and CaSO4
has Ksp = 6.1 x 10-5. Since the Ksp for calcium
sulfate is larger than that for the copper (I) iodide, we
can say that calcium sulfate is more soluble.
But be careful...
Do you see the “problem” here??
can’t compare Ksp to predict solubility for
compounds with different ratios
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Mixing Solutions - Will a Precipitate Form?
p. 581
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Pb(NO3)2 (aq) + K2CrO4 (aq) ⇔ PbCrO4 (s) + 2 KNO3 (aq)
Step 1: Is an insoluble salt formed?
We can see that a double replacement reaction can
occur and produce PbCrO4. Since this salt has a low
solubility (very small Ksp), it may precipitate from the
mixture. The solubility equilibrium is:
If 15 mL of 0.024-M lead (II) nitrate is mixed with 30 mL of
0.030-M potassium chromate - will a precipitate form?
Pb(NO3)2 (aq) + K2CrO4 (aq) ⇔ PbCrO4 (s) + 2 KNO3 (aq)
PbCrO4 (s) ⇔ Pb2+ (aq) + CrO42- (aq)
Ksp = 2 x 10-16 = [Pb2+][CrO42-]
If a precipitate forms, it means the amount of ions in
solution is greater than the equilibrium concentrations
This will happen only if product of the ion concentrations
(Qsp) > Ksp in our mixture.
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Step 2: Find the concentrations of the ions that form the
insoluble salt.
Since we are mixing two solutions in this example,
the concentrations of the Pb2+ and CrO42- will be
diluted. We have to do a dilution calculation!
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Step 3: Calculate IP (or Qsp) for the mixture.
ion product (IP or Qsp) – the product of the molar conc.
in an ionic solution (raised to an appropriate power)
PbCrO4 (s) ⇔ Pb2+ (aq) + CrO42- (aq)
Dilution: M1V1 = M2V2
Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M)
Pb(NO3)2 : Pb2+ and the K2CrO4 : CrO42- ratios are 1:1
therefore:
Qsp = 1.6 x 10-4
[Pb2+] = M1V1 = (0.024M)(15 mL) = 0.0080 Pb2+
V2
(45 mL)
[CrO42-] = M1V1 = (0.030M)(30 mL) = 0.020 CrO42V2
(45 mL)
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Step 4: Compare Qsp to Ksp.
Qsp = 1.6 x 10-4 > Ksp = 2 x 10-16
Summary
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Since Qsp > Ksp, a precipitate will form when the two
solutions are mixed!
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Note: If Qsp = Ksp, the solution is saturated
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If Qsp < Ksp, the solution is unsaturated
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How do you know if a ppt forms?
ion product > Ksp
ppt. forms
ion product = Ksp
ion product < Ksp
ppt. will
not form
Either way, no ppt will form!
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Another ppt. problem with
DILUTION!
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CaCl2 (aq) + Na2SO4 (aq) ⇔ CaSO4 (s) + 2 NaCl (aq)
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Step 1: Calc. The new ion concentrations using the
new volume.
In dilution problems we use: Mi Vi = Mf Vf
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e.g. 50mL of 0.0010M CaCl2 and 50mL of 0.010 M
Na2SO4 were mixed. Will a precipitate of CaSO4 be
formed? Ksp of CaSO4 = 2.4 x 10-5
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The new volume is the combination of the two
solutions i.e. 50 mL + 50 mL = 100 mL
When calculating concentrations volume must be in
litres (L) therefore we have 0.100 L
The new (final) concentration is calculated as
follows:
[CaCl2] = M i Vi = (0.0010)(0.050)= 5.0 x 10-4 M
Vf
0.100
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CaCl2 ⇔ Ca2+ + 2 Cltherefore [Ca2+] = 5.0 x 10-4 M (1:1 ratio)
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[NaSO4] =
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Na2SO4 ⇔ 2 Na+ + SO42therefore [SO42-] = 5.0 x 10-3 M (1:1 ratio)
M i Vi
=
Vf
(0.010)(0.050)
= 5.0 x 10-3M
0.100
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Step 2: calculate ion product
We are looking for the formation of CaSO4
precipitate therefore we use the following:
CaSO4 (s) ⇔ Ca2+ (aq) + SO42- (aq)
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The Common Ion Effect
The presence of a common ion in a solution
will lower the solubility of a salt.
shift left
↑
AgCl (s) ⇔ Ag+ (aq) + Cl- (aq)
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ion product = [Ca2+] [SO42-]
ion product = (5.0 x 10-4) (5.0 x 10-3)
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= 2.5 x 10-6
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The Ksp for CaSO4 is 2.4 x
ion product
2.5 x 10-6
<
Ksp
2.4 x 10-5
(adding NaCl)
LeChatelier’s Principle:
The addition of the common ion will shift the
solubility equilibrium to the left. This means
that salt precipitates in the solution and the
solubility is lowered.
e.g. the addition of NaCl to a AgCl
equilibrium lowers the solubility of the
AgCl ( the common ion is Cl-)
10-5
therefore no
precipitate!
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Common Ion Calc
The Common Ion Effect on Solubility
The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L.
What happens to the solubility if we dissolve the MgF2 in a
solution of NaF, instead of pure water?
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Estimate the solubility of barium sulfate in a
0.020 M sodium sulfate solution. The Ksp 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.
Let "x" represent the barium sulfate that
dissolves in the sodium sulfate solution
expressed in moles per liter.
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Common Ion Calc
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Common Ion Calc
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BaSO4(s)
⇌
Ba2+(aq)
✦
SO42-(aq)
0.020 M
I
All solid
0
C
- x dissolves
+x
+x
E
Less solid
x
0.020 M + x
(from Na2SO4)
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Common Ion Calc
Calculate molar solubility if Ca(IO3)2 in 0.060 mol/L
NaIO3. Ksp Ca(IO3)2 = 7.1 x 10-7
Ans: 2.0 x 10-4 mol/L
Calculate the molar solubility of silver chloride in a
0.20 M sodium chloride solution.
Ksp AgCl = 1.6 x 10-10
Ans: 8.0 x 10 -10 mol/L
Calc the molar solublity of silver chloride in a 1.5 x
10-3 mol/L silver nitrate solution.
Ksp AgCl = 1.6 x 10-10
Ans: 1.3 x 10-3 mol/L
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Substitute into the equilibrium expression and
solve for x.
The assumption is 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 mol/L Reduced Solubility BaSO4
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