solubility problem?

Question

1.1 x 10^-2 g of SrF2 dissolve per 100 mL of aqueous solution at 25.0 degrees C. What is the solubility product for SrF2?

Answer 1

Ksp = [Sr+2]*[F-]²

If you have 0.011 grams of SrF2 in 100 mL, then you've got 0.11 grams/liter.

MW of SrF2 = 125.62 g/mole, so you have

0.11 grams/125.62 g/mole = 8.76x10^-4 moles per liter

SrF2 ===> Sr+2 + 2F-, thus one mole of SrF2 will give one mole of Sr ions and TWO moles of F ions:

Ksp = [8.76x10^-4]*[2*8.76x10^-4]² = 2.69x10^-9

(the Textbook value is 4.33x10^-9)