100 ml of 0.015 M $H C l$ solution is mixed with 100 ml of 0.005 M $H C l$ . What is the $P^{H}$ of the resultant solution?

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answer is C.

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Detailed Solution

$\begin{array}{l} M_{f} = \frac{0.015 \times 100 + 0.005 \times 100}{100 + 100} = \frac{2}{200} = 10^{- 2} \\ P^{H} = - \log 10^{- 2} = 2 \end{array}$

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