Chapter 17: Problem 45
When \(250 \mathrm{mg}\) of \(\mathrm{SrF}_{2},\) strontium fluoride, is added to 1.00 L of water, the salt dissolves to a very small extent. $$\operatorname{SrF}_{2}(s) \rightleftarrows \operatorname{Sr}^{2+}(a q)+2 F^{-}(a q)$$ At equilibrium, the concentration of \(\mathrm{Sr}^{2+}\) is found to be \(1.03 \times 10^{-3} \mathrm{M}\). What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{SrF}_{2} ?\)
Short Answer
Step by step solution
Understanding the Reaction
Writing the Expression for Ksp
Identify Known Concentrations
Calculate the Ksp Value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
- \([\text{Sr}^{2+}]\) represents the equilibrium concentration of strontium ions.
- \([\text{F}^{-}]\) is the equilibrium concentration of fluoride ions, squared because two fluoride ions are produced per formula unit.
Equilibrium Concentration
Dissolution Reaction
- Strontium ions \((\text{Sr}^{2+})\)
- Fluoride ions \((\text{F}^{-})\), with two fluoride ions generated per formula unit of \( \text{SrF}_2 \)
Chemical Equilibrium
- The concentrations of \( \text{Sr}^{2+} \) and \( \text{F}^{-} \) ions at equilibrium are determined by the system's \( K_{sp} \).
- Equilibrium doesn’t mean equal concentrations, but rather that the processes occurring in the solution and solid are balanced.