If the degree of dissociation of water at 90-C is 1-28-10-8- then the ionization constant of water at 90-C is-

The correct option is C 9.07 ×10^ 15 ...

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Introduction and Degree of Hydrolysis

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If the degree of dissociation of water at $90^{\circ} C$ is $1.28 \times 10^{- 8}$ , then the ionization constant of water at $90^{\circ} C$ is:

1.28 \times 10^{- 14}

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12.81 \times 10^{- 15}

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9.07 \times 10^{- 15}

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7.52 \times 10^{- 12}

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