A 50.0 mL volume of 0.15 mol L⁻¹ HBr is titrated with 0.25 mol L⁻¹ KOH. Calculate the pH after the addition of 12.0 mL of KOH.

• A. pH = 1.00
• B. pH = 4.00
• C. pH = 2.00
• D. pH = 3.00

Answer: (D)

Verified step by step guidance

1. Determine the initial moles of HBr in the solution. Use the formula: \( \text{moles of HBr} = \text{volume (L)} \times \text{concentration (mol L}^{-1}\text{)} \). Convert 50.0 mL to liters and multiply by 0.15 mol L⁻¹.
2. Calculate the moles of KOH added. Use the formula: \( \text{moles of KOH} = \text{volume (L)} \times \text{concentration (mol L}^{-1}\text{)} \). Convert 12.0 mL to liters and multiply by 0.25 mol L⁻¹.
3. Determine the moles of HBr remaining after the reaction with KOH. Since HBr and KOH react in a 1:1 ratio, subtract the moles of KOH from the initial moles of HBr.
4. Calculate the concentration of HBr remaining in the solution. Add the initial volume of HBr and the volume of KOH added to find the total volume in liters. Then, divide the remaining moles of HBr by this total volume.
5. Use the concentration of HBr to find the pH. Since HBr is a strong acid, it dissociates completely, and the concentration of HBr is equal to the concentration of \( \text{H}^+ \). Use the formula: \( \text{pH} = -\log[\text{H}^+] \) to find the pH.