How many moles of $N a O H$ must be removed from one litre of its aqueous solution to change its pH from $12$ to $11$ ?

0.009

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0.01

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0.09

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0.1

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Solution

The correct option is A $0.009$
Solution
Initial pH is 12
pOH=14-12=2
$[O H^{-}]_{i n i t i a l} = - l o g [O H^{-}] = 2$
$[O H^{-}]_{i n i t i a l} = 10^{- 2}$
Final pH is 11
pOH=14-11=3
$[O H^{-}]_{f i n a l} = - l o g [O H^{-}] = 3$
$[O H^{-}]_{f i n a l} = 10^{- 3}$
Since volume is 1 litre molarity and no. of moles are equal
No. moles to be removed=1/100-1/1000
=0.009
The correct option is A

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