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
Approach Solution - 1
To find the solubility product (Ksp) of CaF2, follow these steps:
- Calculate Molar Solubility: The solubility of CaF2 in water is given as 2.34 × 10–3 g/100 mL. Convert this to molarity (mol/L).
Molarity = (solubility in g/mL ÷ molar mass) × 1000
Given molar mass of CaF2 is 78 g/mol,
Molarity = (2.34 × 10–3 g/100 mL ÷ 78 g/mol) × 1000 = 3.0 × 10–3 mol/L.
- Write the Dissolution Equation:
CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
- Determine Ion Concentrations:
Let s be the solubility in mol/L. Thus, at equilibrium, [Ca2+] = s and [F–] = 2s.
From molarity, s = 3.0 × 10–3 mol/L, so [Ca2+] = 3.0 × 10–3 mol/L and [F–] = 2 × 3.0 × 10–3 = 6.0 × 10–3 mol/L.
- Apply the Solubility Product Expression:
Ksp = [Ca2+] × [F–]2
Ksp = (3.0 × 10–3) × (6.0 × 10–3)2
Ksp = 3.0 × 10–3 × 36.0 × 10–6 = 1.08 × 10–7 mol3/L3
- Verify the Range:
The calculated Ksp is 1.08 × 10–7, which falls within the expected range interpreted as closely derived from precision measurements.
Approach Solution -2
\(CaF2 ⇋ Ca^2++2F_{2s}^-\)
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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Partial Differential Equations:
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Homogeneous Differential Equations:
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