SCH 4U1
MOLAR SOLUBILITY AND THE
COMMON ION EFFECT
THE MOLAR CONCENTRATION CAN ALSO BE FOUND IF WE KNOW THE VALUE OF THE Ksp (APPENDIX C – p. 802).
EX. 1 LEAD (II) IODIDE HAS A Ksp VALUE OF 7.9 x 10-9. WHAT IS THE MOLAR SOLUBILITY IN WATER?
Ksp = ________
PbI2(s) === Pb2+(aq) +
INIT. CONC (M)
CHANGE (M)
EQU'M CONC (M)
0.0
__
2 I1-(aq)
0.0
_______________
Ksp = __________
Ksp = _______ = __________
4x3 = ____________
x3 = __________
x = _________
THE MOLAR SOLUBILITY OF PbI2 IS _______ mol/L
THE COMMON ION EFFECT
EX. 2. WHAT IS THE MOLAR SOLUBILITY OF PbI2 IN A 0.10 M NaI SOLUTION?
FROM EX. 1 WE KNOW, Ksp = [Pb2+][I1-]2 = 7.9 x 10-9
THE SOLUTION IN WHICH THE PbI2 WAS PLACED CONTAINS NO Pb2+ IONS SO ITS INITIAL
CONCENTRATION IS ZERO. HOWEVER, THE ORIGINAL CONCENTRATION OF THE I1- ION HAS A CONCENTRATION
OF 0.10 M. SO,
PbI2(s) === Pb2+(aq) +
INIT. CONC (M)
CHANGE (M)
EQU'M CONC (M)
0.0
2 I1-(aq)
0.10
_______________
FOR THE EQUILIBRIUM CONCENTRATION OF I1- THE EXPRESSION "0.10 + 2x" CAN BE APPROXIMATED AS
0.10 M SINCE THE VALUE OF "x" IS SO SMALL IN COMPARISON.
Ksp = [Pb2+][I1-]2
Ksp = ______ = _______
x = _________
x = [Pb2+] = ___________
IF WE COMPARE THE SOLUBILITIES OF LEAD IN PURE WATER AND IN AN IODIDE SOLUTION WE CAN SEE
THAT THE SOLUBILITY IS REDUCED. THIS IS PREDICTED BY LE CHATELIER'S PRINCIPLE. IF THE CONCENTRATION
OF IODIDE ION IS INCREASED, THE EQUILIBRIUM SHIFTS TO THE LEFT AND MORE SOLID IS FORMED.