```Solubility
Definition
Q. How do you measure a compound’s solubility?
A. The amount of that compound that will dissolve in a set volume of water.
This is usually measured in gL-1 or molL-1.
What happens as a sparingly
soluble solid dissolves?
E.g. Dissolving CuSO4 in water.
As you add solid CuSO4, it dissolves until it reaches the saturation point. This is
the point at which no more solid will dissolve. You know when you’ve reached
saturation when____________________________.
At this point an equilibrium system is established where the rates of the forward
and reverse reactions are equal. Write an equation for this equilibrium
reaction.
CuSO4 (s)  Cu2+ (aq) + SO42- (aq)
Now write a K expression for this equilibrium.
Ks = [Cu2+ (aq)] [SO42- (aq)]
The Ks is the SOLUBILITY PRODUCT – the product of the concentration of ions
in a saturated solution of a sparingly soluble compound.
Questions:
Calculate the concentration of all the ions in a saturated solution of the following:
1.
2.
3.
4.
5.
Calcium fluoride (solubility = 2.2 x 10-4 molL-1)
Copper (II) sulfide (solubility = 2.5 x 10-18 molL-1)
Calcium hydroxide (solubility = 1.1 x 10-2 molL-1)
Silver sulfide (solubility = 1.2 x 10-17 molL-1)
Silver iodide (solubility = 9.1 x 10-9 molL-1)
1. Ca2+ = 2.2 x 10-4 molL-1, F- = 4.4 x 10-4 molL-1
2. Cu2+ = 2.5 x 10-18 molL-1, S2- = 2.5 x 10 -18 molL-1
3. Ca2+ = 1.1 x 10-2 molL-1, OH- = 2.2 x 10-2 molL-1
4. Ag+ = 2.4 x 10-17 molL-1, S2- = 1.2 x 10-17 molL-1
5. Ag+ = 9.1 x 10-9 molL-1, I- = 9.1 x 10-9 molL-1
Calculating the solubility product
(Ks) given the solubility
1. Calculate the Ks for BaSO4 given its solubility = 3.87 x 10-5 molL-1
1.50 x 10-9
2. Calculate the Ks for CaF2 given its solubility = 2.2 x 10-4 molL-1
4.26 x 10-11
3. Calculate the Ks for PbI2 given its solubility = 4.1 x 10-2 gL-1
2.8 x 10-12
4. Calculate the Ks for silver sulfide given its solubility = 1.2 x 10-17 molL-1 6.91 x 10-51
First decide if the compound is AB, AB2 or A2B type.
Write an equilibrium equation and a Ks expression for the compound.
Decide on the concentration of each ion using the solubility.
Substitute the ion concentrations into the Ks expression and solve.
Write a general formula
Try writing a general formula for calculating Ks from
solubility. Let solubility = x. Your answer will be
different for AB (e.g. CaSO4) compounds than AB2
(e.g. CaF2) and A2B (e.g. Ag2S) compounds.
AB: Ks = x2
AB2 and A2B: Ks = 4x3
Remember:
• The solubility has to be DOUBLED for A
(in A2B compounds) and B (in AB2
compounds. This is because that ion is
present at TWICE the concentration of the
other ion, and equilibrium constants are a
product of the CONCENTRATION of ions!
Going the other direction…
Calculate the solubility of the following compounds given their Ks value:
1.
2.
3.
4.
The Ks for silver chloride is 1.7 x 10-10. Calculate its solubility.
The Ks for lead chloride is 1.7 x 10-5. Calculate its solubility.
The Ks for silver sulfate is 1.22 x 10-5. Calculate its solubility.
The Ks for calcium hydroxide is 1.0 x 10-2. Calculate the pH of a
saturated solution of calcium hydroxide.
1. 1.3 x 10-5 molL-1 (using Ks = x2)
2. 0.0162 molL-1 (using Ks = 4x3)
3. 0.0145 molL-1 (using Ks =4 x3)
4. 13.4 (using Ks = 4x3, then [OH-]=2 x solubility, then pH = 14—log[OH-])
Key points
• The Ks is the SOLUBILITY PRODUCT. It
refers to a saturated solution ONLY – i.e.
dissolving a maximum amount of a solid in
water
• The solubility in molL-1 = X
• X also equals the concentration of the
single ion (A and B in AB compounds, A in
AB2 compounds and B in A2B compounds)
Ionic product
• To be useful, you have to be able to use
solubility data to PREDICT
PRECIPITATION. E.g. What if we added
fluoride ions to our water supply – would
anything precipitate out producing a nasty
solid floater in our drinking water?
• We need to use the IONIC PRODUCT.
Ionic product and Ks
• The ionic product is the product of the ions
in solution, raised to the appropriate power
• It is calculated in the same way as Ks
HOWEVER, THEY ARE DIFFERENT!!!!!
• Ks is calculated when you make a
saturated solution of a sparingly soluble
compound only (i.e. Dissolve stuff in water to saturation)
• IP is calculated from any source of ions
Example questions
1. Calculate the Ks of BaSO4, given that a
saturated solution of BaSO4 contains 3.87
x 10-5 molL-1 of Ba ions. Answer: Ks = [Ba ][SO ], = x
2+
4
2-
= (3.87 x 10-5)2 = 1.5 x 10-9
2. Calculate the IP of BaSO4 if 4.24 x 10-4
molL-1 Ba2+ ions is added to a water
supply containing 2.21 x 10-6 molL-1
sulfate ions.
Answer: IP = [Ba ][SO ], = (4.24
2+
4
2-
x 10-4)( 2.21x10-6) = 9.37x 10-10
2
Using IP
From the previous page: Will a precipitate form in
the water supply?
Compare IP to Ks
If IP > Ks, then concentration of ions exceeds
solubility so a PRECIPITATE FORMS
If IP < Ks, then concentration of ions is less than
solubility so NO PRECIPITATE FORMS
See Pearson power point “ionic product”
Common ion effect
• The presence of a common ion decreases a
compounds solubility e.g. KCl is less soluble in
water than it is in an NaCl solution. This is
because the salt solution contains one of the
ions you are trying to dissolve
KCl ↔ K+ + ClThe presence of Cl- in the solution pushes the
equilibrium to the left, thus precipitating KCl
• See Pearson Powerpoint “common ion effect”
Common ion question
• Calculate the solubility of PbI2 in pure
water and in 0.1 molL-1 KI. Ks(PbI2) =
equilibrium principles.
Complex ions
Remember these?
Ag(NH3)22+
Zn(NH3)42+
Pb(OH)42Zn(OH)42Al(OH)4Cu(NH3)42+
Complex ions – made from a central metal ion
and a number of ligands. These are soluble.
The presence of one of these ligands can allow
complex ions to form, which can INCREASE the
solubility of some sparingly soluble compounds
Example
• Suppose we have a saturated solution of AgBr, with some
AgBr(s) left undissolved at the bottom of the beaker. We next
add some aqueous ammonia to the system. Its molecules are
strong ligands for silver ions, so they begin to form Ag(NH3)2+
ions from the trace amount of Ag+(aq) initially present in
solution.
• Ag+(aq) + 2 NH3(aq) <==> Ag(NH3)2+(aq)
• The ammonia represents an upset to the equilibrium present in
the saturated solution of
• AgBr(s) <===> Ag+(aq) + Br-(aq)
• By pulling Ag+ ions out of this equilibrium, the equilibrium must
shift to the right to replace them as best it can. i.e. more
AgBr(s) must go into solution. The solubility of a slightly
soluble salt increases when one of its ions can be changed to a
soluble complex ion.
The effect of added acid on
basic anions
• The pH also influences the solubility of salts that contain a
basic anion, an anion from a weak acid for example. In general
we would expect the salts of weak acids to be more soluble in
acidic solutions. For example consider CaF2
• CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
• In an acidic solution the F- ions will be removed by reacting
with H3O+ to form HF, thus shifting the equilibrium to the right
and causing more CaF2 to dissolve.
• H3O+ + F- <==> HF + H2O
• The solubilities of salts containing anions that do not
hydrolyze, such as Cl-, Br-, I-, NO3- in other words anions from
strong acids are not affected by pH.
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