Question

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is the molarity of the calcium ion in the original solution?

Answer 1

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M