#K_(sp)#, #"the solubility product"# derives from the solubility expression:
#CaF_2(s) rightleftharpoonsCa^(2+) + 2F^-#
And #K_(eq)# #=# #([Ca^(2+)][F^-]^2)/([CaF_2(s)])#
However, #[CaF_2(s)]#, the concentration of a solid is a meaningless quantity.
Thus: #K_(eq)# #=# #K_(sp)# #=# #[Ca^(2+)][F^-]^2#
Given this expression, we simply fill in the blanks:
#K_(sp)# #=# #(1.24xx10^-3)(2xx1.24xx10^-3)^2#
#=# #4xx(1.24xx10^-3)^3#
#=# #???#
The given #K_(sp)# is calculated for #35# #""^@C#. At lower temperature, would you expect #K_(sp)# to increase or decrease? Give a reason for your answer.