Description
ionic solid SrF2. When dissolved in pure water, its solubility is found to be 8.9×10-4 M.
Give the formula of the anion in this compound.
What is the concentration of this anion in a saturated solution of SrF2?
Determine the value of Ksp for SrF2.
Explanation & Answer
a) Formula of ion = F^-
b) concentration of anion
The solubility is 8.9×10^-4 M. For each mol of SrF2 that goes into solution, there will be two mols of F-.
= 8.9×10^-4 M * 2
= 1.8×10^-3 M
c) Ksp for SrF2
ksp = [Sr^2+] * [F-] ^ 2
= (8.9×10^-4 M) * (1.8×10^-3 M) ^2
= 2.8 ×10^-9
agrees in magnitude with literature value
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