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 (K_sp) and is a temperature dependent value.

E.g. Copper (l) bromide, CuBr   has   Molar K_sp = 4.20 x 10^-8 at 25^oC.

What is solubility of lead [II]chloride at 25^oC, PbCl₂           ; Molar Ksp = 2.40 x 10^-4  at 25^oC.

PbCl_{2   (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^–