# Question #07cb2

Nov 21, 2016

Here's what I got.

#### Explanation:

Molar solubility is all about the number of moles of a given ionic compound that can be dissolved in one liter of saturated aqueous solution.

In other words, a compound's molar solubility tells you how many moles of said compound can be dissolved in exactly one liter of water in order to make a saturated aqueous solution.

Now, zinc carbonate, ${\text{ZnCO}}_{3}$, is considered insoluble in aqueous solution, which means that only very, very small amounts dissolve when placed in water.

When you dissolve zinc carbonate in water, an equilibrium is established between the undissolved solid and the dissolved ions

${\text{ZnCO"_ (3(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + "CO}}_{3 \left(a q\right)}^{2 -}$

Notice that every mole of zinc carbonate that dissociates produces $1$ mole of zinc cations, ${\text{Zn}}^{2 +}$, and $1$ mole of carbonate anions, ${\text{CO}}_{3}^{2 -}$.

The solubility product constant, ${K}_{s p}$, is a measure of the degree of dissociation, i.e. how many ions are produced in solution.

The ${K}_{s p}$ of zinc carbonate is

${K}_{s p} = \left[{\text{Zn"^(2+)] * ["CO}}_{3}^{2 -}\right]$

The expression of the solubility product constant uses equilibrium concentrations.

If you take $s$ to be the molar concentration of the dissolved zinc cations at equilibrium, you can say that the molar concentration of the carbonate anions will also be equal to $s$, since both ions are produced in equal amounts when the salt dissociates.

$\left[{\text{Zn"^(2+)] = ["CO}}_{3}^{2 -}\right] = s$

This means that you have

${K}_{s p} = s \cdot s = {s}^{2}$

Rearrange to solve for $s$

$s = \sqrt{{K}_{s p}}$

Plug in your value to find

$1.46 \cdot {10}^{- 10} = {s}^{2} \implies s = \sqrt{1.46 \cdot {10}^{- 10}} = 1.21 \cdot {10}^{- 5}$

Here $s$ represents the molar concentration of the dissolved ions, which implies that it's also equal to the molar solubility of the salt, i.e. how many moles of the solid dissociate to produce moles of dissolved ions.

Therefore, you have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{s = 1.21 \cdot {10}^{- 5} \text{M}}}} \to$ molar solubility

The answer is rounded to three sig figs.

So, you can say that you can only dissolve $1.21 \cdot {10}^{- 5}$ moles of solid zinc carbonate in $\text{1 L}$ of water, presumably at room temperature. Any solid that you add past this mark will remain undissolved.