Question #b3525

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Question #b3525

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Answer 1 · 2014-12-15T11:24:13

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 $C o {(N O_{3})}_{2}$ $Co(NO_3)_2$ . According to the aforementioned solubility rules, all nitrate ( $N O_{3}^{-}$ $NO_3^-$ ) salts are soluble, so we would get

$C o {(N O_{3})}_{2 (a q)} \to C o^{2 +} (a q) + 2 N O_{3}^{-} (a q)$ $Co(NO_3)_(2(aq)) -> Co^(2+)(aq) + 2NO_3^(-)(aq)$

This means that the concentrations of $C o^{2 +}$ $Co^(2+)$ and $N O_{3}^{-}$ $NO_3^-$ will be

$\frac{0.50 m o l e s C o {(N O_{3})}_{2}}{1 L s o l u t i o n} \cdot \frac{1 m o l e C o^{2 +}}{1 m o l e C o {(N O_{3})}_{2}} = 0.50 M$ $(0.50 mol es Co(NO_3)_2)/(1L solution) * (1 mol e Co^(2+))/(1 mol e Co(NO_3)_2) = 0.50 M$ $C o^{2 +}$ $Co^(2+)$

$\frac{0.50 m o l e s C o {(N O_{3})}_{2}}{1 L s o l u t i o n} \cdot \frac{2 m o l e s N O_{3}^{-}}{1 m o l e C o {(N O_{3})}_{2}} = 1.0 M$ $(0.50 mol es Co(NO_3)_2)/(1L solution) * (2 mol es NO_3^(-))/(1 mol e Co(NO_3)_2) = 1.0 M$ $N O_{3}^{-}$ $NO_3^-$

Notice that the $N O_{3}^{-}$ $NO_3^-$ 's concentration will be twice as big as the concentration of $C o {(N O_{3})}_{2}$ $Co(NO_3)_2$ , since 2 moles of the former are produced for every 1 mole of the latter.

Moving on to $F e {(C l O_{4})}_{3}$ $Fe(ClO_4)_3$ . The solubility rules tell us that all perchlorate ( $C l O_{4}^{-}$ $ClO_4^-$ ) salts are soluble, so we will get

$F e {(C l O_{4})}_{3 (a q)} \to F e^{3 +} (a q) + 3 C l O_{4}^{-} (a q)$ $Fe(ClO_4)_(3(aq)) -> Fe^(3+)(aq) + 3ClO_4^(-)(aq)$

Therefore, the concentrations of the two ions will be

$\frac{1 m o l e F e {(C l O_{4})}_{3}}{1 L s o l u t i o n} \cdot \frac{1 m o l e F e^{3 +}}{1 m o l e F e {(C l O_{4})}_{3}} = 1 M$ $(1 mol e Fe(ClO_4)_3)/(1Lsolution) * (1 mol e Fe^(3+))/(1 mol e Fe(ClO_4)_3) = 1M$ $F e^{3 +}$ $Fe^(3+)$

$\frac{1 m o l e F e {(C l O_{4})}_{3}}{1 L s o l u t i o n} \cdot \frac{3 m o l e s C l O_{4}^{-}}{1 m o l e F e {(C l O_{4})}_{3}} = 3 M$ $(1 mol e Fe(ClO_4)_3)/(1Lsolution) * (3 mol es ClO_4^-)/(1 mol e Fe(ClO_4)_3) = 3M$ $C l O_{4}^{-}$ $ClO_4^-$

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Question #b3525

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