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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,hArr,overset(o+)NH_(4),+,overset(Θ)OH),(1,,0,,0),(1-alpha,,alpha,,alpha):}`
Given `pH = 11`
`:. [H^(o+)] = 10^(-11),:. [overset(Θ)OH] = 10^(-3) = C alpha`
Since, `C = 0.05`
`:. alpha = 10^(-3)//C = 10^(-3)//0.05 = 2 xx 10^(-2) = 2%`
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Knowledge Check

  • Determine the degree of ionisation of 0.05 M NH_(3) at 25^(0)C in a solution pH = 11 (K_(b) = 1.76 xx 10^(-5))

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    B
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