The ionic product of water at 25^° C is 10-14. Itsionic product at 90^° C will be(1) 1 x 10-14(2) 1x 10-16(3) 1 x 10-20(4) 1x 10-12

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25^{o} C

10^{- 14}

90^{o} C

1.96 \times 10^{- 14}

35^{\circ} C .

10^{\circ} C

40^{o} C

4 \times 10^{- 14}

O H^{-}

40^{\circ} C

70^{o} C

1.28 \times 10^{- 8}

At $90^{\circ}$ C pure water has [ ${H_{3} O}^{+}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$ . The value of $K_{W}$ at $90^{\circ}$ C is

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