Transcript Chemical Equilibrium: Solubility Product Constant

Chemical Equilibrium:
Solubility Product Constant
Chapter 12 part 3

Solubility indicates the maximum amount
of solute that can dissolve in a given
amount of liquid

Solubility of a solid is expressed in mol/L

Solubility equilibrium exists when a
saturated solution contains non-dissolved
solute at the bottom of a container

This indicates that the rate of dissolution
is equal to the rate of re-crystallisation

REMEMBER that when solving for the
constant we DO NOT take into
consideration the conc. of SOLIDS

Therefore the constant is calculated as
the product of the conc at equilibrium of
the two products (and thus the name
Ksp)
Example:
X2Y3 (s) ↔ 2 X+ (aq) + 3 Y-(aq)
Ksp = [X+]2 [Y-]3
Example #1
Calculate the conc of barium ions and
sulfate ions in a saturated solution of
barium sulfate given that
Ksp = 1.1 x 10-10
BaSO4 (s) ↔ Ba2+(aq) + SO42-(aq)
Example #2
Calculate the conc of both ions in a
saturated solution of silver sulfate given
that Ksp = 1.5 x 10-4
Ag2SO4 ↔ 2 Ag+(aq) + CO32-(aq)
Example #3
The solubility of Ag2CO3 is 3.6 x 10-3
g/100.0 ml at ambient temperature,
calculate Ksp.
Ag2CO3 (s) ↔ 2 Ag+ (aq) + CO32- (aq)
Hmwk: p 346-348 # 21, 30,
34, 38, 42, 45