Respuesta :
Using Common ion effect and Le-Chatlier's Principle,
Let, conc. of Ca2+ present in solution = a,
SO4(2-) ions are present in solution because of CaSO4 and Na2SO4,
Let conc. of SO4^(2-) ions present in solution because of CaSO4 = b
and conc. of SO4^(2-) ions present in solution because of Na2SO4 = c = 0.1 m (given)
Dissociation of CaSO4 can be represented as follows,
CaSO4 ↔ Ca^2+ + SO4^(2-)
∴ a = b
Given that Ksp of CaSO4 = 2.4 X 10^(-5)
In present system, Ksp = a(b+c) = a(a+ 0.1) = 2.4 X 10^-5
Now, CaSO4 is a sparingly soluble salt, ∴ a <<< 0.1, ∴ a + 0.1 ≈ 0.1
∴ Ksp = 0.1a = 2.4 X 10^(-5)
∴ a = 2.4 X 10^(-4) m
Thus, effective conc. of CaSO4 in solution is 2.4 X 10^(-4) m
Let, conc. of Ca2+ present in solution = a,
SO4(2-) ions are present in solution because of CaSO4 and Na2SO4,
Let conc. of SO4^(2-) ions present in solution because of CaSO4 = b
and conc. of SO4^(2-) ions present in solution because of Na2SO4 = c = 0.1 m (given)
Dissociation of CaSO4 can be represented as follows,
CaSO4 ↔ Ca^2+ + SO4^(2-)
∴ a = b
Given that Ksp of CaSO4 = 2.4 X 10^(-5)
In present system, Ksp = a(b+c) = a(a+ 0.1) = 2.4 X 10^-5
Now, CaSO4 is a sparingly soluble salt, ∴ a <<< 0.1, ∴ a + 0.1 ≈ 0.1
∴ Ksp = 0.1a = 2.4 X 10^(-5)
∴ a = 2.4 X 10^(-4) m
Thus, effective conc. of CaSO4 in solution is 2.4 X 10^(-4) m
The effective conc. or solubility of CaSO₄ in solution is [tex]2.4 \times 10^{-4} \ m[/tex]
Ksp of CaSO₄ = [tex]2.4 \times 10^{-5}[/tex]
concentration of Na₂SO₄ = 0.1 m
It is required to calculate the molar solubility of CaSO₄
What is molar solubility ?
Molar solubility, which is directly related to the solubility product, is the number of moles of the solute that can be dissolved per liter of solution before the solution becomes
Using the Common ion effect
Let, the conc. of Ca²⁺ present in solution = a,
[tex]SO_4^{2-}[/tex] ions are present in solution because of CaSO₄ and Na₂SO₄,
Let conc. of [tex]SO_4^{2-}[/tex] ions present in solution because of CaSO₄ = b
and conc. of [tex]SO_4^{2-}[/tex] ions present in solution because of Na₂SO₄,= c = 0.1 m (given)
Dissociation of CaSO₄ can be represented as follows,
[tex]CaSO_4\to Ca^{2+} + SO_4^{2-}[/tex]
∴ a = b
Given that Ksp of CaSO₄ = [tex]2.4 \times 10^{-5}[/tex]
In present system, Ksp = a(b+c) = a(a+ 0.1) = [tex]2.4 \times 10^{-5}[/tex]
Now, CaSO₄ is a sparingly soluble salt,
∴ a <<< 0.1, ∴ a + 0.1 ≈ 0.1
∴ Ksp = 0.1×a = [tex]2.4 \times 10^{-4}[/tex]
∴ a = [tex]2.4 \times 10^{-4}[/tex] m
Thus, effective conc. or solubility of CaSO₄ in solution is [tex]2.4 \times 10^{-4} \ m[/tex]
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