The pH of an H2SO4 solution is 2

The pH of an H2SO4 solution is 2. The pH of this solution, if an equal volume of 0.2 M NaCl solution is added to it, will be.

The correct answer and explanation is:

When 0.2 M NaCl is added to a sulfuric acid (H₂SO₄) solution, it does not affect the pH significantly because NaCl is a neutral salt that does not undergo hydrolysis in water. Sodium chloride dissociates completely into Na⁺ and Cl⁻ ions, neither of which affects the concentration of H⁺ ions or the acidity of the solution.

Step-by-step breakdown:

1. Initial pH of H₂SO₄ Solution:
   The pH of the original sulfuric acid solution is 2. Sulfuric acid is a strong acid, and it dissociates completely in water, releasing two H⁺ ions per molecule.
   • The concentration of H⁺ ions in a solution of H₂SO₄ can be calculated from the pH value using the formula:
   [H+]=10−pH[\text{H}^+] = 10^{-\text{pH}}
   • Since the pH is 2, the concentration of H⁺ ions is:
   [H+]=10−2=0.01 M[\text{H}^+] = 10^{-2} = 0.01 \, \text{M} This means that the H₂SO₄ solution has a concentration of 0.01 M H⁺ ions.
2. Effect of Adding NaCl Solution:
   When an equal volume of 0.2 M NaCl is added to the H₂SO₄ solution, the volume of the solution doubles, and the total moles of H⁺ ions remain the same. The H⁺ ions from H₂SO₄ will now be spread out in a larger volume, diluting the concentration of H⁺ ions.
   • The new concentration of H⁺ ions after dilution is:
   [H+]=0.01 M×VH₂SO₄2VH₂SO₄=0.012=0.005 M[\text{H}^+] = \frac{0.01 \, \text{M} \times V_{\text{H₂SO₄}}}{2V_{\text{H₂SO₄}}} = \frac{0.01}{2} = 0.005 \, \text{M}
   • The pH can be calculated using the formula:
   pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]
   • So, the new pH is:
   pH=−log⁡(0.005)≈2.3\text{pH} = -\log(0.005) \approx 2.3

Final Result:

The pH of the solution after adding an equal volume of 0.2 M NaCl will be approximately 2.3.