What is the molar solubility of calcium sulfate in pure water?

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

Nov 14, 2016

We assess the following equilibrium:

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

Explanation:

And K_(sp)=([Ca^(2+)][SO_4^(2-)])/[[CaSO_4(s)]

But since $\left[C a S {O}_{4} \left(s\right)\right]$, the concentration of a solid, is a meaningless concept, the equilibrium is governed solely by the ion product, i.e.

${K}_{s p} = \left[C {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] = 2.4 \times {10}^{-} 5$.

If we say the solubility of $C a S {O}_{4} = S$, then,

${K}_{s p} = \left[C {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] = S \times S = {S}^{2} = 2.4 \times {10}^{-} 5$

And thus $S = \sqrt{2.4 \times {10}^{-} 5}$ $=$ $4.90 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

Given that $C a S {O}_{4}$ has a molar mass of $136.14 \cdot g \cdot m o {l}^{-} 1$,

$S = 136.14 \cdot g \cdot \cancel{m o {l}^{-} 1} \times 4.90 \times {10}^{-} 3 \cdot \cancel{m o l} \cdot {L}^{-} 1 \cong 0.7 \cdot g \cdot {L}^{-} 1$

Note that this problem assumes standard conditions of $298 \cdot K$ and $1 \cdot a t m$ pressure. Under non-standard conditions, say at $398 \cdot K$, how do you think the solubility would evolve?