If the molar solubility of #CaF_2# at 35 C is #1.24 * 10^-3# mol/L, what is Ksp at this temperature?

1 Answer
Jun 9, 2016

Answer:

#K_(sp)# #=# #[Ca^(2+)][F^-]^2# #=# #??#

Explanation:

#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.