1 mL of 0.1 N HCl is added to 999 mL solution of NaCl. The pH of the resulting solution will be :
A
7
B
4
C
2
D
1
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem of finding the pH of the resulting solution when 1 mL of 0.1 N HCl is added to 999 mL of NaCl solution, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the components**: We have 1 mL of 0.1 N HCl and 999 mL of NaCl solution. HCl is a strong acid that will completely dissociate into H⁺ and Cl⁻ ions.
2. **Convert Normality to Molarity**: Since HCl is a strong acid with an n-factor of 1 (it donates one H⁺ ion), the normality is equal to molarity. Therefore, 0.1 N HCl is also 0.1 M HCl.
3. **Calculate the number of moles of H⁺ ions**:
- Volume of HCl = 1 mL = 0.001 L
- Molarity of HCl = 0.1 M
- Number of moles of H⁺ = Molarity × Volume = 0.1 mol/L × 0.001 L = 0.0001 moles (or 0.1 mmol).
4. **Determine the total volume of the solution**:
- Volume of NaCl solution = 999 mL = 0.999 L
- Total volume after adding HCl = 999 mL + 1 mL = 1000 mL = 1 L.
5. **Calculate the final concentration of H⁺ ions**:
- Final concentration of H⁺ = Number of moles of H⁺ / Total volume = 0.0001 moles / 1 L = 0.0001 M = 10⁻⁴ M.
6. **Calculate the pH of the solution**:
- pH = -log[H⁺]
- pH = -log(10⁻⁴) = 4.
### Final Answer:
The pH of the resulting solution is **4**.
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