1.

Calculate the percentage hydrolysis in `0.003M` aqueous solution of NaOCN. `(K_(a)for HOCN=3.33xx10^(-4))`

Answer» `NaOCN+H_(2)OhArrNaOH+HCN`
`h= sqrt([(K_(H))/(C)])=sqrt([(K_(w))/(K_(a).C)])=sqrt((10^(-14))/(3.33xx10^(-4)xx0.003))`
`h= 10^(-4)`
`:. %` hydrolysis `= (10^(-4))/(100)=10^(-2)%`


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