Question

The solubility product of CaSO₄·2H₂O is 2.4 × 10⁻⁵. What mass of this salt will dissolve in 1.0 L of 0.010 M SO₄²⁻?

a) 2.4 × 10⁻⁵ g
b) 1.8 × 10⁻³ g
c) 5.0 × 10⁻⁷ g
d) 3.2 × 10⁻⁴ g

Answer 1

To determine the mass of CaSO₄·2H₂O that will dissolve in 1.0 L of 0.010 M SO₄²⁻, we can use the solubility product expression. The mass of CaSO₄·2H₂O is 3.2 × 10⁻⁴ g. Therefore, the correct answer is d).

Step-by-step explanation:

To determine the mass of CaSO₄·2H₂O that will dissolve in 1.0 L of 0.010 M SO₄²⁻, we can use the solubility product expression. The solubility product (Ksp) of CaSO₄·2H₂O is 2.4 × 10⁻⁵. The equilibrium concentrations of Ca²⁺ and SO₄²⁻ ions will be equal to the concentration of SO₄²⁻, which is 0.010 M.

We can substitute these equilibrium concentrations into the solubility product expression:

Ksp = [Ca²⁺][SO₄²⁻]

2.4 × 10⁻⁵ = [Ca²⁺][0.010]

From this equation, we can solve for [Ca²⁺]:

[Ca²⁺] = 2.4 × 10⁻⁵ / 0.010

[Ca²⁺] = 2.4 × 10⁻⁴ M

Since 1 mole of CaSO₄·2H₂O dissociates to produce 1 mole of Ca²⁺ ions, the mass of CaSO₄·2H₂O that will dissolve in 1.0 L of the solution can be calculated using the molar mass of CaSO₄·2H₂O:

Mass of CaSO₄·2H₂O = molar mass × moles dissolved = (136.14 g/mol) × (2.4 × 10⁻⁴ mol/L)

Mass of CaSO₄·2H₂O = 3.26 × 10⁻² g

Therefore, the correct answer is d) 3.2 × 10⁻⁴ g.