Calculate the solubility product of the the sparingly soluble salt CaF2 from the following data: the molar ionic conductances (at - Chemistry - Electrochemistry

Question

Calculate the solubility product of the the sparingly soluble salt
CaF2​ from the following data: the molar ionic
conductances (at infinite dilution) of Ca2+ and
F- ​ions are 104 x 10-4 and 48 x
10-4 S m2 mol- , respectively The
specific conductance of the saturated solution of
CaF​2 at room temperature is 4 25 x
10-3 S m- ​and specific conductance of
water used for preparing the solution is 2 x 10-4 S
m-

Vartika Jain · Accepted Answer

Dear Student,

λ∞= κ×Vand, λ∞=λ°Ca2+ + λ°F- = 104×10-4+48×10-4=152×10-4 Sm2mol-1κ=4
25×10-3 Sm-1Therefore, V=λ∞κ=152×10-4 4
25×10-3 =3 57 ×109 cm33
57×109 cm3 contain CaF2= 1 g equivalent = 78 g of CaF21000 cm3 contain CaF2=78×10003
57×109=21
84×10-6 g/LCaF2 ↔ Ca2+ + 2F-                   s           2sKsp=[Ca2+][F-]2=(s)×(2s)2=4s3s=21
84×10-6 78=2 80×10-7 mol/LKsp= 4(2
80×10-7)3 = 8 780×10-20 mol3/L3