Determine degree of dissociation of `0.05M NH_(3)` at `25^(@)C` in a solution of `pH = 11`.

Answer

2

Explanation

`{:(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%`