# Question #b3525

Dec 15, 2014

The best way to approach such a question is by checking the solubility rules for each of the ionic compounds (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/).

Let's start with $C o {\left(N {O}_{3}\right)}_{2}$. According to the aforementioned solubility rules, all nitrate ($N {O}_{3}^{-}$) salts are soluble, so we would get

$C o {\left(N {O}_{3}\right)}_{2 \left(a q\right)} \to C {o}^{2 +} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right)$

This means that the concentrations of $C {o}^{2 +}$ and $N {O}_{3}^{-}$ will be

$\frac{0.50 m o l e s C o {\left(N {O}_{3}\right)}_{2}}{1 L s o l u t i o n} \cdot \frac{1 m o l e C {o}^{2 +}}{1 m o l e C o {\left(N {O}_{3}\right)}_{2}} = 0.50 M$ $C {o}^{2 +}$

$\frac{0.50 m o l e s C o {\left(N {O}_{3}\right)}_{2}}{1 L s o l u t i o n} \cdot \frac{2 m o l e s N {O}_{3}^{-}}{1 m o l e C o {\left(N {O}_{3}\right)}_{2}} = 1.0 M$ $N {O}_{3}^{-}$

Notice that the $N {O}_{3}^{-}$'s concentration will be twice as big as the concentration of $C o {\left(N {O}_{3}\right)}_{2}$, since 2 moles of the former are produced for every 1 mole of the latter.

Moving on to $F e {\left(C l {O}_{4}\right)}_{3}$. The solubility rules tell us that all perchlorate ($C l {O}_{4}^{-}$) salts are soluble, so we will get

$F e {\left(C l {O}_{4}\right)}_{3 \left(a q\right)} \to F {e}^{3 +} \left(a q\right) + 3 C l {O}_{4}^{-} \left(a q\right)$

Therefore, the concentrations of the two ions will be

$\frac{1 m o l e F e {\left(C l {O}_{4}\right)}_{3}}{1 L s o l u t i o n} \cdot \frac{1 m o l e F {e}^{3 +}}{1 m o l e F e {\left(C l {O}_{4}\right)}_{3}} = 1 M$ $F {e}^{3 +}$

$\frac{1 m o l e F e {\left(C l {O}_{4}\right)}_{3}}{1 L s o l u t i o n} \cdot \frac{3 m o l e s C l {O}_{4}^{-}}{1 m o l e F e {\left(C l {O}_{4}\right)}_{3}} = 3 M$ $C l {O}_{4}^{-}$