Q44 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
$300 \mathrm{ml}$ of an aqueous solution of $\mathrm{NaOH}$ with $\mathrm{pH}$ value of 10 is mixed with $200 \mathrm{ml}$ of an aqueous solution of $\mathrm{HCl}$ with a $\mathrm{pH}$ value of 4. What will be the $\mathrm{pH}$ of the resultant solution at room temperature?
- A11.262
- Correct answerB9.301
- C10.52
- D8.909
Solution
For the NaOH solution, $\text{pH} = 10$, so $\text{pOH} = 14 - 10 = 4$.
$[\text{OH}^{-}] = 10^{-4} \text{ M}$.
Moles of $\text{OH}^{-} = 300 \times 10^{-3} \text{ L} \times 10^{-4} \text{ mol/L} = 3 \times 10^{-5} \text{ mol}$.
For the HCl solution, $\text{pH} = 4$, so $[\text{H}^{+}] = 10^{-4} \text{ M}$.
Moles of $\text{H}^{+} = 200 \times 10^{-3} \text{ L} \times 10^{-4} \text{ mol/L} = 2 \times 10^{-5} \text{ mol}$.
Since $\text{OH}^{-}$ reacts with $\text{H}^{+}$ to form water, the remaining moles of $\text{OH}^{-}$ are $3 \times 10^{-5} - 2 \times 10^{-5} = 1 \times 10^{-5} \text{ mol}$.
Total volume of the mixture $= 300 \text{ ml} + 200 \text{ ml} = 500 \text{ ml} = 0.5 \text{ L}$.
Resultant concentration of $\text{OH}^{-} = \dfrac{1 \times 10^{-5} \text{ mol}}{0.5 \text{ L}} = 2 \times 10^{-5} \text{ M}$.
$\text{pOH} = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.301 = 4.699$.
$\text{pH} = 14 - 4.699 = 9.301$.
Answer: 9.301
$[\text{OH}^{-}] = 10^{-4} \text{ M}$.
Moles of $\text{OH}^{-} = 300 \times 10^{-3} \text{ L} \times 10^{-4} \text{ mol/L} = 3 \times 10^{-5} \text{ mol}$.
For the HCl solution, $\text{pH} = 4$, so $[\text{H}^{+}] = 10^{-4} \text{ M}$.
Moles of $\text{H}^{+} = 200 \times 10^{-3} \text{ L} \times 10^{-4} \text{ mol/L} = 2 \times 10^{-5} \text{ mol}$.
Since $\text{OH}^{-}$ reacts with $\text{H}^{+}$ to form water, the remaining moles of $\text{OH}^{-}$ are $3 \times 10^{-5} - 2 \times 10^{-5} = 1 \times 10^{-5} \text{ mol}$.
Total volume of the mixture $= 300 \text{ ml} + 200 \text{ ml} = 500 \text{ ml} = 0.5 \text{ L}$.
Resultant concentration of $\text{OH}^{-} = \dfrac{1 \times 10^{-5} \text{ mol}}{0.5 \text{ L}} = 2 \times 10^{-5} \text{ M}$.
$\text{pOH} = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.301 = 4.699$.
$\text{pH} = 14 - 4.699 = 9.301$.
Answer: 9.301
Hence, the correct answer is option: 'B'.