Question

The volume of water needed to dissolve 0.0546 grams of calcium sulfite is L. Assume no volume change upon addition of the solid.

ksp of CaSO3

1.3 × 10-8

Answer 1

Given Terms:

Quantity of Calcium Sulfite (CaSO3) = 0.0546 grams

Ksp of calcium sulfite = 1.3 x 10^-8

Now, we know,

Solubility of CaSO4= K = V1.3 X 10-8 = 1.14X10-4M

And Molar Mass of Cas03 = 120.17g/mol

Therefore,

Mass of CaSO3 = 1.14X10-4X(Molar Mass of CaSO3) = 1.14X10-4 X120.179/L

Mass of CaSO3 = 136.99g/L

This means that volume of water needed to dissolve 1 g of CaSO3 will be,

136.99 X 10-4 litres

For calculating volume of water required for 0.0546 g of calcium sulfite is,

0.0546 136.99 X 10-4 = 0.98 Litres

Hence, 3.98 Litres is the answer.