1.

0.16g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of N_(2)H_(4) that has reacted with water in this solution. The K_(b) for N_(2)H_(4) is reacted with water in this solution. The K_(b) for N_(2)H_(4) is 4.0 xx 10^(-6) M

Answer»


SOLUTION :`{:(,N_(2)H_(4)+,H_(2)O hArr,N_(2)H_(5)^(+)+,OH^(-)),("At t = 0",1,,0,0),("At eqm.",1-alpha,,alpha,alpha),(,,,,):}`
`K_(B) = (C alpha^(2))/((1-alpha)) RARR K_(b) = C alpha^(2)`
(`because alpha` is very small, so `1-alpha ~~ 1`)
`[N_(2)H_(4)] = C = (0.16 xx 1000)/(32 xx 500) = 0.01`
`:. alpha^(2) = (4 xx 10^(-6))/(0.01) = 4 xx 10^(-4) "" (because = K_(b) = 4 xx 10^(-6)M)`
`:. alpha = 2 xx 10^(-2) = 0.02` or 2%.


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