If the concentration of SO_4^{2-} in an aqueo | vedclass

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(C) The dissociation of $Al_2(SO_4)_3$ in water is given by the equation: $Al_2(SO_4)_3(aq) \rightarrow 2Al^{3+}(aq) + 3SO_4^{2-}(aq)$ From the stoich

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$0.18$

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(C) The dissociation of $Al_2(SO_4)_3$ in water is given by the equation: $Al_2(SO_4)_3(aq) ightarrow 2Al^{3+}(aq) + 3SO_4^{2-}(aq)$ From the stoichiometry,$1 \ mol$ of $Al_2(SO_4)_3$ produces $2 \ mol$ of $Al^{3+}$ and $3 \ mol$ of $SO_4^{2-}$. Given that the concentration of $SO_4^{2-}$ is $0.27 \ M$. Using the ratio: $\frac{[Al^{3+}]}{[SO_4^{2-}]} = \frac{2}{3}$ $[Al^{3+}] = \frac{2}{3} 	imes [SO_4^{2-}]$ $[Al^{3+}] = \frac{2}{3} 	imes 0.27 \ M = 2 	imes 0.09 \ M = 0.18 \ M$.

If the concentration of $SO_4^{2-}$ in an aqueous solution of $Al_2(SO_4)_3$ is $0.27 \ M$,then the concentration of $Al^{3+}$ ions will be ....... $M$.

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