$100 m L$ of $0.1 M$ $H C l$ is mixed with $100 m L$ of $0.01 M$ $H C l$ . The pH of the resulting solution is:

2.0

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1.0

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1.26

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None of these

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Solution

The correct option is C $1.26$
When you add two different concentrations of same strong acid, the $p H$ contribution will be from both concentrations.

$M_{1} = 0.1 M, V_{1} = 100 m l$
$M_{2} = 0.01 M, V_{2} = 100 m l$

Final volume of solution is $100 + 100 = 200 m l$

The final concentration of mixture will be calculated as
$M V = M_{1} V_{1} + M_{2} V_{2}$

$\Rightarrow M (100 + 100) = 0.1 \times 100 + 0.01 \times 100$

$M = \frac{(10 + 1)}{200} = 0.055 M$

$p H = - l o g [H^{+}] = - l o g (0.055) = 1.259 \sim 1.26$

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