Question

16\. When solutions of aluminum sulfate and sodium hydroxide are mixed, a white gelatinous precipitate forms. (a) Write a balanced net ionic equation for the reaction. (b) What is the mass of the precipitate when \(2.76 \mathrm{~g}\) of aluminum sulfate in \(125 \mathrm{~mL}\) of solution is combined with \(85.0 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{Na} \mathrm{OH}\) ? (c) What is the molarity of the ion in excess? (Ignore spectator ions and assume that volumes are additive.)

Short answer

Answer: The net ionic equation for the reaction is \(\mathrm{2Al^{3+} + 6OH^- \rightarrow 2Al(OH)_3}\). The mass of the precipitate formed is 0.630 grams, and the molarity of the excess hydroxide ions (\(\mathrm{OH^-}\)) is -0.0152 M.

Step-by-step solution

Determine the balanced net ionic equation

First, write the balanced chemical equation for the reaction between aluminum sulfate (\(\mathrm{Al_2(SO_4)_3}\)) and sodium hydroxide (\(\mathrm{NaOH}\)): \(\mathrm{Al_2(SO_4)_3 + 6NaOH \rightarrow 2Al(OH)_3 + 3Na_2SO_4}\) Next, write the total ionic equation by separating all soluble compounds into their respective ions: \(\mathrm{2Al^{3+} + 3(SO_4)^{2-} + 6Na^+ + 6OH^- \rightarrow 2Al(OH)_3 + 6Na^+ + 3(SO_4)^{2-}}\) Identify and remove the spectator ions, which remain unchanged during the reaction: \(\mathrm{Na^+}\) and \(\mathrm{(SO_4)^{2-}}\) Finally, write the net ionic equation by excluding the spectator ions: \(\mathrm{2Al^{3+} + 6OH^- \rightarrow 2Al(OH)_3}\)

Determine the mass of the precipitate with given reactants

To find the mass of the precipitate, we'll first determine the limiting reactant. Given: Mass of aluminum sulfate = \(2.76\,\mathrm{g}\) Volume of sodium hydroxide solution = \(85.0\,\mathrm{mL}\) Concentration of sodium hydroxide solution = \(0.2500\,\mathrm{M}\) Begin by calculating the moles of aluminum sulfate: Moles of \(\mathrm{Al_2(SO_4)_3}\) = \(\frac{2.76\,\mathrm{g}}{342.15\,\mathrm{g/mol}} = 0.00807\,\mathrm{mol}\) Next, find the moles of \(\mathrm{NaOH}\) using the volume and concentration: Moles of \(\mathrm{NaOH}\) = \((85.0\,\mathrm{mL})(0.2500\,\mathrm{M}) = 0.02125\,\mathrm{mol}\) Use the balanced net ionic equation to find the mole ratio between the reactants: \(\frac{0.00807\,\mathrm{mol\,Al^{3+}}}{0.02125\,\mathrm{mol\,OH^-}} = \frac{1}{6}\) Since each mole of \(\mathrm{Al^{3+}}\) reacts with six moles of \(\mathrm{OH^-}\), the limiting reactant is \(\mathrm{Al^{3+}}\) (found in aluminum sulfate). Using the stoichiometry from the balanced net ionic equation, we can determine the moles of precipitate (\(\mathrm{Al(OH)_3}\)) formed: Moles of \(\mathrm{Al(OH)_3}\) = \((0.00807\,\mathrm{mol\,Al^{3+}})(\frac{2\,\mathrm{mol\,Al(OH)_3}}{2\,\mathrm{mol\,Al^{3+}}}) = 0.00807\,\mathrm{mol\,Al(OH)_3}\) Finally, find the mass of the precipitate formed: Mass of \(\mathrm{Al(OH)_3}\) = \((0.00807\,\mathrm{mol})(78.00\,\mathrm{g/mol}) = 0.630\,\mathrm{g}\)

Calculate the molarity of the ion in excess

Since aluminum sulfate is the limiting reactant, sodium hydroxide is in excess. To find the molarity of the remaining hydroxide ions (\(\mathrm{OH^-}\)), start by determining the moles of \(\mathrm{OH^-}\) used in the reaction: Moles of \(\mathrm{OH^-}\) used = \((0.00807\,\mathrm{mol\,Al^{3+}})(\frac{6\,\mathrm{mol\,OH^-}}{2\,\mathrm{mol\,Al^{3+}}}) = 0.0244\,\mathrm{mol}\) Now subtract the moles of \(\mathrm{OH^-}\) used in the reaction from the initial moles of \(\mathrm{OH^-}\): Moles of remaining \(\mathrm{OH^-}\) = \(0.02125\,\mathrm{mol} - 0.0244\,\mathrm{mol} = -0.00319\,\mathrm{mol}\) Calculate the total volume of the resulting solution by assuming that the volumes are additive: Total volume = \(125\,\mathrm{mL} + 85.0\,\mathrm{mL} = 210\,\mathrm{mL}\) Finally, determine the molarity of the remaining hydroxide ions (\(\mathrm{OH^-}\)): Molarity of remaining \(\mathrm{OH^-}\) = \(\frac{-0.00319\,\mathrm{mol}}{0.210\,\mathrm{L}} = -0.0152\,\mathrm{M}\)