Question

When \(25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) iron(II) sulfate is combined with \(35.0 \mathrm{~mL}\) of \(0.332 \mathrm{M}\) barium hydroxide, two different precipitates are formed. (a) Write a net ionic equation for the reaction that takes place. (b) Estimate the mass of the precipitates formed. (c) What are the equilibrium concentrations of the ions in solution?

Short answer

Answer: The mass of precipitates formed is 3.75 g, and the equilibrium concentrations of the ions are [Fe2+] = 0.01467 M, [SO4^2-] = 0.20833 M, [Ba2+] = 0 M, and [OH^-] = 0 M.

Step-by-step solution

Identify the precipitates

When iron(II) sulfate reacts with barium hydroxide, the possible products are iron(II) hydroxide and barium sulfate. Iron(II) hydroxide is insoluble in water and will form a precipitate, while barium sulfate is also insoluble in water and will form a second precipitate. So the reaction occurring can be written as: FeSO4 + Ba(OH)2 → Fe(OH)2 + BaSO4

Write the net ionic equation

To write the net ionic equation, we first dissociate the soluble compounds into their ions: Fe2+ + SO4^2- + Ba2+ + 2OH^- → Fe(OH)2 + BaSO4 Now, we can identify the spectator ions as SO4^2-. The net ionic equation can now be written as: Fe2+ + Ba2+ + 2OH^- → Fe(OH)2 + BaSO4

Calculate the moles of reactants

To find the mass of the precipitates formed, we need to first determine the number of moles of reactants involved. Using volume and concentration, the moles of iron(II) sulfate and barium hydroxide can be calculated as: moles of FeSO4 = volume x concentration = 0.025 L x 0.500 mol/L = 0.0125 mol moles of Ba(OH)2 = volume x concentration = 0.035 L x 0.332 mol/L = 0.01162 mol

Determine the limiting reactant

To find the limiting reactant, compare the ratio of moles of iron(II) sulfate to barium hydroxide to the stoichiometric ratio in the balanced equation. In this case, FeSO4 and Ba(OH)2 react in a 1:1 ratio: 0.0125 mol FeSO4 / 1 = 0.01162 mol Ba(OH)2 / 1 Since the ratio for Ba(OH)2 is less, it is the limiting reactant.

Calculate the mass of precipitates formed

Both reactants form one mole of precipitate per mole of reactant, so the moles of each precipitate can be calculated using the moles of the limiting reactant, barium hydroxide: moles of Fe(OH)2 = moles of Ba(OH)2 = 0.01162 mol moles of BaSO4 = moles of Ba(OH)2 = 0.01162 mol Now, convert moles to mass using the molar masses of each precipitate: mass of Fe(OH)2 = 0.01162 mol * (molar mass of Fe(OH)2) = 0.01162 mol * 89.86 g/mol = 1.04 g mass of BaSO4 = 0.01162 mol * (molar mass of BaSO4) = 0.01162 mol * 233.43 g/mol = 2.71 g The total mass of precipitates formed is 1.04 g + 2.71 g = 3.75 g.

Calculate the equilibrium concentrations of ions

To find the equilibrium concentrations of the ions, begin by calculating the moles of each ion at equilibrium: moles of Fe2+ = moles of FeSO4 - moles of Fe(OH)2 = 0.0125 - 0.01162 = 0.00088 mol moles of SO4^2- = moles of FeSO4 = 0.0125 mol moles of Ba2+ = moles of Ba(OH)2 - moles of BaSO4 = 0.01162 - 0.01162 = 0 mol moles of OH^- = 2 * moles of Ba(OH)2 - 2 * moles of Fe(OH)2 = 2 * 0.01162 - 2 * 0.01162 = 0 mol Next, find the final volume of the solution by adding the volumes of the reactants: final volume = 0.025 L + 0.035 L = 0.060 L Finally, calculate the equilibrium concentrations by dividing the moles of each ion by the final volume: [Fe2+] = moles of Fe2+ / final volume = 0.00088 mol / 0.060 L = 0.01467 M [SO4^2-] = moles of SO4^2- / final volume = 0.0125 mol / 0.060 L = 0.20833 M [Ba2+] = moles of Ba2+ / final volume = 0 mol / 0.060 L = 0 M [OH^-] = moles of OH- / final volume = 0 mol / 0.060 L = 0 M The equilibrium concentrations of the ions are [Fe2+] = 0.01467 M, [SO4^2-] = 0.20833 M, [Ba2+] = 0 M, and [OH^-] = 0 M.