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What is the pH of 1 M solution of acetic acid ? To what volume one litre of this solution be diluted so that pH of the resulting solution will be twice of the original value ? `(K_(a)=1.8xx10^(-5))` |
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Answer» Case I `{:(,CH_(3)COOH,hArr,CH_(3)COO^(Theta)+,H^(oplus)),("Concentration before dissociation",1,,0,0),("Concentration after dissociation",1-alpha,,alpha,alpha):}` `[H^(o+)] = Calpha = C sqrt((K_(a)//C))` `=sqrt((K_(a)C)) = sqrt((1.8 xx 10^(-5)xx1))` `= sqrt((18 xx 10^(-6)) = 4.24 xx 10^(-3) M` `:. pH =- log H^(o+) =- log (4.24 xx 10^(-3)) = 2.3724`or Use formula for the `pH` of weak acid. `pH = (1)/(2) (pK_(a) - log C) = (1)/(2) (4.7447 - 0) = 2.3724` Case II: New `pH` is given as `= 2.3724 xx 2 = 4.7488` Let new concentration be `C_(1)` and dergee of dissociation be `alpha_(1)` `:. -log [H^(o+)] = 4.7448` `:. [H^(o+)] = 1.8 xx 10^(-5)`or `C_(1) alpha_(1) = 1.8 xx 10^(-5)` Now again `K_(a) = ([H^(o+)][CH_(3)COO^(Θ)])/([CH_(3)COOH])` `K_(a) = (C_(1)alpha_(1) xx C_(1)alpha_(1))/(C_(1)(1-alpha_(1))) = (C_(1)alpha_(1)alpha_(1))/((1-alpha_(1)))` `:. 1.8 xx 10^(-5) = (1.8 xx 10^(-5) xx alpha_(1))/((1-alpha_(1))) :. alpha_(1) = 0.5` Now `C_(1)alpha_(1) = 1.8 xx 10^(-5)` `:. C_(1) = (1.8 xx 10^(-5))/(alpha_(1)) = (1.8 xx 10^(-5))/(0.5) = 3.6 xx 10^(-5)M` Let `1L` of concentrated solution be diluted to `VL` Equivalent of dilute solution `=` Equivalent of concentrated solution `3.6 xx 10^(-5) xxV = 1xx1 [M_(1)V_(1) = M_(2)V_(2)]` `:. V = (1)/(3.6 xx 10^(-5)) = 2.77 xx 10^(4)L` Note: iIn `II` case `alpha` comes `0.5` by `K_(a) = (C_(1)alpha_(1)^(2))/((1-alpha_(1)))`, and thus, it is not advisible to assume `(1-alpha_(1)) ~~1`. |
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