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 strong acid-strong base titrations, always focus on the moles of $H^+$ and $OH^-$. The excess ion determines the final pH. Remember to calculate concentrations using the total volume after mixing.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the pH of a solution formed by mixing a strong acid (HCl) and a strong base (NaOH) in specific concentrations and volume ratios.

Step 2: Key Formula or Approach:
1. Calculate the initial moles of $H^+$ from HCl and $OH^-$ from NaOH.
2. Determine the limiting reactant and the excess moles of $H^+$ or $OH^-$.
3. Calculate the total volume of the mixed solution.
4. Calculate the concentration of the excess ion.
5. Calculate pOH (if $OH^-$ is in excess) or pH (if $H^+$ is in excess).
6. Use $pH + pOH = 14$ if needed.

Step 3: Detailed Explanation:
Let the volume ratio of NaOH solution : HCl solution be 2:3.
Assume the volume of NaOH solution is $2V$ liters and the volume of HCl solution is $3V$ liters.
1. Moles of NaOH ($OH^-$ ions):
Moles of NaOH = Concentration \(\times\) Volume = $0.02 \text{ M} \times 2V \text{ L} = 0.04V \text{ mol}$.
2. Moles of HCl ($H^+$ ions):
Moles of HCl = Concentration \(\times\) Volume = $0.01 \text{ M} \times 3V \text{ L} = 0.03V \text{ mol}$.
3. Reaction: NaOH and HCl react in a 1:1 molar ratio ($H^+ + OH^- \rightarrow H_2O$).
Since $0.04V > 0.03V$, NaOH (and thus $OH^-$) is in excess. HCl (and thus $H^+$) is the limiting reactant.
4. Moles of $OH^-$ remaining after reaction:
Moles of $OH^-$ remaining = Initial moles of $OH^-$ - Moles of $H^+$ reacted
= $0.04V \text{ mol} - 0.03V \text{ mol} = 0.01V \text{ mol}$.
5. Total volume of the mixed solution:
Total volume = $2V \text{ L} + 3V \text{ L} = 5V \text{ L}$.
6. Concentration of $OH^-$ in the mixed solution:
$[OH^-] = \frac{\text{Moles of OH}^- \text{ remaining}}{\text{Total volume}} = \frac{0.01V \text{ mol}}{5V \text{ L}} = \frac{0.01}{5} \text{ M} = 0.002 \text{ M}$.
So, $[OH^-] = 2 \times 10^{-3} \text{ M}$.
7. Calculate pOH:
$pOH = -\log[OH^-] = -\log(2 \times 10^{-3}) = -\log 2 - \log 10^{-3} = 3 - \log 2$.
Using $\log 2 \approx 0.301$:
$pOH = 3 - 0.301 = 2.699$.
8. Calculate pH:
At 298 K, $pH + pOH = 14$.
$pH = 14 - pOH = 14 - 2.699 = 11.301$.
Rounding to one decimal place, $pH \approx 11.3$.

Step 4: Final Answer:
The pH of the mixed solution will be 11.3.