Question

200 mL of 0.1 M NaOH is allowed to react completely with 100 mL of 0.1 M HCl and the solution is diluted to 1.0 L by adding water. The pH of the mixture is

• A. 3
• B. 11
• C. 2
• D. 12

Hint:

If base is in excess, calculate $pOH$ first, then subtract from 14 to find $pH$.

Answer 1

The Correct Option is D

Step 1: Concept
Calculate the milli-moles of $H^{+}$ and $OH^{-}$ to find the resultant concentration after neutralization.

Step 2: Meaning
Moles of $NaOH = 200 \times 0.1 = 20$ mmol. Moles of $HCl = 100 \times 0.1 = 10$ mmol.

Step 3: Analysis
Excess $OH^{-} = 20 - 10 = 10$ mmol. Final volume $= 1.0$ L $= 1000$ mL. Resultant $[OH^{-}] = 10 \text{ mmol} / 1000 \text{ mL} = 0.01$ M $= 10^{-2}$ M.

Step 4: Conclusion
$pOH = -\log(10^{-2}) = 2$. $pH = 14 - pOH = 14 - 2 = 12$. Final Answer: (D)