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=(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}}=(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