Solubility

Types of Solution
Solution – homogeneous mixture of two or more
substances of ions or molecules. E.g. NaCl (aq)
1) Solvent = component which is the component in
greater amount.
2) Solute = component which is present in the
smaller amount.
– Gaseous = gases are completely miscible in each other.
– Liquid = gas, liquid or solid solute dissolved in solute.
– Solid = mixture of two solids that are miscible in each other
to form a single phase.


Colloid – appears to be a homogeneous mixture, but
particles are much bigger, but not filterable. E.g. Fog,
smoke, whipped cream, mayonnaise, etc.
Suspension: larger particle sizes, filterable. E.g. mud,
freshly
squeezed orange juice.
Chapter 12-2
John A. Schreifels
Chemistry 212
Dissolving a salt...





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.

At first, the only process
occurring is the dissolving of the
salt - the dissociation of the
salt into its ions.

However, soon the ions floating
in the water begin to collide with
the salt crystal and are “pulled
back in” to the salt.
(precipitation)

Eventually the rate of
dissociation is equal to the rate
of precipitation.

The solution is now “saturated”.
It has reached equilibrium.
Solubility Equilibrium:
Dissociation = Precipitation
Na+ and Cl - ions
surrounded by
water molecules
In a saturated solution,
there is no change in
amount of solid
precipitate at the
bottom of the beaker.
Concentration of the
solution is constant.
NaCl Crystal
Dissolving NaCl in water
An electrolyte is a substance that, when
dissolved in water, results in a solution that can
conduct electricity.
A nonelectrolyte is a substance that, when
dissolved, results in a solution that does not
conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
4.1
Precipitation Reactions


Mix two aqueous
solutions made by
dissolving ionic
compounds in
water.
If a reaction
happens, a
precipitate (solid) is
formed.
Predicting Products of
Precipitation Reactions
1)
2)
3)
4)
Ionic Compounds are Strong
Electrolytes –Determine charge on all
ions of reactants
Using Ion Charges; Predict formula of
products. ( + ion of one reactant forms
compound with – ion of other reactant)
Balance Equation
Determine is product is solid or
aqueous solution
Solubility Rules for Common Ionic Compounds
In water at 250C
Soluble Compounds
Exceptions
Compounds containing alkali
metal ions and NH4+
THESE ARE
INSOLUABLE
NO3-, HCO3-, ClO3Cl-, Br-, ISO4
2-
Halides of Ag+, Hg22+, Pb2+
Sulfates of Ag+, Ca2+, Sr2+, Ba2+,
Hg2+, Pb2+
Insoluble Compounds
Exceptions
CO32-, PO43-, CrO42-, S2-
Compounds containing alkali
metal ions and NH4+
OH-
Compounds containing alkali
metal ions and Ba2+
4.2
Predicting Products of
Precipitation Reactions (Cont)
5)
6)
Determine spectator ions (Ions that
are still dissolved in water in the
product)
Write net ionic equation (Only shows
ions involved in forming solid)
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
precipitate
Pb(NO3)2 (aq) + 2NaI (aq)
PbI2 (s) + 2NaNO3 (aq)
molecular equation
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2 (s) + 2Na+ + 2NO3-
ionic equation
Pb2+ + 2IPbI2
PbI2 (s)
net ionic equation
Na+ and NO3- are spectator ions
4.2
Write the net ionic equation for the reaction of silver
nitrate with sodium chloride.
AgNO3 (aq) + NaCl (aq)
Ag+ + NO3- + Na+ + ClAg+ + Cl-
AgCl (s) + NaNO3 (aq)
AgCl (s) + Na+ + NO3AgCl (s)
4.2
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon
dioxide gas
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
4.3
Acids


Produce H+ (proton) or (H3O)+ when
dissolved in water
Proton donor
H2O
HNO3 (aq)  H+ (aq) + (NO3)- (aq)
HNO3 (aq) + H2O (l)  H3O+ (aq) + (NO3)- (aq)
Monoprotic acids; Produce one H+ when dissolved
in water
HNO3
H+ + NO3-
Strong electrolyte, strong acid
Diprotic acids; Produce two H+ when dissolved
in water
H2SO4
2 H+ + SO4-2
Strong electrolyte, strong acid
Triprotic acids; Produce three H+ when dissolved
in water
H3PO4
3 H+ + PO4-3
Weak electrolyte, weak acid
4.3
Bases
•Produce (OH)- when dissolved in water
•Proton (H+) acceptor
H2O
Na(OH) (s) -----> Na+ (aq) + (OH)- (aq)
F- (aq) + H2O (l) <-> HF (aq) + (OH)- (aq)
Neutralization Reaction
Acid + Base -> Salt + H2O
Acid + Carbonate ->
Salt + CO2(g) + H2O (l)
Carbonate; Contains
(CO3)-2 or (HCO3)Chalk; Ca(CO3)
Displacement Reactions – Metal
Displaces H from acid or water

Metal + Acid -> Salt + H2 (g)

Metal + Water -> Base + H2(g)

Use Activity Series to Know if a
Reaction Will Happen
Solution Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
M KI
volume KI
500. mL x
moles KI
1L
1000 mL
x
2.80 mol KI
1 L soln
M KI
x
grams KI
166 g KI
1 mol KI
= 232 g KI
4.5
4.5
Acid/Base Titrations


Experimental technique that determines
the concentration (in Molarity) of an acid
(or base)
This is based upon an acid/base
neutralization reaction.
– ACID +BASE -> SALT + H2O

Base (or acid) is added until there is
the same amount (same # moles) of
base and acid.
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7
Fig. 4.17a,b
Acid-Base Titrations
Base; (OH)-
Acid + Base -> Salt + H2O
Acid; H+
Introductory Chemistry 2/e by N Tro,
Prentice Hall, 2006, pg 480
At the endpoint of an acid/base
titration….

Moles acid = Moles base
(MV)acid = (MV)base

Note

– If solid; moles = mass/ MM
– If aqueous solution; moles = MV
What volume of a 1.420 M NaOH solution is
Required to titrate 25.00 mL of a 4.50 M H2SO4
solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH
M
volume acid
25.00 mL x
acid
2H2O + Na2SO4
rx
moles acid
4.50 mol H2SO4
1000 mL soln
x
coef.
M
moles base
2 mol NaOH
1 mol H2SO4
x
base
volume base
1000 ml soln
1.420 mol NaOH
= 158 mL
4.7
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
4.5
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Vi =
MfVf
Mi
0.200 x 0.06
=
= 0.003 L = 3 mL
4.00
3 mL of acid + 57 mL of water = 60 mL of solution
4.5
Dissolving silver sulfate, Ag2SO4, in water

When silver sulfate dissolves it dissociates into ions.
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
equilibrium expression for the reaction:
Ksp = [Ag+]2[SO42-]
Notice that the Ag2SO4 is left out of the expression! Why?
Since K is always calculated by just multiplying concentrations, it is
called a “solubility product” constant - Ksp.
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.
Iron (III) hydroxide, Fe(OH)3
Nickel sulfide, NiS
Silver chromate, Ag2CrO4
Zinc carbonate, ZnCO3
Calcium fluoride, CaF2
Try Problems 1 - 8
Some Ksp Values
Note:
These are experimentally determined, and may
be slightly different on a different Ksp table.
Calculating Ksp of Silver Chromate

A saturated solution of silver chromate, Ag2CrO4, has
[Ag+] = 1.3 x 10-4 M. What is the Ksp for Ag2CrO4?
Ag2CrO4 (s)  2 Ag+ (aq) + CrO42- (aq)
----
----
1.3 x 10-4 M
Ksp = [Ag+]2[CrO42-]
Ksp = (1.3 x 10-4 )2 (6.5 x 10-5) = 1.1 x 10-12
Calculating the Ksp of silver sulfate

The solubility of silver sulfate is 0.014 mol/L. This
means that 0.0144 mol of Ag2SO4 will dissolve to
make 1.0 L of saturated solution. Calculate the value
of the equilibrium constant, Ksp for this salt.
Ag2SO4 (s)  2 Ag+ (aq) + SO42- (aq)
--+ 2s
2s
--+s
s
Ksp = [Ag+]2[SO42-] = (2s)2(s) = (4s2)(s) = 4s3
We know: s = 0.0144 mol/L
Ksp = 4(0.0144)3 = 1.2 x 10-5
Calculating solubility, given Ksp

The Ksp of NiCO3 is 1.4 x 10-7 at 25°C. Calculate its
molar solubility.
NiCO3 (s)  Ni2+ (aq) + CO32- (aq)
---
+s
s
---
+s
s
Ksp = [Ni2+][CO32-]
1.4 x 10-7 = s2
s = 1.4 x 107 = 3.7 x 10-4 M
Other ways to express solubility...

We just saw that the solubility of nickel (II) carbonate
is 3.7 x 10-4 mol/L. What mass of NiCO3 is needed to
prepare 500 mL of saturated solution?
3.7 x 10 4 mol NiCO3 0.500 L
118.72 g
x
x
 0.022 g
1L
1mol NiCO3
0.022 g of NiCO3 will dissolve to make 500 mL solution.
Try Problems 9 - 26
Calculate the solubility of MgF2 in water. What
mass will dissolve in 2.0 L of water?
MgF2 (s)  Mg2+ (aq) + 2 F- (aq)
---+s
s
---+ 2s
2s
Ksp = [Mg2+][F-]2 = (s)(2s)2 = 4s3
Ksp = 7.4 x 10-11 = 4s3
s = 2.6 x 10-4 mol/L
2.6 x 10 4 mol MgF2 2.0 L
62.31g
x
x
 0.032 g MgF2
1L
1mol MgF2
Solubility and pH

Calculate the pH of a saturated solution of silver
hydroxide, AgOH. Refer to the table in your booklet
for the Ksp of AgOH.
AgOH (s)  Ag+ (aq) + OH- (aq)
---+s
s
---+s
s
Ksp = 2.0 x 10-8 = [Ag+][OH-] = s2
s = 1.4 x 10-4 M = [OH-]
pOH = - log (1.4 x 10-4) = 3.85
pH = 14.00 - pOH = 10.15
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?
Calculate the solubility of MgF2 in a
solution of 0.080 M NaF.
MgF2 (s)  Mg2+ (aq) + 2 F- (aq)
---+s
s
0.080 M
+ 2s
0.080 + 2s
Ksp = 7.4 x 10-11 = [Mg2+][F-]2 = (s)(0.080 + 2s)2
Since Ksp is so small…assume that 2s << 0.080
7.4 x 10-11 = (s)(0.080)2
s = 1.2 x 10-8 mol/L
Explaining the Common Ion Effect
The presence of a common ion in a solution will
lower the solubility of a salt.

LeChatelier’s Principle:
The addition of the common ion will shift the
solubility equilibrium backwards. This means
that there is more solid salt in the solution
and therefore the solubility is lower!
Ksp and Solubility

Generally, it is fair to say that salts with very small
solubility product constants (Ksp) are only sparingly
soluble in water.

When comparing the solubilities of two salts,
however, you can sometimes simply compare the
relative sizes of their Ksp values.

This works if the salts have the same number of ions!

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...
Salt
Ksp
Solubility
(mol/L)
CuS
8.5 x 10-45
9.2 x 10-23
Ag2S
1.6 x 10-49
3.4 x 10-17
Bi2S3
-73
-15
1.1 x 10
1.0 x 10
Do you see the “problem” here??
Mixing Solutions - Will a Precipitate Form?
If 15 mL of 0.024-M lead 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)
Pb(NO3)2 (aq) + K2CrO4 (aq)  PbCrO4 (s) + 2 KNO3 (aq)
Step 1: Is a sparingly soluble salt formed?
We can see that a double replacement reaction can
occur and produce PbCrO4. Since this salt has a
very small Ksp, it may precipitate from the mixture.
The solubility equilibrium is:
PbCrO4 (s)  Pb2+ (aq) + CrO42- (aq)
Ksp = 2 x 10-16 = [Pb2+][CrO42-]
If a precipitate forms, it means the solubility
equilibrium has shifted BACKWARDS.
This will happen only if Qsp > Ksp in our mixture.
Step 2: Find the concentrations of the ions that form the
sparingly soluble 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!
Dilution: C1V1 = C2V2
[Pb2+]
C1V1 (0.024 M)(15 mL)
2


0.0080
M
Pb
=
V2
(45 mL)
C1V1 (0.030 M)(20 mL)
2

0.020
M
CrO
24
[CrO4 ] = V
(45
mL)
2
Step 3: Calculate Qsp for the mixture.
Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M)
Qsp = 1.6 x 10-4
Step 4: Compare Qsp to Ksp.
Since Qsp >> Ksp, a precipitate will form when
the two solutions are mixed!
Note: If Qsp = Ksp, the mixture is saturated
If Qsp < Ksp, the solution is unsaturated
Either way, no ppte will form!