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