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Auto Protolysis of Water

The ionic of ...

Question

The ionic of water at $25^{o} C$ is $10^{- 14}$ . The ionic product at $90^{o} C$ will be:

1 \times 10^{- 20}

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1 \times 10^{- 12}

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1 \times 10^{- 14}

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1 \times 10^{- 16}

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Solution

The correct option is B $1 \times 10^{- 12}$
When increasing the temperature the value of ionic product also increase.
Option B is correct.

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