# Given K_"sp" ZnCO_3=1.46xx10^-10, what is the "ppm" concentration of a saturated solution?

Jul 5, 2017

$S = 1.21 \times {10}^{-} 5 M = 1.51 \times {10}^{-} 3 \left(\frac{g}{L}\right)$

#### Explanation:

$Z n C {O}_{3} r i g h t \le f t h a r p \infty n s Z {n}^{+ 2} + C {O}_{3}^{2 -}$ is a 1:1 ionization ratio. The solubility can be determined using $S = \sqrt{K s p}$

$S = \sqrt{K s p} = \sqrt{1.46 \times {10}^{-} 10} M = 1.21 \times {10}^{-} 5 M$

$1.21 \times {10}^{-} 5 \frac{m o l}{L} \times \frac{125 g}{\text{mol}} = 1.51 \times {10}^{-} 3 \left(\frac{g}{L}\right)$

Jul 5, 2017

$\text{Solubility"_(ZnCO_3) = "1 ppm}$

#### Explanation:

We consider the equilibrium reaction.........

$Z n C {O}_{3} \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s Z {n}^{2 +} + C {O}_{3}^{2 -}$

And ${K}_{s p} = \left[Z {n}^{2 +}\right] \left[C {O}_{3}^{2 -}\right] = 1.46 \times {10}^{-} 10$

We let the ${\text{Solubility}}_{Z n C {O}_{3}} = x \cdot m o l \cdot {L}^{-} 1$

And thus ${\text{Solubility}}_{Z n C {O}_{3}} = \sqrt{1.46 \times {10}^{-} 10}$

$= 1.21 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

And thus the $\text{gram solubility}$ of $\text{zinc carbonate}$

$\equiv 1.21 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1 \times 125.4 \cdot g \cdot m o {l}^{-} 1 \cong 1 \cdot m g \cdot {L}^{-} 1$, i.e. $\text{ppm}$ concentrations.