0.1 M HCl and 0.1 M H 2 SO 4 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

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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).

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$\log 2 - \log 3$

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$\log 3 - \log 2$

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

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The Correct Option is B

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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