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The P^(H) of 0.005 M Ba (OH)(2) is...

The `P^(H)` of 0.005 M Ba `(OH)_(2)` is

A

`2.301`

B

`11.699`

C

`12`

D

`7`

Text Solution

Verified by Experts

The correct Answer is:
C
`0. 00 5 M " Ba " (OH)_2 , N = 5 xx 10 ^(-3) xx 2 = 10 ^(-2) `
` [OH^(-) ] = 10 ^(-2) rArr [H^(+) ] = 10 ^(-12) rArr pH = 12`
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AAKASH SERIES-IONIC EQUILIBRIUM -LECTURE SHEET (EXERCISE -II LEVEL -I (MAIN) (Straight Objective Type Questions))
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