Question

50 mL of 0.05 M $BaC{l}_2$ (aq) solution and 100 mL of 0.1 M of KCl (aq) solutions are mixed. Calculate the molarity of $C{l}^{-}$ ions in the resulting solution.

• A. 0.33 M
• B. 0.10 M
• C. 0.66 M
• D. 1.33 M

Answer 1

The correct option is B 0.10 M

We have,
50 mL of 0.05 M $BaC{l}_2$ solution and 100 mL of 0.1 M KCl solution.
Millimoles of $C{l}^{-}$ ions from $BaC{l}_2\left(n_1\right)=2\times M_1{V}_1{n}_1=2\times 0.05\times 50=5millimole$
Since, 1 mol of $BaC{l}_2$ gives 2 mol of $C{l}^{-}$ ions.
Where, $M_{1}and{V}_1$ are the molarity and volume in mL of the given $BaC{l}_2$ solution.
Similarly,
Millimoles of $C{l}^{-}$ ions from $KCl\left(n_2\right)$ = $1\times M_2{V}_2$
$n_2=1\times 0.1\times 100=10millimole$
Since, 1 mol of KCl gives 1 mol of $C{l}^{-}$ ions.
Where, $M_{2}and{V}_2$ are the molarity and volume in mL of the given KCl solution.
Total millimoles of $C{l}^{-}$ ions = 5+10 = 15 millimoles
Total volume of solution = 50 + 100
= 150 mL
Now, the molarity is given by:
$M=\frac{molesofsolute}{volumeofsolution\left(mL\right)}\times 1000$
$M\left(C{l}^{-}\right)=\frac{15\times {10}^{-3}mol}{150\left(mL\right)}\times 1000=0.1M$
Therefore, the molarity of $C{l}^{-}$ ion will be 0.10 M