# If a solution is 1.1*"ppm" with respect to calcium ion, what are [Ca^(2+)], and [Cl^-] if the solution is prepared from calcium chloride?

Sep 1, 2017

You gots $C a C {l}_{2}$.........and the concentration of $C {l}^{-}$ is $2 \cdot \text{ppm}$.

#### Explanation:

And in aqueous solution, calcium chloride speciates to give.....

$C a C {l}_{2} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 C {l}^{-}$

Now if it is $1.1 \cdot \text{ppm}$ with respect to $C {a}^{2 +}$, there are $1.1 \times 1 \cdot m g \cdot {L}^{-} 1$ of solution WITH RESPECT to the calcium ion (and at these concentrations we really don't have to worry about density change).

And so....

$\left[C {a}^{2 +}\right] = \frac{1.1 \times {10}^{-} 3 \cdot g}{40.08 \cdot g \cdot m o {l}^{-} 1} = 2.745 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

And necessarily (why), $\left[C {l}^{-}\right] = 5.489 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$. Why so.....?

This corresponds to a mass concentration of $5.489 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1 \times 35.45 \cdot g \cdot m o {l}^{-} 1 = 1.945 \times {10}^{-} 3 \cdot g \cdot {L}^{-} 1$

$\equiv 2 \cdot \text{ppm}$