Voltage of the cell `Pt, H_(2) (1atm)|HOCN(1.3 xx 10^(-3)M)||Ag^(+) (0.8M) |Ag(s)` is `0.982V`. Calculate the `K_(a)` for `HOCN`, neglect `[H^(+)]` because of oxidation of `H_(2)(g)`
`Ag^(+) +e^(-) rarr Ag(s) = 0.8V`

Answer

`K_(a) = 6.74 xx 10^(-4)`

Explanation

`(e^(-)+Ag^(+)rarrAg)/((1)/(2)H_(2)+Ag^(+)rarrH^(+)+Ag_((s)))`
`0.982 =0.8 - (0.059)/(1) log.([H^(+)])/([Ag^(+)])`
`HOCN hArr H^(+) +OCN^(-)`
`K_(a) = ([H^(+)][OCN^(-)])/([HOCN])`