# If given the following solubilities, how do you calculate the K_(sp) for each compound?

## (a) $C u S C N$, 5.0 mg/L (b) $S n S$, 2.0 x ${10}^{-} 5$ g/L (c) $C o {\left(O H\right)}_{2}$, 3,2 x ${10}^{3}$ g/L (d) $A {g}_{2} C r {O}_{4}$, 3.4 x ${10}^{-} 2$ g/L?

Jun 2, 2018

I will do you ONE example.....$d .$

#### Explanation:

We examine the solubility equilibrium...

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

For which we write the solubility expression....

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

But if $S = \text{solubility of silver chromate...}$, then $\left[A {g}^{+}\right] = 2 S$, and $\left[C r {O}_{4}^{2 -}\right] = S$...thus ${K}_{\text{sp}} = {\left(2 S\right)}^{2} S = 4 {S}^{3}$...

Now $S = \frac{\frac{3.4 \times {10}^{-} 2 \cdot g}{331.73 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = 1.0025 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...

${K}_{\text{sp}} = 4 \times {\left(1.0025 \times {10}^{-} 4\right)}^{3} = 4.03 \times {10}^{-} 12$.

When you do the others perhaps you might post the solutions in this thread?