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The pK(b) of CN^(Theta) is 4.7. The pH i...

The `pK_(b)` of `CN^(Theta)` is `4.7`. The `pH` is solution prepared by mixing `2.5 mol` of `2.5 mol` of `KCN` of `2.5 mol` of `HCN` in water and making the total volume upto `500mL` is

A

`10.3`

B

`9.3`

C

`8.3`

D

`4.7`

Text Solution

Verified by Experts

The correct Answer is:
B
It forms basic buffer
`pOH = pK_(b) + log ((2.5//500)/(2.5//500)) = 4.7`
`pH = 14 - 4.7 = 9.3`.
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