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Determine the degree of ionisation and pH of a 0.05 M ammonia solution. Also calculate ionisation constant of the conjugate acid ammonia. Given that `K_(b)` for `NH_(3)" is "1.77 xx 10^(-5)`. |
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Answer» Calculation of degree of ionisation `(alpha)`. The ionisation of `NH_(3)` may be represented as `NH_(3) +H_(2)OhArr NH_(4)^(+) +OH^(-)` The dissociation constant for the base may be given as : `K_(b) =(Calpha^(2))/(1-alpha)=Calpha^(2)` or " "`alpha = ((K_(b))/(C))^(1//2) =((1.77xx10^(-5))/(0.05))^(1//2)=0.018` (II) Calculation of pH of solution `[OH^(-)] =Calpha = 0.05 xx0.018 =9.4 xx 10^(-4) M` `[H^(+)]=(K_(w))/[[OH^(-)]] = (10^(-14))/(9.4 xx10^(-4)) =1.06 xx 10^(-11)` `pH =- log [H^(+)] =- log (1.06 xx 10^(-11)) = 10.97` (III). Caculation of ionisation constant of conjugate acid `NH_(4)^(+)` ions We know that for conjugate base-acid `K_(a) xxK_(b) =K_(w) or K_(a)=(K_(w))/(K_(b)) , K_(a) =(10^(-14))/(1.77xx10^(-5)) =5.64 xx 10^(-10)` |
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