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: Calculate Millimoles: - HCl: pH=2 \(\implies [H^+] = 0.01\) M. \( n_{H^+} = 0.01 \times 200 = 2 \) mmol. - NaOH: pH=12 \(\implies [H^+] = 10^{-12} \implies [OH^-] = 0.01\) M. \( n_{OH^-} = 0.01 \times 300 = 3 \) mmol.
Step 2: Neutralization: \( 2 \) mmol \( H^+ \) neutralizes \( 2 \) mmol \( OH^- \). Remaining \( OH^- = 3 - 2 = 1 \) mmol.
Step 3: Calculate Final Concentration: Total Volume = 1.0 L = 1000 mL. \( [OH^-]_{final} = \frac{1 \text{ mmol}}{1000 \text{ mL}} = 10^{-3} \) M.
Step 4: Calculate pH: \( \text{pOH} = -\log(10^{-3}) = 3 \). \( \text{pH} = 14 - 3 = 11.0 \).