SOLUBILITY AND COMMON ION EFFECT
SOLUBILITY
Solubility (s) is the amount of substance that dissolves in a solvent at a particular temperature.
Soluble salt = dissolves completely.
Insoluble salt= does not dissolve completely.
A solution consists of a solute and solvent. Dissolved soluble solute usually exist as solvated ions or solvated molecules. However, some substances are sparingly or insoluble in water. This could be due to poor solvation by water, or the chemical bonds are too strong to be overcome by solvation. NaCl is soluble in water (solvated Na+ + Cl– ions). AgNO3 is soluble; (solvated Ag+ and NO3– ions) . However, AgCl is insoluble or sparingly soluble.
SOLUBILITY CALCULATIONS
Concentration in molality = s
SOLUBLE SALTS
Water solubility of Sodium chloride 1.0 g in 1 kilogram of water.
NaCl ____> Na+ + Cl– 100% dissociation to ions, therefore solubility is same as concentration of ions.
Moles of NaCl = 1.00/[23 + 35.5] = 0.0171mol
Solubility of NaCl = 0.0171mol/1kg = 0.0171 molal solution
What is solubility of Sodium ions in this solution?
NaCl ____> Na+ + Cl– 100% dissociation to ions, therefore solubility is same as concentration of ions.
Concentration of NaCl=Concentration of sodium ions = concentration of chloride ions=[s] ;
[Na+] x [Cl–] = [0.0171]2
[s]2 = [0.0171]2
[s]=0.0171m
Exercise: Convert the value of 0.0171 molal to Molarity.
INSOLUBLE SALTS:
Do not dissociate completely, therefore equilibrium is reached between solid compound and dissolved ions in solution.
AB(s) <= > A+aq + B–aq
Keq = [A+aq][B–aq]/ [AB(s)] ; SOLID (s) is not in solution and therefore has to be ignored, therefore.
Keq = [A+aq][B–aq]
This product of ion concentrations is called SOLUBILITY PRODUCT (Ksp) and is a temperature dependent value.
E.g. Copper (l) bromide, CuBr has Molar Ksp = 4.20 x 10-8 at 25oC.
What is solubility of lead [II]chloride at 25oC, PbCl2 ; Molar Ksp = 2.40 x 10-4 at 25oC.
PbCl2 (S) <=> Pb2+ (aq) + 2Cl– (aq)
s x [2s]2 = 4s3 = solubility product of PbCl2 = 2.40 x 10-4
s = [(1/4) x 2.40 x 10-4 ]1/3
s = 0.03915 M
COMMON ION EFFECT
Addition of a salt containing one of the ions already present in solution of original salt will change the solubility of the common ion.
EXAMPLE
Calculate solubility of the above solution if 3.0g of NaCl is added to 1.0L of the solution of PbCl2.
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Moles NaCl = [1mol/58.5g] x 3.0g = 0.0513mol = 0.0513 mol
Mol NaCl = mol Na+ = mol Cl– = 0.0513mol
The common ion for NaCl and PbCl2 is Cl– , Therefore 0.0513 mol Cl– has been added.
Therefore, Solubility product will be;
[Pb2+] ([2Cl–]2 +[added Cl–]) = 2.40 x 10-4
we get (s x [(2s)2 + 0.0513]) = 2.40 x 10-4
s [4s2 + 0.0513] =2.40 x 10-4
4s3 + 0.0513s = 2.40 x 10-4
PbCl2 is very insoluble implies solubility s is very small compared to 0.0513 from soluble NaCl , therefore 4s2 + 0.0513 is approximately equal to 0.0513 ,
therefore, equation becomes 0.0513s = 2.40 x 10-4,
s = 0.004678 M
This is good approximation since 4 x [0.004678]3 = 4.09x 10-7 is very small compared to 0.0513.
The value of solubility s of PbCl2 has decreased from 0.03915 M to 0.004678 M due to addition of common ion Cl–
