Question

0.1 M HCl and 0.1 M H₂SO₄ each of volume 2 mL are mixed and the volume is made up to 6 mL by adding 2 mL of 0.01 N NaCl solution. The pH of the resulting mixture is

• A. $1.17$
• B. $1.0$
• C. $0.3$
• D. $\log 2 - \log 3$
• E. $\log 3 - \log 2$

Hint:

The total $H^+$ concentration determines pH; for strong acids, assume full dissociation.

Answer 1

The Correct Option is B

Step 1: The total concentration of $H^+$ ions is calculated as: \[ {HCl contribution} = 0.1 M \times \frac{2}{6} = \frac{0.2}{6} \] \[ {H}_2{SO}_4 { contribution} = 2 \times (0.1 M \times \frac{2}{6}) = \frac{0.4}{6} \] Step 2: Total $H^+$ concentration: \[ [{H}^+] = \frac{0.2}{6} + \frac{0.4}{6} = \frac{0.6}{6} = 0.1 M \] Step 3: pH calculation: \[ {pH} = -\log [H^+] \] \[ {pH} = -\log(0.1) = 1.0 \] Step 4: Therefore, the correct answer is (B).