First slide

Ionic product of water

Question

At certain temperature, $p^{H}$ of 0.001 N solution of NaOH is 9. Ionic product of water at that temperature will be

Difficult

Solution

$Temperature influences K_{w}; 0.001 N NaOH, [{OH}^{-}] = 10^{- 3}; p^{OH} = 3; Given p^{H} = 9; p^{K_{w}} = 9 + 3 = 12; K_{w} = 10^{- 12}$

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