Question

Solid iron(III) hydroxide is added to \(625 \mathrm{~mL}\) of \(0.280 \mathrm{M}\) HCl. The resulting solution is acidic and titrated with 238.2 \(\mathrm{mL}\) of \(0.113 \mathrm{M} \mathrm{NaOH}\). What mass of iron(III) hydroxide was added to the HCl?

Short answer

Answer: The mass of iron(III) hydroxide added to the hydrochloric acid solution was 5.28 g.

Step-by-step solution

Write the balanced reaction equation

For solid iron(III) hydroxide reacting with hydrochloric acid, the balanced equation is: \(Fe(OH)_3 + 3HCl \rightarrow FeCl_3 + 3H_2O\)

Determine the moles of HCl initially present

Using the given volume (625 mL) and concentration (0.280 M) of HCl, we'll convert these values into moles. Since 1 L = 1000 mL, \(625 \mathrm{~mL} = 0.625 \mathrm{~L}\) Using the formula for the concentration (M), Moles of HCl = Volume of HCl (in L) × Concentration of HCl (in M) Moles of HCl = \(0.625 \mathrm{~L} \times 0.280 \mathrm{M} = 0.175 \mathrm{mol}\)

Calculate the moles of HCl remaining after titration

First, we need to calculate the moles of NaOH used in the titration. Using the given volume (238.2 mL) and concentration (0.113 M) of NaOH, we'll convert these values into moles. Since 1 L = 1000 mL, \(238.2 \mathrm{~mL} = 0.2382 \mathrm{~L}\) Moles of NaOH = Volume of NaOH (in L) × Concentration of NaOH (in M) Moles of NaOH = \(0.2382 \mathrm{~L} \times 0.113 \mathrm{M} = 0.02693 \mathrm{mol}\) Since we know that every mole of \(HCl\) neutralized by a mole of \(NaOH\) in the titration, we can calculate moles of HCl remaining after the titration: Moles of HCl_remaining = Moles of HCl_initial - Moles of NaOH Moles of HCl_remaining = \(0.175 \mathrm{mol} - 0.02693 \mathrm{mol} = 0.14807 \mathrm{mol}\)

Calculate moles of iron(III) hydroxide

Looking at the balanced equation, we see that 3 moles of HCl react with 1 mole of iron(III) hydroxide. So, we can use the stoichiometry of the reaction to find the moles of iron(III) hydroxide. Moles of iron(III) hydroxide = (Moles of HCl_remaining) / 3 Moles of iron(III) hydroxide = \(0.14807 \mathrm{mol} / 3 = 0.04936 \mathrm{mol}\)

Convert moles of iron(III) hydroxide to grams

Now that we have moles of iron(III) hydroxide, we can convert it into grams using its molar mass. Molar mass of iron(III) hydroxide: \(Fe = 55.85 \mathrm{g/mol}, O = 16.00 \mathrm{g/mol}, H = 1.01 \mathrm{g/mol}\) Molar mass of \(Fe(OH)_3 = 55.85 + 3 \times (16.00 + 1.01) = 106.87 \mathrm{g/mol}\) Mass of iron(III) hydroxide = Moles of iron(III) hydroxide × Molar mass of iron(III) hydroxide Mass of iron(III) hydroxide = \(0.04936 \mathrm{mol} \times 106.87 \mathrm{g/mol} = 5.28 \mathrm{g}\) So, the mass of iron(III) hydroxide added to the HCl solution is 5.28 g.