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The [OH^(-)] of 0.005 M is H(2)SO(4)...

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

A

`2xx10^(-12)M`

B

`5xx10^(-12)M`

C

`10xx10^(-12)M`

D

`10xx10^(-14)M`

Text Solution

Verified by Experts

The correct Answer is:
C
`H_(2)SO_(4) to 2H^(+)+SO_(4)^(-2)2xx0.005`
`K_(w)=[H^(+)][OH^(-)`
`[OH^(-)]=(1xx10^(-14))/(2xx0.005)=(1xx10^(-14))/(0.01)=(10^(-14))/(10^(-2))`
`:.[OH^(-)]=10^(-12)`
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