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

First, we need to calculate the initial moles of Barium Hydroxide (Ba(OH)₂) in the solution using the given concentration and volume: Moles of Ba(OH)₂ = Molarity × Volume Moles of Ba(OH)₂ = 0.100 mol/L × 0.080 L = 0.008 mol

For each volume of HCl added, we will calculate the moles of HCl using its concentration and the given volume. We will do that with the following formula: Moles of HCl = Molarity × Volume For example, for volume a (0.0 mL), we have: Moles of HCl = 0.400 mol/L × 0 mL = 0 mol Repeat this for each additional volume of HCl provided.

Next, we will create and fill in a table to compare the moles of Ba(OH)₂ and HCl at different stages of the titration. | Volume of HCl added | Moles of Ba(OH)₂ | Moles of HCl added | Moles of reacted Ba(OH)₂ | Moles of unreacted Ba(OH)₂ | Moles of remaining HCl | |---------------------|------------------|--------------------|-------------------------|---------------------------|-----------------------| | a. 0.0 mL | 0.008 | 0 | 0 | 0.008 | 0 | | b. 20.0 mL | 0.008 | 0.008 | 0.008 | 0 | 0 | | c. 30.0 mL | 0.008 | 0.012 | 0.008 | 0 | 0.004 | | d. 40.0 mL | 0.008 | 0.016 | 0.008 | 0 | 0.008 | | e. 80.0 mL | 0.008 | 0.032 | 0.008 | 0 | 0.024 | We will now explain in detail how to fill in each row of the table. - 1. Write the moles of Ba(OH)₂ at the start - 2. Write the moles of HCl added in each scenario - 3. Calculate the moles of reacted Ba(OH)₂ by taking the minimum of the moles of Ba(OH)₂ and moles of HCl added - 4. Calculate the moles of unreacted Ba(OH)₂ by subtracting the moles of reacted Ba(OH)₂ from the initial moles of Ba(OH)₂ (if any) - 5. Calculate the moles of remaining HCl by subtracting the moles of reacted Ba(OH)₂ from the moles of HCl added (if any)

Lastly, based on the results in the table, we will calculate the pH of the solution at each stage of the titration. It is important to analyze in which phase the solution is, whether it is before the equivalence point, at the equivalence point, or after the equivalence point. This can be established through the comparison of moles of unreacted Ba(OH)₂ and remaining HCl in the table: a. pH = 14 - pOH Since we have unreacted Ba(OH)₂, pOH=-log[OH⁻], calculate the concentration of OH⁻ ions from unreacted Ba(OH)₂ and find the pH. b. At equivalence point, the pH will be neutral (pH = 7) as the strong acid fully reacts with the strong base. c,d,e. Since there is excess HCl in the solution, directly calculate the pH using the H⁺ ion concentration from the remaining moles of HCl: pH = -log[H⁺] By following these steps, we can analyze the titration and find the pH at different stages of the process.