# If 3.06*g "strontium chromate" is added to 1/2*L of water, what is the UNDISSOLVED mass of salt?

Apr 10, 2017

We interrogate the equilibrium:

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

I get approx. $2.5 \cdot g$ strontium chromate as the undissolved mass.

#### Explanation:

Now we have the volume of the solvent, and we can calculate the molar quantity of strontium chromate. We need another parameter, ${K}_{s p}$ for strontium chromate.......

This site gives ${K}_{s p}$ $S r C r {O}_{4} = 2.2 \times {10}^{-} 5$ (which is higher than I would have expected, I hope the source is kosher).

Given ${K}_{s p}$, $\left[S {r}^{2 +}\right] \left[C r {O}_{4}^{2 -}\right] = 2.2 \times {10}^{-} 5$

If $S = \text{solubility of strontium chromate}$, then ${S}^{2} = {K}_{s p}$;

and $S = {\sqrt{K}}_{s p} = \sqrt{\left[S {r}^{2 +}\right] \left[C r {O}_{4}^{2 -}\right]} = 4.69 \times {10}^{- 3} \cdot m o l \cdot {L}^{-} 1$.

Which leads to a dissolved mass of............

$= \frac{1}{2} \cdot L \times 203.61 \cdot g \cdot m o {l}^{-} 1 \times 4.69 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

$= 0.478 \cdot g$ with respect to $\text{strontium chromate}$.

And thus, for a saturated solution, the mass of strontium chromate remaining is $\left(3.06 - 0.478\right) \cdot g = 2.58 \cdot g$