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The dissociation constant of HCN is 5 xx...

The dissociation constant of HCN is `5 xx 10^(-10)`. The pH of the solution prepared by mixing `1.5` mole of HCN and `0.15` moles of KCN in water and making up the total volume to `0.5 dm^(3)` is

A

7.302

B

9.302

C

8.302

D

10.302

Text Solution

Verified by Experts

The correct Answer is:
C
`pH = -log K_(a) + log.([KCN])/([HCN])`
`pH = -log [5 xx 10^(-10)] + log((0.5)/(1.5)) = 8.302`
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