Calculate the pH of the solution resulting from the mixing of 175.0 mL of 0.250 M HNO₃ with 75.0 mL of 0.200M Ba(OH)₂.

• A. pH = 12.74
• B. pH = 0.94
• C. pH = 1.26
• D. pH = 13.06

Answer: (C)

Verified step by step guidance

1. Determine the moles of HNO3 by using the formula: \( \text{moles of HNO3} = \text{volume (L)} \times \text{molarity (M)} \). Convert 175.0 mL to liters and multiply by 0.250 M.
2. Determine the moles of Ba(OH)2 using the formula: \( \text{moles of Ba(OH)2} = \text{volume (L)} \times \text{molarity (M)} \). Convert 75.0 mL to liters and multiply by 0.200 M.
3. Recognize that Ba(OH)2 is a strong base that dissociates completely in water to produce 2 moles of OH⁻ ions per mole of Ba(OH)2. Calculate the total moles of OH⁻ ions produced.
4. Calculate the moles of H⁺ ions remaining after the neutralization reaction between HNO3 and OH⁻ ions. Use the stoichiometry of the reaction: \( \text{HNO3} + \text{OH}^- \rightarrow \text{H2O} + \text{NO3}^- \). Subtract the moles of OH⁻ from the moles of HNO3.
5. Calculate the pH of the resulting solution using the formula: \( \text{pH} = -\log[\text{H}^+] \), where \([\text{H}^+]\) is the concentration of H⁺ ions remaining in the solution. Divide the moles of H⁺ by the total volume of the solution in liters to find the concentration.