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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
`K_(a)(HCN) = 5 xx10^(-10)`
`[HCN] = 1.5//0.5 "moles/L" = 3M `
`[KCN] = 0.15//0.5 "moles/L" =0.3 M `
`pH = pK_(a)+ log. (["salt"])/(["acid"])`
` = - log ( 5 xx 10^(-10)) +log (0.3//3)`
`= 10 - log 5-1 = 8.301 `
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