If $\pu{50.0 mg}$ of $\ce{CO3^2-}$ and $\pu{50.0 mg}$ of $\ce{Ca^2+}$ are present in one liter of water, what will be the final (equilibrium) concentration of $\ce{Ca^2+}$ expressed in moles per liter?


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  • 2
    $\begingroup$ This is a weird question. It's not like you can just add 50 mg of a cation and 50 mg of an anion. They have to come with counterions. If $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ are the only ions in solution, then they need to be in equimolar amounts. Now, if we added 50 mg of $\ce{Na2CO3}$ as a source of carbonate and 50 mg of $\ce{Ca(NO3)2}$ as a source of calcium, then we have a more viable question. Also, have you cover solubility product constants yet? $\endgroup$ – Ben Norris Jun 3 '15 at 10:52

The premise of your question is weird. $\pu{50 mg}$ of calcium ion and $\pu{50 mg}$ of carbonate ion are not equimolar. However, let's assume that they have appropriate counterions somewhere. This problem can be approached two ways.

Limiting reactant

We have a precipitation reaction between the calcium and the carboncate ions, as calcium carbonate is nearly insoluble in water: $$\ce{Ca^2+ (aq) + CO3^2- (aq) -> CaCO3 (s) v}$$

Convert your masses of ions to amount of substance and figure out which ion is in excess. If calcium ion is in excess, then use the amount of substance in excess to determine the remaining calcium concentration.
For example: Let's say you determined that your solution has $\pu{0.1663 mol}$ of $\ce{Ca^2+}$ and $\pu{0.1221 mol}$ of $\ce{CO3^2-}$. The maximum amount of $\ce{CaCO3}$ that could precipitate is $\pu{0.1221 mol}$, because we would run out of carbonate ion. Then there would be $\pu{0.0442 mol}$ of calcium ion left in $\pu{1 L}$ of solution for a concentration of $\pu{0.0442 M}$.

Solubility product

Calcium carbonate is not completely insoluble. It's solubility product constant is $4.8\times 10^{-9}$. Thus we can set this up as an ICE equilibrium problem. I'll use the same fake amount of substances from the previous example:

\begin{align} \ce{CaCO3 (s) &<=> Ca^2+ (aq) + CO3^2- (aq)}\\ K_\mathrm{sp}&=[\ce{Ca^2+}][\ce{CO3^2-}] \end{align} \begin{array}{lcc} \hline & \text{calcium}/\pu{mol} & \text{carbonate}/\pu{mol}\\ \hline \text{Initial} & 0.1663 & 0.1221 \\ \text{Change} & -x & -x \\ \text{Equilibrium} & 0.1663 - x & 0.1221 - x \\ \hline \end{array} \begin{align} 4.8\times 10^{-9}&=(0.1663-x)(0.1221-x) \end{align}

Solving for $x$ gives you $x=0.1221,0.1663$ and only the first number makes sense. In this case the final calcium amount in solution would still be $\pu{0.0442 mol}$. The carbonate amount in solution is not actually zero. It is $\pu{1.09E-7 M}$, but that is for another day.

  • $\begingroup$ I agree that the premise of the question is weird, but I think the question means that you have 50 mg of $\ce{CO3^{2-}}$ from an appropriate carbonate source (e.g. 88.33 mg of anhydrous $\ce{Na2CO3}$) and 50 mg of $\ce{Ca^{2+}}$ from maybe 273.3 mg of $\ce{CaCl2.6H2O}$. On a side note, if we were very particular about accuracy then we should probably worry about the $\ce{HCO3^-}$/$\ce{CO3^{2-}}$ equilibrium, but that is also for another day :) $\endgroup$ – orthocresol Jun 3 '15 at 12:32

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