Question

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 k?

Answer 1

Answer:

CaSO4 <=> Ca²+ + SO4²-

Let s is the solubility of CaSO4

then, we know, Ksp = [Ca²+][SO4²-]

Ksp = s.s = s²

A/C to question,

Ksp = 9.1 × 10^-6

so, 9.1 × 10^-6 = s²

s = √{9.1 × 10^-6} = 3.017 × 10^-3 M

so, solubility of CaSO4 = 3.017 × 10^-3 M

= 3.017 × 10^-3 mol/L

= 3.017 × 10^-3 × 136g/L [ weight = molecule weight × mole . and molar weight = 136g/mol]

= 410.3 × 10^-3 g/L

= 0.4103 g/L

it means 0.4103 g CaSO4 is dissolved in 1 L

Therefore, 1g CaSO4 is dissolved in 1/0.4103

= 2.437 L