Question

The molar solubility of CaSO4 in water is 0 .67 gram per liter of solution. Calculate the Ksp.

Answer 1

Answer:

Ksp = 2.4 * 10⁻⁵

Explanation:

The equation for the dissolution and dissociation of CaSO4 in water is given as: CaSO₄ ---> Ca²⁺ + SO₄²⁻

The expression for the Ksp of the dissociation equation above is:

Ksp = [Ca²⁺] [SO₄²⁻]

The molar ratio of the dissociated ions and the solute is 1 : 1, this means that every 1 mole of CaSO₄ that dissolves produces 1 mole of Ca²⁺ and SO₄²⁻ each

The molar solubility of a substance is the number of moles that dissolve per liter of solution.

molar solubility of CaSO₄ = number of moles /liter of solution

number of moles of CaSO₄ = mass/molar mass

molar mass of CaSO₄ = 136 g/mol

number of moles of CaSO₄ = 0.67/136 = 0.0049 moles

molar solubility of CaSO₄ = 0.0049 mol/ 1 L =  4.9 * 10⁻³ moles per liter

therefore, molar solubility of CaSO₄ = 4.9 * 10⁻³

Ksp = (4.9 * 10⁻³) (4.9 * 10⁻³)

Ksp = 2.4 * 10⁻⁵

Answer 2

Answer:

2.4 * 10^-5

Explanation:

Molar solubility of CaSO4  = 0 .67 g/L/136 g/mol = 4.9 * 10^-3 Mol/L

Given that;

CaSO4(s) -------> Ca^2+(aq) + SO4^2-(aq)

Hence, Ksp = [Ca^2+] [SO4^2-]

Where  [Ca^2+] =[SO4^2-]=s

Ksp = s^2

since s = 4.9 * 10^-3 Mol/L

Ksp = (4.9 * 10^-3 Mol/L)^2

Ksp = 2.4 * 10^-5