A 65-mL solution of \(0.40 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is mixed with \(125 \mathrm{~mL}\) of \(0.17 \mathrm{M}\) iron(II) nitrate. Solid sodium hydroxide is then added without a change in volume. (a) Which will precipitate first, \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Fe}(\mathrm{OH})_{2} ?\) (b) What is \(\left[\mathrm{OH}^{-}\right]\) when the first compound begins to precipitate?

First, we need to calculate the initial concentrations of Al³⁺ and Fe²⁺ ions in the mixture. The initial concentration of aluminum ions can be found by: \([\mathrm{Al^{3+}}]_{initial}=\frac{(\mathrm{Molarity \ of \ Al(NO_3)_3})(\mathrm{Volume \ of \ Al(NO_3)_3})}{\mathrm{Total \ volume}}\) The initial concentration of iron(II) ions can be found by: \([\mathrm{Fe^{2+}}]_{initial}=\frac{(\mathrm{Molarity \ of \ Fe(NO_3)_2})(\mathrm{Volume \ of \ Fe(NO_3)_2})}{\mathrm{Total \ volume}}\) Plug in the given values to find the concentrations: \([\mathrm{Al^{3+}}]_{initial}=\frac{(0.40 \ \mathrm{M})(65 \ \mathrm{mL})}{65 \ \mathrm{mL} + 125 \ \mathrm{mL}} = 0.146 \ \mathrm{M}\) \([\mathrm{Fe^{2+}}]_{initial}=\frac{(0.17 \ \mathrm{M})(125 \ \mathrm{mL})}{65 \ \mathrm{mL} + 125 \ \mathrm{mL}} = 0.0925 \ \mathrm{M}\)

In this step, we need to find the solubility product constants (\(K_{sp}\)) for \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{2}\). These constants can be found in a reference source or given by the problem. For this example, we will use the following values: \(K_{sp}(\mathrm{Al}(\mathrm{OH})_{3}) = 3.0 \times 10^{-34}\) \(K_{sp}(\mathrm{Fe}(\mathrm{OH})_{2}) = 8.0 \times 10^{-16}\)

In this step, we will calculate the concentrations of hydroxide ions needed for respective metal ions to precipitate as their hydroxides. For \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{2}\), we have: \(K_{sp} = [\mathrm{Al^{3+}}][\mathrm{OH^-}]^3\) \(3.0 \times 10^{-34} = (0.146)[\mathrm{OH^-}]^3\) \(K_{sp} = [\mathrm{Fe^{2+}}][\mathrm{OH^-}]^2\) \(8.0 \times 10^{-16} = (0.0925)[\mathrm{OH^-}]^2\) Now, we need to solve for the values of \([\mathrm{OH^-}]\) in each case: \([\mathrm{OH^-}]_{Al} = \sqrt[3]{\frac{3.0 \times 10^{-34}}{0.146}} = 6.07 \times 10^{-12}\) \([\mathrm{OH^-}]_{Fe} = \sqrt{\frac{8.0 \times 10^{-16}}{0.0925}} = 9.17 \times 10^{-8}\) Comparing these two values, we see that \([\mathrm{OH^-}]_{Al} < [\mathrm{OH^-}]_{Fe}\).

Since \([\mathrm{OH^-}]_{Al} < [\mathrm{OH^-}]_{Fe}\), aluminum hydroxide will precipitate first. Therefore, the concentration of hydroxide ions needed for the first compound to precipitate is: \([\mathrm{OH^-}] = 6.07 \times 10^{-12} \ \mathrm{M}\) (a) \(\mathrm{Al}(\mathrm{OH})_{3}\) will precipitate first (b) The concentration of hydroxide ions when the first compound begins to precipitate is \(6.07 \times 10^{-12} \ \mathrm{M}\).