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Q.

$p^{H}$ of 0.005M Barium hydroxide solution at 298 K will be

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a

2.3

b

12

c

11.7

d

2

answer is B.

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Detailed Solution

$[{OH}^{-}] = Normality of Ba {(OH)}_{2}; [{OH}^{-}] = 2 \times 5 \times 10^{- 3}; [{OH}^{-}] = 10^{- 2}; p^{OH} = 2; p^{H} = 12$

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