Question

The precipitate of CaF2 (Ksp=1.7x10-10) is obtained when equal volumes of the following are mixed a)10^-4 M Ca2+ + 10^-4 M F - b)10^-2 M Ca2+ + 10^-3 M F- C)10^-5 M Ca2+ + 10^-3 M F-

Answer 1

equation;CaF2(s)→Ca2+(aq)+2F-(aq)
ksp=(Ca2+)(F-)^2=1.7×10^-10
When combining the two solutions,the concentration of each ion will be halved from its initial concentration.So,in solution 1,the final concentration of the Ca2+ will be 5×10^-5,& the concentration of F- will also be 5×10^-5.So,
(5×10^-5)(5×10^-5)^2=1.25×10^-13.
since this is below ksp,no precipitate will form.
for sln 2:
5×10^-3×(5×10^-4)^2=1.25×10^-9
since this is larger than ksp,a precipitate will form.
Hence the precipitate of CaF2(KSP=1.7×10^-10)Is obtained when equal volumes are mixed. 

Answer 2

Answer: Option B

Explanation:

Precipitation takes place when the ionic product of a salt is greater than the solubility product. We are given that the solubility product of CaF2 is 1.7 X 10-10. So, the ionic product of CaF2 should be greater than this value. The ionic product of CaF2 can be calculated as follows :

= [Ca2+] X [F-]2

on using the values given in (a), the ionic product of CaF2 comes out to be 1 X 10-12, while using the values given in (c), we get 1 X 10-13 as the answer. However, on using the values given in (b), the ionic product comes out to be

= [10-2] X [10-3]2

= 10-2 X 10-6

= 10-8

Thus, in this case, the ionic product of CaF2 is greater than solubility product. Hence the correct answer is (b) 10-2 M Ca2+ and 10-3 M F-