Question

20 ml of 0.2 M $A{l_2}{\left( {S{O_4}} \right)_3}$ is mixed 20 ml of 6.6 M $BaC{l_2}$. the concentration of Cl ion in solution is ?
A. 0.2 M
B. 6.6 M
C. 0.02 M
D. 0.06 M

Answer 1

To solve this question we need to know the basic concept of molarity. We also need to understand how moles of an element work. We are required to find the chloride ion concentration. $BaC{l_2}$will be responsible to contribute the chloride ions in the solution.

Complete step by step solution:
- Volume of $A{l_2}{\left( {S{O_4}} \right)_3}$ = 20 mL
- Volume of $BaC{l_2}$= 20 mL
- Molarity of $A{l_2}{\left( {S{O_4}} \right)_3}$ = 0.2 M
- Molarity of $BaC{l_2}$ = 6.6 M

In this solution, $BaC{l_2}$ will produce the required chloride ions. 1 mole of $BaC{l_2}$ produces 2 moles of Cl ions. So if we calculate the number of moles of $BaC{l_2}$, we could also be able to determine the number of moles of Cl ion in solution. Then we can calculate its concentration in terms of Molarity as all the options are in terms of molarity.
We know
$Molarity = \frac{{n \times 1000}}{{V(mL)}}$ , where n is the number of moles of solute.
- Calculating the moles of $BaC{l_2}$ from this formula:
$\begin{gathered} 6.6 = \frac{{n \times 1000}}{{20}} \\\ n = 0.132moles \\\ \end{gathered}$
- Then the number of moles of Cl ion in solution=
$2 \times n = 2 \times 0.132 = 0.264moles$
- Now calculation the concentration/molarity of these ions:
$Molarity = \frac{{n \times 1000}}{{V(mL)}}$
$\begin{gathered} M = \frac{{0.264 \times 1000}}{{(20 + 20)}} \\\ M = \frac{{264}}{{40}} = 6.6M \\\ \end{gathered}$
- The solution contains 6.6M Cl ions.
Hence the correct option is B, 6.6 Molar.

Note: In the last step, we have taken the volume as (20+20), this is because while calculating the molarity of a solute, the volume of the entire solution is taken into consideration as the denominator. In this case, the volume of both the given compounds formed the solution.