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Calculate the $P^{H}$ of 0.08 M solution of hypochlorous acid $(H O C l)$ . The ionization constant of the acid is $2.5 \times 10^{- 5}$ . (Consider $0.08 \mp x = 0.08$ and $\log (1.41) = 0.15$ )

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6.5

2.80

2.85

11.15

answer is B.

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Detailed Solution

$k_{a} = \frac{x^{2}}{0.08 - x} = 2.5 \times 10^{- 5} x = 1.41 \times 10^{- 3} = [H^{+}] P^{H} = - \log (1.41 \times 10^{- 3}) = 2.85$

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