Engineering Chemistry S Datta

IONIC EQUILIBRIUM 193
[Ca +2 ] = 0.0002 g ion l –1 . [F – ] = 2 × 0.0002 g ion l –1 .
K sp
= [Ca +2 ] [F – ] 2 = 0.0002 × (0.0004) 2 = 3.2 × 10 –11 .
Effect of Common Ion on Solubility
We know that, for a saturated solution of a sparingly soluble salt BA,
K sp
= [B + ] [A – ]
If we add to the solution a substance, which has an ion common to the above salt then
concentration of that ion will increase. As a result, the concentration of other ion will decrease
otherwise the constant K sp
will change. Therefore, B + and A – will combine to give the solid
BA(s) and the solubility of BA will decrease.
Highlights:
• It is almost invariably true that addition of a common ion at first diminishes the
solubility of a sparingly soluble salt but when relatively large amounts are added
the solubility starts to increase.
• The presence of a neutral salt without a common ion often results in an increase of
solubility of a sparingly soluble salt.
Example. 3. (a) At 25° the solubility of BaSO 4
is 0.00233 gl – . Calculate the solubility
product of the salt, taking complete dissociation.
(b) Calculate the solubility of BaSO 4
in a solution of (NH 4
) 2
SO 4
containing 13.2 gl – at
25°C. [Ba = 137]
Sol. (a) [Ba ++ ] = [SO = 4
] = 0 . 00233 = 10
137 + 96
–5 g.ion l –1
∴ K sp
= [Ba ++ ] . [SO = 4
] = 10 –5 × 10 –5 = 10 –10 .
(b) [(NH 4
) 2
SO 4
] = 13 . 2 = 0.1 (M)
132
∴ [SO = 4
] = 0.1 g.ion l –1 .
Let, the changed solubility of BaSO 4
= S′.
∴ K sp
= S′ × (S′ + 0.1) = 10 –10 .
−10
∴ S′ = 10
[S′ is negligible compared to 0.1]
−1
10
= 10 –9 mol. l –1 .
∴ S′ = 2.33 gl –1 .
Example 4. A litre of a solution contains 0.01 mole NH 4
OH and 0.001 g.ion Mg +2 . To
arrest the precipitation of Mg(OH) 2
. What will be concentration of NH
+
4
ion from NH 4
Cl? Given
= 1.8 × 10 –5 and K sp Mg(OH)2
= 1.12 × 10 –11 .
K bNHOH
4
Sol. We know that K sp
= [Mg +2 ] [OH – ] 2 = 1.12 × 10 –11
∴ [OH – ] =
L
N
M
112 . × 10
0.
001
O
Q
P
−11 1/2
= 1.1 × 10 –4 g.ion l –1 .
So, [OH – ] should not exceed 1.1 × 10 –4 g.ion l –1 to arrest the precipitation of Mg(OH) 2
.
For doing so, we are to suppress the dissociation of NH 4
OH by adding common ion NH 4 + .
So, K b
=
+ −
[NH 4 ] [OH ]
= 1.8 × 10
[NH OH]
–5 .
4