# What volume of water is required to dissolve a 6.7*g mass of "calcium sulfate" if K_(sp)*CaSO_4=2.4xx10^-5?

May 29, 2017

A volume of $10 \cdot L$ is required........

#### Explanation:

We assess the equilibrium reaction.......

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

Where ${K}_{\text{eq"=K_"sp}} = \left[C {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right]$.

And if we put $\left[C {a}^{2 +}\right] = \left[S {O}_{4}^{2 -}\right] = x$..............

.......then $x = \sqrt{{K}_{\text{sp}}} = \sqrt{2.4 \times {10}^{-} 5} = 0.005 \cdot m o l \cdot {L}^{-} 1$

And so ((6.7*g)/(136.14*g*mol^-1))/(??*L)=0.005*mol*L^-1

i.e. $\frac{\frac{6.7 \cdot g}{136.14 \cdot g \cdot m o {l}^{-} 1}}{0.005 \cdot m o l \cdot {L}^{-} 1} \cong 10 \cdot L$