Question

$50\text{mL}$ of $0.05\text{M}$ $BaC{l}_2$ (aq) solution and $200\text{mL}$ of $0.5\text{M}$ of $KCl$ (aq) solutions are mixed. Calculate the molarity of $C{l}^{-}$ ions in the resulting solution.

• A. $0.33\text{M}$
• B. $0.06\text{M}$
• C. $0.60\text{M}$
• D. $0.03\text{M}$

Answer 1

The correct option is B $0.06\text{M}$

We have,
$50\text{mL}$ of $0.05\text{M}$ $BaC{l}_2$ solution and $200\text{mL}$ of $0.5\text{M}$ $KCl$ solution.
Millimoles of $C{l}^{-}$ ions from $BaC{l}_2,\left(n_1\right)=2\times M_1{V}_1\therefore n_1=2\times 0.05\times 50=5\text{millimole}$
Since, $1$ mol of $BaC{l}_2$ gives $2$ mol of $C{l}^{-}$ ions.
Where, $M_1\text{and}{V}_1$ are the molarity and volume in mL of the given $BaC{l}_2$ solution.
Similarly,
Millimoles of $C{l}^{-}$ ions from $KCl,\left(n_2\right)=1\times M_2{V}_2$
$\therefore n_2=1\times 0.5\times 200=100\text{millimole}$
Since, $1$ mol of $KCl$ gives $1$ mol of $C{l}^{-}$ ions.
Where, $M_2\text{and}{V}_2$ are the molarity and volume in mL of the given $KCl$ solution.
Total millimoles of $C{l}^{-}$ ions $=5+100=15\text{millimoles}$
Total volume of solution $=(50+200)\text{mL}=250\text{mL}$
Now, the molarity 0f $C{l}^{-}$ can be calculated:
$M\left(C{l}^{-}\right)=\frac{15\times {10}^{-3}\text{mol}}{250\left(\text{mL}\right)}\times 1000=0.06\text{M}$
Therefore, the molarity of $C{l}^{-}$ ion will be $0.06\text{M}$