The [OH]^(-)] of 0.005 M H_(2)SO_(4) is

1.	The [OH]^(-)] of 0.005 M H_(2)SO_(4) is
Answer» `2xx 10^(-12)M` ` 5xx10 ^(-3) M` ` 10^(-2) M` ` 10^(-12) M` Solution :` [H^(+) ] =( 5 xx 10 ^(-3)) 2 ,[H^(+) ][OH^(-) ] = 10 ^(-14) ` ` 10 ^(-2)=[OH^(-) ] = 10 ^(-14)rArr [OH^(-)] = 10 ^(-14)rArr [OH^(-) ]10 ^(-12)M `

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