JEE Main Chemistry question
The solubility of a saturated solution of calcium fluoride is 2×10⁻⁴ moles per litre. Its solubility product is:
Answer: (D)
Solution
Ksp=(2×10⁻⁴)(4×10⁻⁴)²=32×10⁻¹².
The solubility of a saturated solution of calcium fluoride is 2×10⁻⁴ moles per litre. Its solubility product is:
Ksp=(2×10⁻⁴)(4×10⁻⁴)²=32×10⁻¹².