Question

1. (a) Write the expression for solubility
product (Ksp) of a sparingly soluble salt
CaF2 and calculate its Ksp. Given that
solubility of CaF2 is 2-2x10-2 2-1
1+2-​

Answer 1

Answer:

3.4×10  

−9

M

The dissociation equilibrium of calcium fluoride is as given below.

CaF  

2

​

⇌  

x

Ca  

2+

 

​

+  

2x

2F  

−

 

​

 

The dissociation of NaF is as shown below.

NaF→  

0.1

Na  

+

 

​

+  

0.1

F  

−

 

​

 

Thus total fluoride ion concentration is 2x+0.1 M but 2x<<<0.1.  

Hence, 2x is neglected and the total fluoride ion concentration is 0.1 M.

The expression for the solubility product is as follows.

K  

sp

​

=[Ca  

2+

][F  

−

]  

2

 

Substitute values in the above expression:

3.4×10  

−11

=x(0.1)  

2

 or x=3.4×10  

−9

 M.