Question

Given $20\;mL$ of $0.2M\;A{l_2}{\left( {S{O_4}} \right)_3}$ is mixed with $20\;mL$ of $6.6M\;BaC{l_2}$, the concentration of $C{l^ - }$ ion in the solution is:
A. $0.2M$
B. $6.6M$
C. $0.02M$
D. $0.06M$

Answer 1

For the given question, $BaC{l_2}$ will react with $A{l_2}{(S{O_4})_2}$ to form a precipitate of barium sulphate while the aluminium chloride formed on reaction will remain dissociated as aluminium and chloride ions. So, the number of chloride ions will remain unchanged in the solution and thus, we can calculate the concentration of chloride ions by simply calculating the number of moles of chloride ions in $BaC{l_2}$.

Complete answer:
When barium chloride reacts with aluminium sulphate then the formation of aluminium chloride takes place along with the formation of barium sulphate as a precipitate. The reaction takes place as follows:
$A{l_2}{(S{O_4})_3} + 3BaC{l_2} \to 2AlC{l_3} + 3BaS{O_4}(ppt.)$
As the number of chloride ions remain same in the solution after reaction, so calculating the number of moles of chloride ions in $BaC{l_2}$ as follows:
Dissociation of $BaC{l_2}$ takes place as per following reaction:
$BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - }$
We know that, number of moles $=$ molarity $\times$ Volume in litres.
So, the number of moles of $BaC{l_2}$ in reaction$= 6.6 \times \dfrac{{20}}{{1000}}$
Therefore, the number of moles of chloride ion ${n_{C{l^ - }}} = 2 \times 6.6 \times \dfrac{{20}}{{1000}}$
Total volume of the solution after mixing barium chloride with aluminium sulphate ${V_t}$ $= 40\;mL$
Hence, the concentration of chloride ion after reaction $\left[ {C{l^ - }} \right] = \dfrac{{{n_{C{l^ - }}}}}{{{V_t}}}$
Substituting values:
$\Rightarrow \left[ {C{l^ - }} \right] = \dfrac{{2 \times 6.6 \times \dfrac{{20}}{{1000}}}}{{\dfrac{{40}}{{1000}}}}$
$\Rightarrow \left[ {C{l^ - }} \right] = 6.6\;M$
Thus, the concentration of $C{l^ - }$ ion in the solution is $6.6M$.
So, option (B) is the correct answer.

Note:
Remember that molarity or concentration of any solution is equal to the number of moles of solute dissolved in one litre of solution. It is important to note that for the given reaction in the question, no concentration will be obtained for $B{a^{2 + }}$ or $SO_4^{2 - }$ in the solution after the reaction because $BaS{O_4}$ gets precipitated.