The `pH` of `0.1 M` solution of cyanic acid `(HCNO)` is `2.34`. Calculate the ionization constant of the acid and its degree of ionisation in the solution.

Explanation

HCNO ionizes as :
`HCNO hArr H^(+)+ CNO^(-)`
The dissociation constant
`K_a=([H^+][CNO^-])/([HCNO])`
`[H^+]=[CNO^-]`
Now pH=2.34
- log `[H^+]` =2.34
log `[H^+]=-2.34 =bar(3).66`
`[H^+]` = Antilog `(bar(3).66)`
`=4.57xx10^(-3)` M
`therefore [H^+]=[CNO^-]=4.57xx10^(-3)` M and [HCNO]=0.1 M
`K_a=((4.57xx10^(-3))xx(4.57xx10^(-3)))/0.1`
`=2.09xx10^(-4)`
Degree of dissociation =`alpha = sqrt(K_a/c)`
`=sqrt((2.09xx10^(-4))/0.1)`
`=4.57xx10^(-2)` or =0.0457.