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The `pH` of `0.005 M` codine `(C_(18)H_(21)NO_(3))` solution is `9.95`. Calculate its ionisation constant and `pK_(b)`. |
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Answer» Codenine is `W_(B)`: `BOH hArr B^(o+)+overset(Θ)(O)H` `pH=9.95 :. pOH=14-9.95=4.05` `pOH_(W_(B))=1/2(pK_(b)-logC)` `4.05xx2=(pK_(b)-log 5xx10^(-3))` `8.1=(pK_(b)-0.7+3), pK_(b)=8.1-2.3=5.8` `K_(b)=Antilog (-5.8)=` Antilog `(-5-0.8+1-1)` =Antilog `bar(6).2=1.588xx10^(-16)` `:. K_(b)=1.588xx10^(-6)` |
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