Engineering Chemistry S Datta

192 ENGINEERING CHEMISTRY
Therefore, when 1.738 g-moles of CH 3
COONa is mixed with 1 g-mole of CH 3
COOH and
the volume of the solution is made upto one litre, the desired buffer solution of pH 5 is obtained.
Solubility Product
The law of mass action cannot be successfully applied to the process of electrolytic
dissociation except to the cases of weak electrolytes. But it can be successfully applied to the
saturated solutions of sparingly soluble salts, because a heterogeneous equilibrium is set up as
follows.
BA(s) BA (solution) B + + A –
Applying law of mass action,
+ −
[B ] [A ]
K =
[BA( s)]
or K[BA(s)] = [B + ][A – ]
The concentration of solid salt [BA (s)] is constant. By convention it is unity. So
[BA (s)] = 1.
Therefore, [B + ] [A – ] = constant = K sp
.
The constant K sp
is called the solubility product of the salt BA. The above equation
signifies that the product of concentrations of B + and A – of a salt is constant independent of the
individual concentrations of B + and A – ions when temperature is constant.
If S is the solubility of BA, then [B + ] = S g.mol l –1 and [A – ] = S g.mol.l –1 .
K sp
= [B + ] [A – ] = S 2
For a salt like B x
A y
:
B x
A y
(s) B x
A y
(Solution) xB + + yA –
If the solubility is S mol l –1 then [B + ] = xS, [A – ] = yS
∴ K sp
= [B + ] x [A – ] y = (xS) x (yS) y = x x y y .S x+y
It is a universally applicable equation.
Highlights:
• When the solubility of Mg 3
(PO 4
) 2
is C mol. l –1
Mg 3
(PO 4
) 2
(s) Mg 3
(PO 4
) 2
(Solution) 3Mg +2 + 2PO
–3
4
its K sp
is [Mg +2 ] 3 [PO –3 4
] 2 = (3C) 3 × (2C) 2 = 108 C 5 .
• The solubility product of a sparingly soluble binary electrolyte is simply the square
of its solubility in pure water expressed in mol l –1 .
• It is customary to express the concentrations of the ions in g. ion l –1 i.e., formula
weight per litre and not in g. equiv. per litre.
The solubility product of a salt may be defined as the greatest possible product of
concentrations of its constituent ions in its saturated solution expressed in g.ion l –1 , each
concentration term being raised to a power representing the number of gm. ions of each type
formed by dissociation of one molecular formula weight of the salt.
Example 1. The solubility of chalk is 0.0305 gl –1 . Calculate its solubility product. (mol.
wt. of CaCO 3
= 100)
Solubility S = 0.0305 gl –1 = 3.05 × 10 –4 mol l –1
K sp
= S 2 = (3.05 × 10 –4 ) 2 = 9.3 × 10 –8 .
Example 2. The solubility of CaF 2
is 0.0002 mol l –1 . Calculate K sp
of CaF 2
.
CaF 2
Ca +2 + 2F –