I have a precipitation reaction of this form:

$\ce{Ca^{2+} + CO_{3}^{2-} → CaCO3}$

How many mole of $\ce{Ca^{2+}}$ ion is contained in a 200ml solution?

I'd like to know how to calculate this, thanks for the help!

  • $\begingroup$ Please see the link below in the close reason for information on asking your homework questions on the site. $\endgroup$ – jonsca Dec 5 '13 at 0:23

Carbonates are usually insoluble in water except for when combined with group 1 elements and and ammonium ions, however, as you may already know, almost everything is slightly soluble in water.

Generally in order to see a precipitate form, a solution of sodium carbonate and aqueous calcium chloride are mixed. The calcium ions and carbonate ions are then in contact and can form the calcium carbonate precipitate. The other product is sodium chloride which is known to be soluble in water.

Please keep in mind that $ M = \frac{moles \; solute}{liter \; solution}$

Which is expressed as mol per liter.

The solubility of calcium carbonate is very dependent upon the pH of its environment. Assuming that this reaction occurs in water, the pH will be very close to 7. Tap water generally contains many ions, chloride for instance that will interfere with our calculation in the moles of calcium ions that we have.

First, we need the solubility constant product denoted as $K_{sp}$.


Taken from the UC Davis ChemWiki "the Solubility Product Constant, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. A more a substance dissolves, the higher the Ksp value it has."

The expression for the solubility product of calcium carbonate is $$[Ca^{2+}][CO_3^{2-}] = 8.7 * 10^{-9} $$

The concentration of each ion is 1:1 as shown in our equation, so there is no need to square the concentrations.

If S mol of $CaCO_3$ dissolved in 1L, the equilibrium concentration of $Ca^{2+}$ will be $[Ca^{2+}] = S$. The concentration of $CO_3^{2-}$ ions will be $[CO_3^{2-}] = S$ again, because each mol of $CaCO_3$ each produces one mol of $Ca^{2+}$ ions and one mol of $CO_3^{2-}$ ions. Therefore,

$$[Ca^{2+}][CO_3^{2-}] = S * S = S^2 = K_sp = 8.7 * 10^{-9}$$

Solving for $S^2$ gives

$$ S^2 = 8.7 * 10^{-9}$$

Taking the square root on each side of this equation gives

$$ \sqrt{S^2} = \sqrt{8.7 * 10^{-9}}$$

$$ S = 9.327*10^{-5}$$

The equilibrium concentrations are therefore

$$ [Ca^{2+}] = S = 9.327*10^{-5} M $$ $$ [CO_3^{2-}] = S = 9.327*10^{-5} M $$

Since we have our concentration of $Ca^{2+}$ ions, we can then calculate the number of moles of ions that we have.

$$ M = \frac{moles \; solute}{liter \; solution}$$

$$9.327*10^{-5} M = \frac{moles \; solute}{1 L}$$

$$ mol = 9.327*10^{-5} M * L^{-1} * \frac{1L}{1000mL} * 200 = 1.8654*10^{-5} \; mol \; of \; Ca^{2+} ions $$

For completeness sake, lets take a look of the number of moles in grams.

Because the molar mass of $CaCO_3$ is $100.0869 g/mol$, using our our definition of molarity from above, we can calculate the gram solubility which is

$$ gram \; solubility = (9.327*10^{-5} \; mol \; L^{-1})(100.0869 \; g/mol) = 0.009335 \; g/L$$

Converting this to our 200mL solution,

$$ 0.009335 \frac{g}{L} * \frac{1L}{1000mL} * 200mL = 0.001867 g$$

Last note:

When $$CaCO_3(s) \rightarrow Ca(aq) + CO_3(aq)$$

some of the carbonate hydrolyses to produce HCO3.

$$CO_3(aq) + H_2O(l) \rightarrow HCO_3(l) + OH(aq)$$

In this instance, the hydrolysis of carbonate decreases the concentration of carbonate, which pulls the solubility equilibrium to the right making calcium carbonate more soluble.


Please take a look at the link provided, as the actual number of moles will vary due to the pH of the environment. They nicely derive the formula at eq. 11 which you can then find the number of moles of $Ca^{2+}$ ions.

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  • $\begingroup$ Thanks so much for your extensive answers! However, it is best to prompt the user asking the question with hints before offering a full answer. If you see questions of this type (just a homework question with no effort towards a solution), please flag them as "off-topic". $\endgroup$ – jonsca Dec 5 '13 at 0:23
  • $\begingroup$ @jonsca My apologies. I am a little new here. :) $\endgroup$ – Jun-Goo Kwak Dec 5 '13 at 1:24

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