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Determine degree of dissociation of 0.05...

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

Text Solution

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`NH_(4)OH Leftrightarrow NH_(4)^(+)+OH^(-)`
`"1 0 0"`
`c(1-alpha)" "c alpha" "c alpha`
Given pH =11
`therefore [H^(+)]=10^(-11) [? [H^(+)] [OH]=10^(-14)]`
`[OH^(-)]=10^(-3) =c alpha`
Since, c=0.05
`therefore alpha =(10^(-3))/(c)=(10^(-3))/(0.005) =2 xx 10^(-2) or 2%`
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