Consider the titration of \(80.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by \(0.400 M\) HCl. Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\)

To calculate the initial moles of \(\mathrm{Ba(OH)}_{2}\) and moles of \(\mathrm{HCl}\) added at different volumes, we use the formula: Moles = Molarity × Volume For \(\mathrm{Ba(OH)}_{2}\): Initial moles = \(0.100 \mathrm{M}\) × \(80.0 \mathrm{~mL}\) = \(0.100 \mathrm{M}\) × \(0.0800 \mathrm{~L}\) = \(0.00800 \mathrm{~mol} \) For \(\mathrm{HCl}\), we will calculate the moles for each part (a-e) as the volume changes.

Compare the moles of \(\mathrm{Ba(OH)}_{2}\) and \(\mathrm{HCl}\) to determine the limiting reactant. Since 1 mole of \(\mathrm{Ba(OH)}_{2}\) reacts with 2 moles of \(\mathrm{HCl}\), we will compare the moles of \(\mathrm{HCl}\) with two times the moles of \(\mathrm{Ba(OH)}_{2}\).

Based on the limiting reactant, calculate the moles of the remaining reactant and evaluate the concentration of the ions in the solution.

Use the equilibrium constant of the remaining reactant or the concentration of the ions to calculate the pH of the solution. Now, let's perform these steps for each part: a) \(0.0 \mathrm{~ml}\) of \(\mathrm{HCl}\) added: In this case, no \(\mathrm{HCl}\) is added, therefore we only have the Barium hydroxide in the solution. We can directly compute the pOH and then find pH by subtracting it from 14. b) \(20.0 \mathrm{~ml}\) of \(\mathrm{HCl}\) added: Moles of \(\mathrm{HCl} = 0.400 \mathrm{M}\) × \(20.0 \mathrm{~mL}\) = \(0.400 \mathrm{M}\) × \(0.0200 \mathrm{~L}\) = \(0.00800 \mathrm{~mol}\) MoRAR (Moles of Reactant Available for Reaction) = Moles of \(\mathrm{HCl}\) - 2 × Moles of \(\mathrm{Ba(OH)}_{2}\) MoRAR = 0.00800 - (2 × 0.00800) = -0.00800 Since the MoRAR is negative, it indicates that \(\mathrm{HCl}\) is a limiting reactant. Now, we can find the remaining moles of \(\mathrm{Ba(OH)}_{2}\) and calculate its concentration. c) \(30.0 \mathrm{~ml}\) of \(\mathrm{HCl}\) added: Calculate the moles of \(\mathrm{HCl}\) and compute MoRAR as described above. Evaluate the pH of the solution by considering the MoRAR value. d) \(40.0 \mathrm{~ml}\) of \(\mathrm{HCl}\) added: Perform the same steps as described above to find the MoRAR and calculate the pH of the solution. e) \(80.0 \mathrm{~ml}\) of \(\mathrm{HCl}\) added: Finally, perform these steps for \(80.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added and calculate the final equilibrium pH of the reaction mixture.