Q4 / 180 · COMEDK 2024 (Afternoon Shift) · Chemistry
$1 \mathrm{~cm}^3 \text { of } 0.01 \mathrm{~M} \mathrm{~HCl} \text { is added to } 1 \mathrm{~L} \text { of } \mathrm{NaCl} \text { solution. The } \mathrm{pH} \text { of the resulting solution is: }$
- A3
- Correct answerB5
- C2
- D1
Solution
The volume of the HCl solution added is $V_1 = 1 \text{ cm}^3 = 1 \times 10^{-3} \text{ L}$.
The concentration of the HCl solution is $M_1 = 0.01 \text{ M} = 10^{-2} \text{ M}$.
The total volume of the resulting solution is $V_2 = 1 \text{ L} + 0.001 \text{ L} \approx 1 \text{ L}$.
The number of moles of $\text{H}^{+}$ ions added is $n = M_1 \times V_1 = 10^{-2} \text{ mol/L} \times 10^{-3} \text{ L} = 10^{-5} \text{ mol}$.
The concentration of $\text{H}^{+}$ ions in the final solution is $[\text{H}^{+}] = \dfrac{n}{V_2} = \dfrac{10^{-5} \text{ mol}}{1 \text{ L}} = 10^{-5} \text{ M}$.
The pH of the solution is given by $\text{pH} = -\log[\text{H}^{+}] = -\log(10^{-5}) = 5$.
Answer: 5
The concentration of the HCl solution is $M_1 = 0.01 \text{ M} = 10^{-2} \text{ M}$.
The total volume of the resulting solution is $V_2 = 1 \text{ L} + 0.001 \text{ L} \approx 1 \text{ L}$.
The number of moles of $\text{H}^{+}$ ions added is $n = M_1 \times V_1 = 10^{-2} \text{ mol/L} \times 10^{-3} \text{ L} = 10^{-5} \text{ mol}$.
The concentration of $\text{H}^{+}$ ions in the final solution is $[\text{H}^{+}] = \dfrac{n}{V_2} = \dfrac{10^{-5} \text{ mol}}{1 \text{ L}} = 10^{-5} \text{ M}$.
The pH of the solution is given by $\text{pH} = -\log[\text{H}^{+}] = -\log(10^{-5}) = 5$.
Answer: 5
Hence, the correct answer is option: 'B'.