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For a weak base BOH,K(b)=10^(-4) .Calcul...

For a weak base `BOH,K_(b)=10^(-4)` .Calculate `pH` of `10^(-4)BOH` solution. (Take `log6.2=0.79)`

A

`10`

B

`9.79`

C

`8`

D

None of these

Text Solution

Verified by Experts

`{:(,BOH hArr ,B^(+),+,OH^(+)),(t=0,10^(-4),0,,0),(t=eq,10^(-4)alpha,10^(-4)alpha,,10^(-4)alpha):}`
`K_(b)=([B^(+)][OH^(-)])/([BOH])=(10^(-4)alpha10^(-4)alpha)/(10^(-4)(1-alpha)).....(i)`
Expecting `alphaltlt1,10^(-4)alpha^(2):. alpha=1` (not negligible)
So the solve quardic(1):
`10^(-4)=(10^(-4)alpha,10^(-4)alpha)/(10^(-4)(1-alpha))rArr alpha^(2)+alpha-1=0`
`alpha=(-1+-sqrt(1+4))/(2)rArr alpha=(sqrt(5)-1)/(2)~~0.62`
`:.[OH^(-)]=Calpha=10^(-4)xx0.62=6.2xx10^(-5)M`
`:.pOH=4.21:.pH=9.79`
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