# Ksp=2.4x10^-5 for calcium sulfate. What is the molar solubility of calcium sulfate in pure water? What is the mass solubility of calcium sulfate in pure water, expressed in g/L?

## Ksp=2.4x10^-5 for calcium sulfate. What is the molar solubility of calcium sulfate in pure water? What is the mass solubility of calcium sulfate in pure water, expressed in g/L?

Sep 7, 2016

(a)

$\textsf{4.9 \times {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$

(b)

$\textsf{0.67 \textcolor{w h i t e}{x} \text{g/l}}$

#### Explanation:

(a)

Calcium sulfate is sparingly soluble. In a saturated solution the aqueous ions are in equilibrium with the insoluble salt:

$\textsf{C a S {O}_{4 \left(s\right)} r i g h t \le f t h a r p \infty n s C {a}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -}}$

For which

$\textsf{{K}_{s p} = \left[C {a}_{\left(a q\right)}^{2 +}\right] \left[S {O}_{4 \left(a q\right)}^{2 -}\right] = 2.4 \times {10}^{- 5} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{-} 2}$

The solubilty $\textsf{s}$ of the salt is equal to $\textsf{\left[C {a}_{\left(a q\right)}^{2 +}\right]}$.

Since $\textsf{\left[C {a}_{\left(a q\right)}^{2 +}\right] = \left[S {O}_{4 \left(a q\right)}^{2 -}\right]}$ we can write:

$\textsf{{K}_{s p} = {s}^{2} = 2.4 \times {10}^{- 5}}$

$\therefore$$\textsf{s = \sqrt{2.4 \times {10}^{- 5}} = 4.9 \times {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$

This is the molar solubility.

(b)

To find the mass solubility we multiply by the mass of 1 mole. I will assume we are using the anhydrous salt $\textsf{C a S {O}_{4}}$ for which the $\textsf{{M}_{r} = 136.1}$

$\therefore$ Mass solubility = $\textsf{4.9 \times {10}^{- 3} \times 136.1 = 0.67 \textcolor{w h i t e}{x} \text{g/l}}$