# Question 4359c

##### 1 Answer
May 18, 2017

$\left[C {l}^{-}\right] = 0.667 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

This question has a sting in the tail. It expects you (i) to do the tedious arithmetic involved in the percentage mass; and (ii) it tries to distract students who know that $B a S {O}_{4}$ is insoluble in aqueous solution. Of course, the insolubility of $B a S {O}_{4}$ is a distractor.

$B {a}^{2 +} + S {O}_{4}^{2 -} \rightarrow B a S {O}_{4} \left(s\right) \downarrow$

But we only need to assess the concentration of chloride ion WHICH REMAINS IN SOLUTION.

$\text{Moles of}$ BaCl_2=(50*mLxx20.8%*g*mL^-1)/(208.23*g*mol^-1)# $= 0.050 \cdot m o l$.

And thus there are $0.100 \cdot m o l$ with respect to $\text{chloride ion}$. And of course this is present in a COMBINED VOLUME of $100 \cdot m L + 50 \cdot m L = 150 \cdot m L = 0.150 \cdot L$

Now concentration is simply $\text{moles of solute"/"volume of solution} = \frac{0.100 \cdot m o l}{150 \times {10}^{-} 3 L}$

$\left[C {l}^{-}\right] = 0.667 \cdot m o l \cdot {L}^{-} 1$