At 310 K, the solubility of CaF2 in water is 2.34 ×10–3 g/100 mL. The solubility product of CaF2 is _____ × 10–8 (mol/L)3.
(Given molar mass : CaF2 = 78 g mol–1).
(Given molar mass : CaF2 = 78 g mol–1).
Correct Answer: 0
Solution and Explanation
\(CaF2 ⇋ Ca^2++2F_{2s}^-\)
Ksp = s(2s)2
= 4s3
Solubility(s) = 2⋅34 × 10–3 g/100 mL
= \(\frac{2.34×10-3×10}{78} \) mole/lit
= 3×10-4 mole/lit
∴Ksp = 4×(3×10-4)3
= 108×10-12
= 0.0108×10-8(mole/lit)3
∴ x ≈ 0
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Concepts Used:
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Ordinary Differential Equations is an equation that indicates the relation of having one independent variable x, and one dependent variable y, along with some of its other derivatives.
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A partial differential equation is a type, in which the equation carries many unknown variables with their partial derivatives.
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Homogeneous Differential Equations:
When the degree of f(x,y) and g(x,y) is the same, it is known to be a homogeneous differential equation.
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