Question

What is the concentration of chloride ions in a solution whose concentration is $ 6 \times {10^{ - 3}}mol.{L^{ - 1}} $ with respect to calcium chloride?

Answer 1

Hint: It is said that the concentration is $ 6 \times {10^{ - 3}}mol.{L^{ - 1}} $ with respect to calcium chloride. This means that the Molarity of the Calcium Chloride solution is $ 6 \times {10^{ - 3}}mol.{L^{ - 1}} $ . We can solve this question by dissociating the salt calcium chloride into constituent ions and determining the concentration of each ion.

Complete Step By Step Answer:
The molecular formula of Calcium Chloride is $ CaC{l_2} $ . It constitutes 1 atom of Calcium and 2 atoms of chlorine. The molarity of calcium chloride indicates the no. of moles of calcium chloride per unit volume of the solution. Molarity is also known as concentration and has the SI unit of mol/L or $ mol/d{m^3} $ . The formula for molarity can be given as: $ Molarity = \dfrac{n}{V} $
Where, n is the no. of moles and V is the volume of the solution ( in litres).
The salt calcium chloride is sparingly soluble in water, hence it partially dissociates into its constituent ions. The dissociation can be shown as:
 $ CaC{l_2} \rightleftharpoons C{a^{ + 2}} + 2C{l^ - } $
After dissociation it leads to the formation of one calcium cation and two chloride ions. i.e. $ 1mol{\text{ }}CaC{l_2} = 1\;mol{\text{ }}C{a^{ + 2}} + 2mols{\text{ }}C{l^ - } $
Therefore, $ 1mol{\text{ }}CaC{l_2} = 2mols{\text{ }}C{l^ - } $
Since the volume of solution isn’t given, we will consider it as 1L itself. The no. of moles of Calcium Chloride solution thus will be equal to the concentration = $ 6 \times {10^{ - 3}}mol $
Now, $ 6 \times {10^{ - 3}}mol $ of $ CaC{l_2} = 2 \times 6.0 \times {10^{ - 3}}moles $ of Chloride ions.
Hence the no. of moles of Chloride ions will be $ = 12 \times {10^{ - 3}}mol = 1.2 \times {10^{ - 2}}mols $
The concentration will also be the same as the no. of moles as the volume considered here is 1L. The concentration of chloride ions will be $ 1.2 \times {10^{ - 2}}mol.{L^{ - 1}} $
This is the required answer.

Note:
If we are given a salt that dissociates as $ A \rightleftharpoons B + 3C $ then the concentration of C ions will be three times the concentration of A. For this purpose, always remember to balance the chemical equation to obtain the correct stoichiometries.