# Question 17a7b

Oct 20, 2017

Here's my take on this.

#### Explanation:

The idea here is that you need to figure out the mass of calcium carbonate that must be dissolved per liter of water in order to produce a concentration of

${\text{45 mg L"^(-1) = 45 color(white)(.)color(red)(cancel(color(black)("mg")))/"L" * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = "0.045 g L}}^{- 1}$

of calcium cations, ${\text{Ca}}^{2 +}$, in the solution.

It's important to keep in mind that calcium carbonate is actually poorly soluble in water, meaning that, at ${25}^{\circ} \text{C}$, you can only dissolve about $\text{0.013 g}$ of calcium carbonate per liter of water $\to$ see here color(darkorange)((!)).

This implies that the mass of calcium carbonate that can be dissolved in $\text{1 L}$ of water cannot exceed $\text{0.013 g}$ at ${25}^{\circ} \text{C}$.

Use the molar mass of the calcium cations, which is essentially equal to the molar mass of elemental calcium, to calculate the number of moles of calcium cations present in the solution.

0.045 color(red)(cancel(color(black)("g"))) * "1 mole Ca"^(2+)/(40color(red)(cancel(color(black)("g")))) = "0.001125 moles Ca"^(2+)

Now, the dissociation equilibrium for calcium carbonate looks like this

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

As you can see, every mole of calcium carbonate that dissociates produces $1$ mole of calcium cations in aqueous solution.

This implies that $0.001125$ moles of calcium carbonate must be dissolved in $\text{1 L}$ of water in order to produce that many moles of calcium cations.

Use the molar mass of calcium carbonate to convert this to grams

0.001125 color(red)(cancel(color(black)("moles CaCO"_3))) * ((40 + 12 + 3 * 16)color(white)(.)"g")/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "0.1125 g"

Notice that this value exceeds the value listed in color(darkorange)((!)), which implies that you cannot dissolve enough calcium carbonate in $\text{1 L}$ of water at ${25}^{\circ} \text{C}$ in order to get a concentration of ${\text{45 mg L}}^{- 1}$ of calcium cations.

In fact, you can start from the solubility of calcium carbonate and figure out the maximum concentration of calcium cations that can be produced in $\text{1 L}$ of water at ${25}^{\circ} \text{C}$.

0.013 color(red)(cancel(color(black)("g"))) * "1 mole CaCO"_3/((40 + 12 + 3 * 16)color(red)(cancel(color(black)("g")))) = "0.00013 moles CaCO"_3

This implies that you will get $0.00013$ moles of calcium cations in the solution, the equivalent of

0.00013 color(red)(cancel(color(black)("moles Ca"^(2+)))) * "40 g"/(1color(red)(cancel(color(black)("mole Ca"^(2+))))) = "0.0052 g"#

You can thus say that the maximum concentration of calcium cations that can be obtained by dissolving calcium carbonate in $\text{1 L}$ of water at ${25}^{\circ} \text{C}$ is equal to ${\text{5.2 mg L}}^{- 1}$.