The value of K_(sp) for SrSO_4 is 2.8 times 10^-7. What is the solubility of SrSO_4 in moles per liter?

Sep 5, 2017

$\text{Solubility}$ $\cong$ 100*mg*L^-1?

Explanation:

Now we interrogate the solubility equilibrium.....

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

And ${K}_{\text{sp}} = \left[S {r}^{2 +}\right] \left[S {O}_{4}^{2 -}\right]$

We write $S = \text{solubility of strontium sulfate.}$ And thus $S = \left[S {r}^{2 +}\right]$, and $S = \left[S {O}_{4}^{2 -}\right]$...and we substitute these values into the solubility expression......

${K}_{\text{sp}} = \left[S {r}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] = S \times S = {S}^{2}$

And thus..............................

$S = \sqrt{{K}_{\text{sp}}} = \sqrt{2.8 \times {10}^{-} 7} = 5.29 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$.

And the gram solubility is thus.....

5.29xx10^-4*mol*L^-1xx183.68*mol*L^-1=??

If we attempted to dissolve the slat up in say $0.100 \cdot m o l \cdot {L}^{-} 1$ $\text{sodium sulfate}$, do you think the solubility would increase or decrease? Why?