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Calculate the [OH^(-)] in 0.01M aqueous ...

Calculate the `[OH^(-)]` in `0.01M` aqueous solution of `NaOCN(K_(b)` for `OCN^(-)=10^(-10)) :`
(a)`10^(-6)` M
(b)`10^(-7)` M
(c)`10^(-8)` M
(d)None of these

A

`10^(-6)M`

B

`10^(-7)M`

C

`10^(-8)M`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
1
`K_(a(HOCN))=(10^(-14))/(10^(-10))=10^(-4)`
`K_(h)=K_(b(OCN^(-)))=10^(-10)rArr" "alpha=sqrt((K_(h))/(C))=sqrt((10^(-10))/(0.01))" ":'alphalt lt0.1," ":.K_(h)=calpha^(2)`
`[OH^(-)]=calpha,[OH^(-)]=0.01xx10^(-4)=10^(-6)M`
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