The pH of the solution at $25$ which has twice as many hydroxide ion as in pure water at $25$ , will be:

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6.699

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7.301

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Solution

The correct option is C 7.301
Given,
$T e m p = 25^{0} C$
So, No of hydroxide in pure water is $10^{- 7}$
A/C to this condition no of hydroxide ion is $2 \times 10^{- 7}$
So, $p o H = - l o g_{10} [2 \times 10^{- 7}]$
$7 + 0.301 = 7.301$

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