Question

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10^–6).

Answer 1

\[\ce{CaSO_{4(s)} <=> Ca^{2+}_{(aq)} + SO_{4(aq)}^{2-}}\]

`"K"_("sp") = ["Ca"^(2+)]["SO"_4^(2-)]`

Let the solubility of CaSO₄ be s.

Then `"K"_("sp") = "s"^2`

`9.1 xx 10^(-6) = "s"^2`

`"s" = 3.02 xx 10^(-3) "mol/L"`

Molecular mass of CaSO₄ = 136 g/mol

Solubility of `"CaSO"_4` in gram/L = 3.02 × 10^–3 × 136

= 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄

Therefore, to dissolve 1g of CaSO₄ we require = `1/0.41 "L" = 2.44 " L"` of water.