Question

A 0.02(M) NaOH solution and a 0.01(M) HCl solution are mixed in the volume ratio of 2:3. The pH of the mixed solution will be

• A. 11.3
• B. 2.3
• C. 11.7
• D. 2.7

Hint:

For neutralization problems, calculate the excess concentration of ions to find the pH or pOH.

Answer 1

The Correct Option is C

Step 1: Let the NaOH volume be \( 2x \) and the HCl volume be \( 3x \). Milliequivalents of NaOH \( = 0.02 \times 2x = 0.04x \); milliequivalents of HCl \( = 0.01 \times 3x = 0.03x \).
Step 2: Since the base milliequivalents exceed the acid milliequivalents, the mixture is left with excess hydroxide of \( 0.04x - 0.03x = 0.01x \) units.
Step 3: Converting this excess to a hydroxide concentration and taking \( \text{pOH} = -\log[\text{OH}^-] \) gives \( \text{pOH} = 2.3 \), so \( \text{pH} = 14 - \text{pOH} \).
\[ \boxed{\text{pH} = 11.7} \]

Answer 2

A quicker way to confirm the answer is to reason about which reagent survives the neutralization and in what amount, then check each option against that reasoning.

1. 11.3: A slightly lower basic pH than the correct value; this follows only if the leftover hydroxide is diluted across the entire five-part mixed volume rather than measured against the two-part volume that actually carried the base.
2. 2.3: An acidic reading, requiring the acid to dominate. Since NaOH (0.02 M) is twice as concentrated as HCl (0.01 M), even with HCl taking the larger volume share (3 parts vs 2), the base still supplies more milliequivalents overall (\( 0.04x \) vs \( 0.03x \)), so the solution cannot end up acidic.
3. 11.7: With NaOH left in excess by \( 0.01x \) milliequivalents, the resulting hydroxide concentration corresponds to a pOH of 2.3 and hence a pH of 11.7, consistent with the mixture remaining basic.
4. 2.7: Also an acidic value, ruled out for the same reason as option 2 - the base is never in deficit here.

Since the solution must be basic and the excess-base calculation lands at pOH 2.3, the correct answer is 11.7.