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100 ml of `0.2 M H_(2)SO_(4)` is added to 100 ml of 0.2 M NaOH. The resulting solution will beA. AcidicB. BasicC. NeutralD. Slightly basic |
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Answer» Correct Answer - 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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