A solid sample of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is added to $0.500 \mathrm{~L}\( of \)0.250 \mathrm{M}\( aqueous \)\mathrm{H}_{2} \mathrm{SO}_{4}$. The solution that remains is still acidic. It is then titrated with $0.500 \mathrm{M} \mathrm{NaOH}\( solution, and it takes \)12.5 \mathrm{~mL}$ of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) was added to the $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution?

To calculate the moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the \(0.500\mathrm{~L}\) of \(0.250\mathrm{M}\) solution, we can use the formula: Moles = Molarity × Volume So, Moles of \(\mathrm{H}_{2} \mathrm{SO}_{4} = 0.250 \mathrm{M} \times 0.500 \mathrm{~L} = 0.125\mathrm{~mol}\)

To find the moles of unreacted \(\mathrm{H}_{2} \mathrm{SO}_{4}\), we need to find the moles of \(\mathrm{OH}^{-}\) ions present in the \(12.5\mathrm{~mL}\) of \(0.500\mathrm{M}\) \(\mathrm{NaOH}\). Moles of \(\mathrm{OH}^{-} = 0.500 \mathrm{M} \times (12.5 \times 10^{-3}) \mathrm{~L} = 0.00625\mathrm{~mol}\) Since the ratio of \(\mathrm{H}^{+}\) ions to \(\mathrm{OH}^{-}\) ions in the neutralization reaction is 1:1, there must be the same number of moles of \(\mathrm{H}^{+}\) ions provided by the unreacted \(\mathrm{H}_{2}SO_{4}\). Moles of unreacted \(\mathrm{H}_{2} \mathrm{SO}_{4} = 0.00625\mathrm{~mol}\)

In the reaction between \(\mathrm{Fe}(\mathrm{OH})_{3}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the ratio of \(\mathrm{Fe}(\mathrm{OH})_{3}\) to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 2:3. We initially had \(0.125\mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and there are \(0.00625\mathrm{~mol}\) left unreacted. So, the reacted moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are: Reacted moles of \(\mathrm{H}_{2} \mathrm{SO}_{4} = 0.125 - 0.00625 = 0.11875\mathrm{~mol}\) Now, we can calculate the moles of \(\mathrm{Fe}(\mathrm{OH})_{3}\) that reacted with the initial \(\mathrm{H}_{2} \mathrm{SO}_{4}\) using the 2:3 ratio: Moles of \(\mathrm{Fe}(\mathrm{OH})_{3} = (\frac{2}{3}) \times 0.11875\mathrm{~mol} = 0.07917\mathrm{~mol}\)

To calculate the mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\), we use the formula: Mass = Moles × Molar mass The molar mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is \(106.87\mathrm{g/mol}\). So, Mass of \(\mathrm{Fe}(\mathrm{OH})_{3} = 0.07917\mathrm{~mol} \times 106.87\mathrm{g/mol} = 8.46\mathrm{~g}\) Therefore, the mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) added to the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is \(8.46\mathrm{~g}\).