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)=6.74xx10^(-4)`

Explanation

`H_(2)+2H^(+)(C)+2e^(-)`
`underline(2Ag^(+)(0.8M)+2e^(-)to2Ag).
`underline(H_(2)(g)+2Ag^(+)(0.8M)to2H^(+))(C)+2Ag," "E^(@)=0+0.8=0.8"volt"`
`0.982=0.8-(0.0591)/(2)log(([H^(+)]^(2))/(1xx(0.8)^(2)))` …(i)
`K_(a(HOCN))=([H^(+)]^(2))/(1.3xx10^(-3)-[H^(+)])` ...(ii)
From equation (i) and (ii)
`K_(a(HOCN))=6.74xx10^(-4)`