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Choose the correct answer: At 40^(@)C ...

Choose the correct answer:
At `40^(@)C` the ionic product of water being `2.92xx10^(-14)`, its pH is-

A

`lt7.0`

B

`gt7.0`

C

`7.0`

D

`14.0`

Text Solution

Verified by Experts

For pure water, `pH=(1)/(2)pK_(w)=-(1)/(2)log_(10)K_(w)`.
At `40^(@)C,K_(w)=2.92xx10^(-14)` for pure water.
`therefore pH=-(1)/(2)log_(10)(2.92xx10^(-14))=6.76`.
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