SOLUBILITY AND SOLUBILITY PRODUCT
[MH 5; 16.1 & 16.2]
• In this section we are going to consider the solubility of ionic solids
in water.
• “Solubility” may be considered to be an equilibrium; the equilibrium
is between solid and ions in solution.
• Any ionic solid is 100% ionized in aqueous solution; once it actually
dissolves.
• The term solubility always refers to the amount of solid (either in
moles or grams) that actually does dissolve in solution, producing
ions; this amount can be calculated for a particular solid.
• The term solubility may also be used in a qualitative sense; there is
no “magic point” above which a salt can be described as “soluble”, and
below which a salt can be described as “insoluble”.
Very roughly........
“Soluble” - at least 0.1 mol L—1 dissolves
“Slightly Soluble” - maybe 0.001 to 0.1 mol L—1
“Insoluble” - less than 0.001 mol L—1
EXAMPLE 1: Bi2S3(s) º 2 Bi3+(aq) + 3 S2—(aq)
What is the form of the equilibrium constant for this particular
equilibrium?
– 142 –
,EXAMPLE 2:
For a solution of AgCR in water, the equilibrium is:
AgCR(s) º Ag+(aq) + CR— (aq)
• When the equilibrium is set up all three components must be
present, to be sure that the system is at equilibrium.
• The equilibrium constant K =
• We call this type of K a Solubility Product Constant, symbolized by
Ksp.
• So, Ksp(AgCR) is the solubility product constant for AgCR, when it
dissolves (or tries to!) according to the above equation.
• Like most Keq’s, its value depends on T; this is why more solid usually
dissolves in a solution at a higher temperature.
Calculations Using Ksp
EXAMPLE 1:
Calculate the solubility, in mol L—1, of AgCR(s). [Ksp (AgCR) = 1.8 × 10—10]
– 143 –
,EXAMPLE 2:
Calcium carbonate, CaCO3, has a solubility in water of 0.0180 g L —1
at 25 EC. Calculate the Ksp for CaCO3. [MM of CaCO3 = 100.1 g mol —1]
– 144 –
, EXAMPLE 3:
Calcium fluoride, CaF2, dissolves in water to the extent of 0.00170 g
per 100 mL. What is the Ksp for (CaF2) ? [MM CaF2 = 78.1 g mol —1]
– 145 –
[MH 5; 16.1 & 16.2]
• In this section we are going to consider the solubility of ionic solids
in water.
• “Solubility” may be considered to be an equilibrium; the equilibrium
is between solid and ions in solution.
• Any ionic solid is 100% ionized in aqueous solution; once it actually
dissolves.
• The term solubility always refers to the amount of solid (either in
moles or grams) that actually does dissolve in solution, producing
ions; this amount can be calculated for a particular solid.
• The term solubility may also be used in a qualitative sense; there is
no “magic point” above which a salt can be described as “soluble”, and
below which a salt can be described as “insoluble”.
Very roughly........
“Soluble” - at least 0.1 mol L—1 dissolves
“Slightly Soluble” - maybe 0.001 to 0.1 mol L—1
“Insoluble” - less than 0.001 mol L—1
EXAMPLE 1: Bi2S3(s) º 2 Bi3+(aq) + 3 S2—(aq)
What is the form of the equilibrium constant for this particular
equilibrium?
– 142 –
,EXAMPLE 2:
For a solution of AgCR in water, the equilibrium is:
AgCR(s) º Ag+(aq) + CR— (aq)
• When the equilibrium is set up all three components must be
present, to be sure that the system is at equilibrium.
• The equilibrium constant K =
• We call this type of K a Solubility Product Constant, symbolized by
Ksp.
• So, Ksp(AgCR) is the solubility product constant for AgCR, when it
dissolves (or tries to!) according to the above equation.
• Like most Keq’s, its value depends on T; this is why more solid usually
dissolves in a solution at a higher temperature.
Calculations Using Ksp
EXAMPLE 1:
Calculate the solubility, in mol L—1, of AgCR(s). [Ksp (AgCR) = 1.8 × 10—10]
– 143 –
,EXAMPLE 2:
Calcium carbonate, CaCO3, has a solubility in water of 0.0180 g L —1
at 25 EC. Calculate the Ksp for CaCO3. [MM of CaCO3 = 100.1 g mol —1]
– 144 –
, EXAMPLE 3:
Calcium fluoride, CaF2, dissolves in water to the extent of 0.00170 g
per 100 mL. What is the Ksp for (CaF2) ? [MM CaF2 = 78.1 g mol —1]
– 145 –
