SrF2 ==> Sr^2+ + 2F^-
Ksp = (Sr^2+)(F^-)^2 = 4.3E-9
For A substitute 0.013 for (Sr^2+) in the Ksp expression and solve for (F^-), then take 1/2 to determine solubility of SrF2. Note that 1 mol SrF2 produces 2 mols F^-.
For B substitute 0.015 for (F^-) in the Ksp expression and solve for (Sr^2+) to obtain solubility of SrF2.
Calculate the molar solubility of SrF2 in the following substances. Ksp=4.3×10−9
A. 1.3×10−2 M Sr(NO3)2
B. 1.5×10−2 M NaF
1 answer
4 answers