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Calculate the percentage dissociation ...

Calculate the percentage dissociation of `0.5` M `NH_(3)` at `25^(@)C` in a solution of pH = 12

Text Solution

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`{:(NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-)),(C,,0,,0),(C(1-alpha),,Calpha,,Calpha ):}`
`pH = 12,[H^(+)] = 10^(-12) , or [OH] = 10^(-2)`
`[ OH] = C alpha = 10^(-2)`
`alpha = (10^(-2))/C = (10^(-2))/(0.5)`
`= 2 xx 10^(-2) ` or 2 %
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