Question

Calculate the molar solubility (s) for the same slightly soluble salt AX₂ in a 0.20 M solution of KX₂ (assume KX₂ is a soluble potassium salt of the anion X). The Ksp for the salt is 7.3 x 10^-7 .

Answer 1

Solution:

Dissociation of soluble salt KX₂ :

KX₂ is a soluble salt and it will completely dissociate into its constituent ions, as per the following chemical reaction:

KX₂ (aq.) K⁺(aq.) + 2X^- (aq.)

	KX2 (aq.)	K+(aq.)	X- (aq.)
Initial Concentration	0.20 M	0 M	0 M
Change in Concentration	-0.20 M	+ 0.20 M	+ 0.40 M
Final Concentration	0 M	0.20 M	0.40 M

Dissociation of slighlty soluble salt AX₂ :

Let the molar solubility of slighlty soluble salt AX₂ be S mol/litre.

AX₂ dissociates in water as:

AX₂ (aq.) A²⁺ (aq.) + 2X^- (aq.)

The concentration of AX₂ is S mol/litre.

Therefore, concentration of A²⁺(aq.) and X^-(aq.) at equilibrium will be S mol/litre and 2S mol/litre respectively.

Final Concentration of X^- ions in the solution = X^- ions released from salt AX₂ + X^- ions released from salt KX₂

= 2S mol/litre + 0.40 mol/litre

Expression of K_sp for salt AX₂ :

K_sp = [A²⁺][X^-]²

7.3 * 10^-7 = (S) * (2S + 0.40)²

Since, K_sp of salt AX₂ is very small, and AX₂ is a slightly soluble salt, the concentration of X^- from AX₂ salt will be very less as compared to 0.40 M from KX₂ salt.

i.e. 2S + 0.40 0.40

Therefore,

7.3 * 10^-7 = (S) * (0.40)²  

or

S = ( 7.3 * 10-7 ) / ( 0.40 )²

S = 45.625 * 10^-7

or

S = 4.6 * 10^-6 mol/litre   

Hence, the molar solubility of AX₂ salt will be 4.6 * 10 ^-6 mol/litre. (Ans.)