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100ml of 0.2 M H(2)SO(4) is added to 100...

`100ml` of `0.2 M H_(2)SO_(4)` is added to `100 ml` of `0.2 M NaOH`. The resulting solution will be

A

Acidic

B

Basic

C

Neutral

D

Slightly basic

Text Solution

Verified by Experts

The correct Answer is:
A
M.eq. of 0.2 M `H_(2)SO_(4) = (2 xx 0.2 M)/(1000) xx 100 = 0.04 m//l`
M.eq. of 2M NaOH `= (0.2)/(1000) xx 100 = 0.02 m//l`
left `[H^(+)] = .04 - .02 = .02`
Total volume `= 200 = (.02)/(200) =.0001 = 10^(-4)M`
pH =4.
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