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.

Answer 2

This is a strong acid-strong base mixing problem, so the first job is to work out which one is in excess, then check each of the four given pH values against that result.

1. 11.3: Taking 2 parts NaOH solution and 3 parts HCl solution, moles of \(OH^-\) supplied are \(0.02 \times 2 = 0.04\) (in concentration-times-volume-part units), while moles of \(H^+\) supplied are \(0.01 \times 3 = 0.03\). Since \(0.04 > 0.03\), the base is in excess by \(0.01\) part-units, spread over a total of \(5\) parts. That gives \([OH^-] = \frac{0.01}{5} = 2 \times 10^{-3}\) M, so \(pOH = -\log(2 \times 10^{-3}) \approx 2.7\) and \(pH = 14 - 2.7 = 11.3\). This matches the calculation exactly.
2. 2.3: A pH this low would mean the solution ends up strongly acidic, which would only happen if HCl were in large excess. Since NaOH actually supplies more moles here, this value does not fit.
3. 11.7: This value is close to the correct one but slightly too high; it would arise if someone mistakenly used the full \(0.04\) part-units of excess \(OH^-\) without subtracting the \(0.03\) part-units consumed by the acid, overstating how much base is left over.
4. 2.7: This is the correct pOH value, not the pH. Someone who forgot the final step of converting pOH to pH using \(pH + pOH = 14\) would mistakenly report this number as the answer.

Working through the excess-base calculation carefully, the mixture is mildly basic with a pH just above 11.

Therefore, the correct answer is 11.3.