Question 5.1: Calculating Ion Concentrations in a Solution of a Strong Ele......

Analyze

The solute is a strong electrolyte. Thus, it dissociates completely in water. First, we write a balanced chemical equation for the dissociation of Al_{2}(SO_{4})_{3}(aq), and then set up stoichiometric factors to relate Al^{3+}  and  SO_{4}^{2−} to the molarity of Al_{2}(SO_{4})_{3}.

Solve

The dissociation of Al_{2}(SO_{4})_{3} is represented by the equation below.

Al_{2}(SO_{4})_{3}(aq) → 2  Al^{3+}(aq) + 3  SO_{4}^{2−}(aq)

The stoichiometric factors, shown in blue in the following equations, are derived from the fact that 1 mol Al_{2}(SO_{4})_{3} produces 2 mol Al^{3+}  and  3  mol  SO_{4}^{2−}.

[Al^{3+}] = \frac{0.0165  mol  Al_{2}(SO_{4})_{3}}{1  L} × \frac{2  mol  Al^{3+}}{1  mol  Al_{2}(SO_{4})_{3}} = \frac{0.0330  mol  Al^{3+}}{1  L} = 0.0330 M

[SO_{4}^{2−}] = \frac{0.0165  mol  Al_{2}(SO_{4})_{3}}{1  L} × \frac{3  mol  SO_{4}^{2−}}{1  mol  Al_{2}(SO_{4})_{3}} = \frac{0.0495  mol  SO_{4}^{2−}}{1  L} = 0.0495 M

Assess

For a strong electrolyte, the concentrations of the ions will always be integer multiples of the electrolyte molarity. For example, in 0.0165 M MgCl_{2}, we have [Mg^{2+}] = 1 × 0.0165 M and [Cl^{−}] = 2 × 0.0165 M.