5. 1 Definition
Solubility product is the product of the concentrations of
each ion in a saturated solution of a sparingly soluble salt
at 298 K, raised to the power of their relative
concentrations.
Solubility is affected by temperature and hence the value
of Ksp depends on temperature.
5. 2 Ksp calculation example 1
The solubility product of a salt can be obtained from its
solubility. A saturated solution of silver chloride contains
−3
−3
1.46 × 10 g dm at 18 °C. What is the Ksp?
Step 1: Find the solubility of salt in
-3
moldm
The solubility of silver chloride at 18°C = 1.46 ×
−3
dm
−3
10 g
5. 3 Ksp calculation example 1
Step 1: Find the solubility of salt in
-3
moldm
The solubility of silver chloride = 1.46 ×
Molar mass = 143.5 g
—1
mol
−3
10 g
−3
dm
5. 4 Ksp calculation example 1
Step 2: Use the ICE table to find the
from the amount that dissolves.
AgCl(s)
Initial
Change
Equiibrium
⇌
+
Ag (aq)
+
+
[Ag ]
and
—
Cl (aq)
—
[Cl ],
5. 5 Ksp calculation example 1
Step 3: Place the values of
expression.
+
[Ag ]
and
—
[Cl ]
in Ksp
5. 6 Ksp calculation example 2
A saturated solution of magnesium fluoride, MgF2, has a
–3
–3
solubility of 1.22 × 10 mol dm . Calculate the Ksp?
MgF2(s)
Initial
Change
Equiibrium
⇌
2+
Mg (aq)
+
—
2F (aq)
5. 7 Ksp calculation example 2
5. 8 Skill check
Calculate the Ksp of a saturated aqueous solution of
–11
–3
cadmium sulfide, CdS (solubility = 1.46 × 10 mol dm )
5. 9 Skill check
Calculate the Ksp of a saturated aqueous solution of
–3
calcium fluoride, CaF2, containing 0.0168 g dm CaF2
5. 10 Calculating solubility from Ksp 1
Solubility products can be used to calculate the solubility
of compounds.
Calculate the solubility of copper(II) sulfide in mol
–36
2
–6
(Ksp for CuS = 6.3 × 10 mol dm )
Step 1: Write down the equilibrium equation.
CuS(s)
⇌
Cu2+(aq)
+ S2—(aq)
–3
dm
5. 11 Calculating solubility from Ksp 1
Step 2: Find equilibrium concentrations.
CuS(s)
⇌
Cu2+(aq)
+ S2—(aq)
Initial
Change
Equiibrium
If x is the amount of CuS that dissolves, then
2—
[S ]
=x
2+
[Cu ]
= x and
5. 12 Calculating solubility from Ksp 1
Step 3: Subsititute the value of Ksp and x into the Ksp
expression.
5. 13 Calculating solubility from Ksp 2
Given that the solubility product constant of Ni(OH)2 is
−18
3
−9
6.5 × 10 mol dm at 298 K, calculate the solubility of
nickel(II) hydroxide in water.
5. 14 Skill check
Calculate the solubility in mol
–23
2
–6
(Ksp =1.6 × 10 mol dm )
–3
dm
of zinc sulfide, ZnS.
5. 15 Skill check
Calculate the solubility of silver carbonate, Ag2CO3.
–12
3
–9
(Ksp = 6.3 × 10 mol dm )
5. 16 Predicting precipitation
The solubility product can be used to predict whether
precipitation will occur when two solutions are mixed.
For example: will we get a precipitate when we mix a
solution of barium chloride, BaCl2, with a very dilute
solution of sodium carbonate?
5. 17 Predicting precipitation
Both barium chloride and sodium carbonate are soluble
salts, but barium carbonate is relatively insoluble. We
must consider the equilibrium for the insoluble salt
dissolving in water:
BaCO3 (s)
⇌
Ba2+ (aq) + CO32— (aq)
5. 18 Predicting precipitation
The solubility product is given by:
Ksp =
2+
[Ba ]
2–
[CO3 ]
2+
[Ba ]
2–
[CO3 ]
2+
[Ba ]
2–
[CO3 ]
= 5.5 ×
–10
10
2
mol
If
is greater than 5.5 ×
precipitate will form.
If
is less than 5.5 ×
precipitate will form.
–6
dm
–10
10
–10
10
2
mol
2
mol
–6
dm
–6
dm
no
a
5. 19 Predicting precipitation example
Will a precipitate form if we mix equal volumes of
–4
–3
–
solutions of 1.00 × 10 mol dm Na2CO3 and 5.00 × 10
5 mol dm–3 BaCl2?
2+
[Ba ]
= 1/2 x 5.00 ×
2–
[CO3 ]
–5
10 =
= 1/2 x 1.00 ×
–4
10
2+
[Ba ]
x
2.50 ×
–5
10
= 5.00 ×
2–
[CO3 ]
mol
–5
10
=?
–3
dm
mol
–3
dm
5. 20 Predicting precipitation example
2+
[Ba ]
x
2–
[CO3 ]
= 1.25 ×
—9
2
—6
10 mol dm
This value is greater than the solubility product, so a
precipitate of barium carbonate forms.
5. 21 Common ion effect
Adding a common ion, (one which is already present in the
solution), will result in the precipitation of a sparingly soluble
ionic compound.
Definition: The common ion effect is the reduction in the
solubility of a dissolved salt achieved by adding a
solution of a compound which has an ion in common
with the dissolved salt. This often results in
precipitation.
5. 22 Common ion effect
Adding NaCl to a saturated solution of AgCl will result in
the precipitation of AgCl.
AgCl(s)
⇌
+
Ag (aq)
+
—
Cl (aq)
Adding NaCl to a solution of AgCl increases the
—
—
concentration of Cl (aq). Cl (aq) is a common ion as it is
already in solution.
5. 23 Common ion effect
NaCl(s)
→
Na+(aq) + Cl—(aq)
—
Cl
The extra
ions means that the solubility product is
+
—
exceeded. To reduce the value of [Ag ][Cl ] below the
Ksp, some ions are removed from solution by
precipitating.
Ksp = [Ag+][Cl—] = 1 × 10–10 mol2 dm–6
5. 24 Common ion effect
The effect of increasing or decreasing the concentration
of one of the ions in equilibrium with the sparingly soluble
salt may be predicted qualitatively using Le Chatelier’s
principle.
AgCl(s)
⇌
Ag+(aq) + Cl—(aq)
5. 25 Common ion effect
AgCl(s)
⇌
Ag+(aq) + Cl—(aq)
+
Ag (aq)
—
Cl (aq)
The addition of either of
or
into the above
system in equilibrium would displace the equilibrium to
the left causing the precipitation of AgCl.
5. 26 Common ion effect
+
M (aq)
—
X (aq)
In other words when either
or
is added the
ion product will exceed the solubility product and MX will
be precipitated until the value of the Ksp is restored.
5. 27 Common ion effect example 1
Since the
+
Ksp [Ag ]
—
[Cl ]
=
—10
2
—6
1x10
mol dm
[Ag+] = [Cl—] = 1x10—5 moldm-3
When [Cl—] is raised to 0.01 moldm-3 by adding either NaClthe [Ag+]
must decrease to 1x10–8 moldm-3
as Ksp = [Ag+][Cl—]
= 1x10—8 x 0.01 = 1x10—10
5. 28 Common ion effect
In calculations of solubility in solutions containing common ion, the
concentration of the ion from the sparingly soluble salt is usually
negligible in comparison to the concentration of the common ion.
5. 29 Common ion effect example 2
Solubility of BaSO4 in 0.1 mol
Ksp(BaSO4) = 1 ×
−10
10
2
mol
−3
dm
Na2SO4(aq)
−6
dm
Solubility of BaSO4 in 0.1 M Na2SO4(aq) = w
BaSO4 (s)
⇌
Ba2+ (aq) + SO42— (aq)
5. 30 Common ion effect example 2
Na2SO4(s)
[Ba2+] = w
→
2Na+(aq) + SO42—(aq)
&
[SO42—] = (w + 0.1)
but as w is much less than 0.1
2—
[SO4 ]
Ksp =
= (w + 0.1) ≅ 0.1
2+
[Ba ]
2—
[SO4 ]
= w x 0.1 = 1 ×
−10
10
2
mol
−6
dm
5. 31 Common ion effect example 3
How is the solubility of an ionic compound affected when
the compound is dissolved in a solution that already
contains one of its ions? For example, what is the
solubility of CaF2 in a solution that is 0.100 M in NaF?
Ksp of CaF2 = 1.46 x
—10
10
3
—9
mol dm
5. 32 Common ion effect example 3
Step 1: Begin by writing the reaction by which solid
CaF2 dissolves into its constituent aqueous ions.
Write the corresponding expression for Ksp.
CaF2(s)
⇌
2+
Ca (aq)
+
—
2F (aq)
5. 33 Common ion effect example 3
Step 2: Use the ICE table to find the concentrations of
the ions from the sparingly soluble salt (CaF2).
5. 34 Common ion effect example 3
Step 3: State the total concentrations of the ions from
the sparingly soluble salt (CaF2) accounting for the
common ion (F from NaF)
5. 35 Common ion effect example 3
Step 4: Plug in the values in the Ksp expression to find
the value of the solubility of CaF2 in NaF.
5. 36 Skill check
Calculate the mass of calcium hydroxide that will
3
-3
dissolve in 100 cm of 0.1 moldm sodium hydroxide
o
-4
3
—9
solution at 25 C. Ksp of Ca(OH)2 = 2.12 x10 mol dm
5. 37 Skill check
–6
3
—9
10 mol dm .
Ksp of PbBr2 is 6.6 x
An excess of PbBr2
-3
is stirred with 0.4 moldm KBr solution, and excess
2+
PbBr2 filtered off. Calculate the [Pb ] in the remaining
solution.
5. 38 Skill check
—3
gdm
o
25 C.
The solubility of lead chloride is 4.75
at
What mass of lead chloride will be precipitated if 5.85 g
3
of sodium chloride is added to 1 dm of saturated
solution of lead chloride?
5. 39 Skill check
The Ksp for Ca(OH)2 at 298K is 4.78×
2+
[Ca ] in the saturated solution.
3
dm
–5
10 . Calculate
the
5
of a saturated solution of was made. To this was
-3
added 0.1 moldm of NaOH. What mass of Ca(OH)2 will
precipitate?