Question

200 mL of an aqueous solution of HCl (pH = 2) is mixed with 300 mL of aqueous solution of NaOH (pH = 12) and is diluted to 1.0 L. The pH of the resulting solution is

• A. 10.3
• B. 11.0
• C. 11.3
• D. 11.7

Hint:

For mixing strong acids and bases, always work with moles. Calculate the initial moles of H⁺ and OH⁻, find the moles of the excess ion, and then divide by the final total volume to get the final concentration before calculating the pH.

Answer 1

The Correct Option is B

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 = \(-\log_{10}[H^+]\), so \([H^+] = 10^{-\text{pH}}\).
2. pH + pOH = 14, and \([OH^-] = 10^{-\text{pOH}}\).
3. Moles = Molarity \(\times\) 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, \(V_{acid} = 200 \text{ mL} = 0.2 \text{ L}\).
- pH = 2, so the concentration of H⁺ is \([H^+] = 10^{-2} \text{ M}\).
- Moles of H⁺ = \([H^+] \times V_{acid} = 10^{-2} \text{ M} \times 0.2 \text{ L} = 0.002 \text{ mol}\).
Calculate moles of OH⁻ from NaOH solution:
- Volume of NaOH solution, \(V_{base} = 300 \text{ mL} = 0.3 \text{ L}\).
- pH = 12. Since this is a basic solution, we first find the pOH.
- pOH = 14 - pH = 14 - 12 = 2.
- The concentration of OH⁻ is \([OH^-] = 10^{-\text{pOH}} = 10^{-2} \text{ M}\).
- Moles of OH⁻ = \([OH^-] \times V_{base} = 10^{-2} \text{ M} \times 0.3 \text{ L} = 0.003 \text{ mol}\).
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 = \(0.003 - 0.002 = 0.001 \text{ mol}\).
Calculate the final pH:
- The final mixture is diluted to a total volume of \(V_{final} = 1.0 \text{ L}\).
- The final concentration of OH⁻ in the solution is:
\[ [OH^-]_{final} = \frac{\text{Moles of OH}^- \text{ remaining}}{V_{final}} = \frac{0.001 \text{ mol}}{1.0 \text{ L}} = 10^{-3} \text{ M} \] - Now, calculate the pOH of the final solution:
\[ \text{pOH} = -\log_{10}([OH^-]_{final}) = -\log_{10}(10^{-3}) = 3 \] - Finally, calculate the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \] Step 4: Final Answer:
The pH of the resulting solution is 11.0. Therefore, option (B) is correct.