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Determine degree of dissociation of `0.05M NH_(3)` at `25^(@)C` in a solution of `pH = 11`. |
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Answer» Correct Answer - 2 `{:(NH_(4)OHhArr,NH_(4)^(+)+,OH^(-)),(1,0,0),((1-alpha),alpha,alpha):}` Given `pH =11` `:. [H^(+)]=10^(-11)` `:. [OH^(-)]=10^(-3)=c.alpha` Since `c=0.05` `alpha= (10^(-3))/(c )=(10^(-3))/(0.05)=0.02=2%` |
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