Ionic Equilibrium In Solution

Step 1: Understanding the Concept:
This problem involves the neutralization reaction between a strong acid (HCl) and a strong base (NaOH). We need to calculate the initial moles of H⁺ and OH⁻, determine which is in excess after they react, and then calculate the final concentration and pH in the new total volume.
Step 2: Key Formula or Approach:
1. pH = , so .
2. pH + pOH = 14, and .
3. Moles = Molarity Volume (in L).
4. The reaction is H⁺ + OH⁻ → H₂O.
Step 3: Detailed Explanation:
Calculate moles of H⁺ from HCl solution:
- Volume of HCl solution, .
- pH = 2, so the concentration of H⁺ is .
- Moles of H⁺ = .
Calculate moles of OH⁻ from NaOH solution:
- Volume of NaOH solution, .
- pH = 12. Since this is a basic solution, we first find the pOH.
- pOH = 14 - pH = 14 - 12 = 2.
- The concentration of OH⁻ is .
- Moles of OH⁻ = .
Determine the excess reagent after neutralization:
- Initial moles of H⁺ = 0.002 mol.
- Initial moles of OH⁻ = 0.003 mol.
- The acid (H⁺) is the limiting reagent. It will be completely consumed.
- Moles of OH⁻ remaining = (Initial moles of OH⁻) - (Initial moles of H⁺)
- Moles of OH⁻ remaining = .
Calculate the final pH:
- The final mixture is diluted to a total volume of .
- The final concentration of OH⁻ in the solution is:
- Now, calculate the pOH of the final solution:
- Finally, calculate the pH:
Step 4: Final Answer:
The pH of the resulting solution is 11.0. Therefore, option (B) is correct.