# What is chloride ion concentration when....?

## .... a $20 \cdot m L$ volume of $\text{aluminum sulfate}$ at $0.20 \cdot m o l \cdot {L}^{-} 1$ concentration is mixed with a $20 \cdot m L$ volume of $\text{barium chloride}$ at $6.6 \cdot m o l \cdot {L}^{-} 1$ concentration?

May 31, 2017

Well we need a stoichiometric reaction........and it turns out that chloride is a spectator ion. We get $\left[C {l}^{-}\right] = 6.6 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

$A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right) + 3 B a C {l}_{2} \left(a q\right) \rightarrow 3 B a S {O}_{4} \left(s\right) \downarrow + 2 A l C {l}_{3} \left(a q\right)$

Note that barium sulfate is insoluble in aqueous solution. This is an experimental fact that simply must be known.

$\text{Moles of aluminum sulfate} = 20 \times {10}^{-} 3 \cdot L \times 0.20 \cdot m o l \cdot {L}^{-} 1 = 4.0 \times {10}^{-} 3 \cdot m o l .$

$\text{Moles of barium chloride} = 20 \times {10}^{-} 3 \cdot L \times 6.6 \cdot m o l \cdot {L}^{-} 1 = 0.132 \cdot m o l .$

And thus $\left[C {l}^{-}\right] = \frac{2 \times 0.132 \cdot m o l}{20 \times {10}^{-} 3 \cdot L + 20 \times {10}^{-} 3 \cdot L}$

=??*mol*L^-1.

Why did I double the numerator in the calculation?