# Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693c: 55

$K_{sp} (SrF_2) = (2.686 \times 10^{-9})$

#### Work Step by Step

1. Calculate the molar mass: 87.62* 1 + 19* 2 = 125.62g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.011}{ 125.62}$ $n(moles) = 8.757\times 10^{- 5}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 8.757\times 10^{- 5}}{ 0.1}$ $C(mol/L) = 8.757\times 10^{- 4}$ 4. Write the $K_{sp}$ expression: $SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2F^{-}(aq)$ $K_{sp} = [Sr^{2+}]^ 1[F^{-}]^ 2$ 5. Determine the ion's concentrations: $[Sr^{2+}] = [SrF_2] * 1 = [8.757 \times 10^{-4}] * 1 = 8.757 \times 10^{-4}$ $[F^{-}] = [SrF_2] * 2 = 1.750 \times 10^{-3}$ 6. Calculate the $K_{sp}$: $K_{sp} = (8.757 \times 10^{-4})^ 1 \times (1.750 \times 10^{-3})^ 2$ $K_{sp} = (8.757 \times 10^{-4}) \times (3.067 \times 10^{-6})$ $K_{sp} = (2.686 \times 10^{-9})$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.