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Multiple Choice
What is the minimum concentration of sulfate ions (SO₄²⁻) required to cause precipitation of calcium ions (Ca²⁺) from a 9.0×10⁻² M CaI₂ solution, given that the Ksp of CaSO₄ is 2.4×10⁻⁵?
A
3.6×10⁻³ M
B
4.8×10⁻² M
C
1.5×10⁻³ M
D
2.7×10⁻⁴ M
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Verified step by step guidance
1
Identify the relevant chemical reaction: The precipitation of calcium sulfate can be represented by the equation: Ca²⁺(aq) + SO₄²⁻(aq) ⇌ CaSO₄(s).
Understand the concept of solubility product (Ksp): The Ksp is the equilibrium constant for the dissolution of a sparingly soluble compound. For CaSO₄, Ksp = [Ca²⁺][SO₄²⁻] = 2.4×10⁻⁵.
Determine the concentration of calcium ions: Since the solution is 9.0×10⁻² M in CaI₂, and each formula unit of CaI₂ provides one Ca²⁺ ion, the concentration of Ca²⁺ is 9.0×10⁻² M.
Set up the expression for Ksp: Substitute the known concentration of Ca²⁺ into the Ksp expression: 2.4×10⁻⁵ = (9.0×10⁻²)[SO₄²⁻].
Solve for the concentration of sulfate ions: Rearrange the equation to find [SO₄²⁻] = 2.4×10⁻⁵ / 9.0×10⁻². This will give you the minimum concentration of sulfate ions required to start precipitation.