Question

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. \(80.0 \mathrm{~mL}\)

Short answer

a) For \(0.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added, the pH = 13.74. b) For \(20.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added, the pH = 12.74. c) For \(30.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added, the pH = 7.00. d) For \(40.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added, the pH = 1.26. e) For \(80.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) added, the pH = 0.593.

Step-by-step solution

Calculate the initial moles of Ba(OH)₂ and moles of HCl added

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.

Determine the limiting reactant

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 \(\mathrm{Ba(OH)}_{2}\).

Calculate moles of the remaining reactant and concentration of ions

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

Calculate the pH of the resulting 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.

Key concepts

Barium Hydroxide

Barium hydroxide, known by its chemical formula \(\mathrm{Ba(OH)}_2\), is a strong base. It readily dissolves in water to release barium ions \(\mathrm{Ba}^{2+}\)and hydroxide ions \(\mathrm{OH}^-\). This dissociation is complete, similar to how hydrochloric acid dissociates entirely in an aqueous solution.
Barium hydroxide solutions are highly basic, and the hydroxide ions present raise the pH significantly. They are used in titration procedures due to their well-defined stoichiometry and clear endpoints.
In this exercise, we start with \(80.0\,\mathrm{mL}\) of 0.100 M barium hydroxide. This solution contains 0.00800 moles of Ba(OH)₂, calculated using the formula:

• Moles = Molarity \(\times \) Volume

This value will be key in further calculations to determine the pH after titration with hydrochloric acid.

Hydrochloric Acid

Hydrochloric acid, represented by the chemical formula \(\mathrm{HCl}\), is one of the most commonly used strong acids. Like barium hydroxide, it fully dissociates in water to produce hydrogen ions \(\mathrm{H}^{+}\) and chloride ions \(\mathrm{Cl}^-\).
This complete dissociation is crucial for precise pH calculations in titration. The strong acidic nature of HCl makes it very effective in neutralizing bases like barium hydroxide.
In our example, we have an HCl solution with a molarity of 0.400 M. As different volumes of HCl are added to the solution of barium hydroxide, the HCl reacts with available hydroxide ions, impacting the solution's pH.
By knowing the initial moles of HCl added at each step in the exercise, we can determine the limiting reactant, which is essential for predicting the outcome of the titration.

pH Calculation

The pH level of a solution is an indicator of its acidity or alkalinity. It is calculated using the concentration of hydroxide ions in a basic solution or hydrogen ions in an acidic solution. The fundamental equation relating to pH and pOH is:

• \(\mathrm{pH} + \mathrm{pOH} = 14\)

This makes it possible to calculate the pH if the pOH is known and vice versa. For acidic solutions, pH is calculated directly from hydrogen ion concentration:

• \(\mathrm{pH} = -\log [\mathrm{H}^+]\)

For basic solutions resulting from the dissociation of barium hydroxide, we first calculate the pOH:

• \(\mathrm{pOH} = -\log [\mathrm{OH}^-]\)
• Then, \(\mathrm{pH} = 14 - \mathrm{pOH}\)

By adding different amounts of HCl to the existing base, the hydroxide ions get neutralized. The remaining concentration of ions, after accounting for the limiting reactant, is used to find the resulting pH.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product that can be formed. Identifying the limiting reactant is crucial for stoichiometric calculations, such as determining pH in a titration.
For this titration exercise, we leverage the balanced equation:

• \(\mathrm{Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O}\)

This tells us that one mole of barium hydroxide requires two moles of hydrochloric acid to reach full neutralization. Thus, the moles of HCl provided must be compared to twice the amount of Ba(OH)₂ present.
When performing the titration:

• If \(\mathrm{2 \times [Ba(OH)_2]}\) is greater than \([HCl]\), the barium hydroxide is in excess.
• Otherwise, HCl becomes the limiting reactant.

Each scenario affects the concentration of ions left in solution, which is directly used to compute the pH, informing us whether the solution remains basic, becomes neutral, or turns slightly acidic.