Equal volume of `0.1 M` potassium hydroxide and `0.1 M` sulphuric acid are mixed. The `P^(H)` of resulting solution is

Correct Answer - D
normality of acid `gt` normality of base, so resultant solution is acidic in nature
`(N_(2)V_(2)- N_(1)V_(1))/(V_(1) + V_(2)) = (V[0.2 - 0.1])/(2V) = 0.1/2 = 0.05N`
`"acid" lt 7`

Equal volume of `0.1 M` potassium hydroxide and `0.1 M` sulphuric acid are mixed. The `P^(H)` of resulting solution is

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