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InterviewSolution
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Caculate the pH value of 0.20M solution of methyl amine `(CH_(3)NH_(2))` at 298 K, given that its ionisation constant `(K_(b))` is `4.4 xx 10^(-5).` |
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Answer» Let `alpha` be the degree of ionisation of methylamine. The concentration of the various species before the reaction and at the equilibrium point are: `{:(,CH_(3)NH_(2) +H_(2)O(l) ,hArr, CH_(3)NH_(3)^(+) (aq) ,+,OH^(-) (aq)),("Initial molar con".,0.2,,0,,0),("Equilibrium molar conc".,(0.2-alpha),,alpha,,alpha):}` Applying Law fo chemical equilibrium, `K_(b) =[[CH_(3)NH_(3)^(+)][OH^(-)]]/[[CH_(3)NH_(2)]] , 4.4 xx 10^(-5) = (alphaxx alpha)/((0.2 -alpha))` As methylamine is a weak base, `alpha` is very small. It can be negtected. Therefore ,0.2.This, `4.4 xx 10^(-5) (alpha^(2))/(0.2) or alpha = (0.2 xx 4.4 xx10^(-5))^(1//2) = 2.97 xx 10^(-3)` `[OH^(-)] = 2.97 xx 10^(-3) M` `[H_(3)O^(+)]=(K_(2))/[[OH^(-)]]=((1.0xx10^(-14)M^(2)))/((2.97xx10^(-3)M))=3.37 xx 10^(-12) M` `pH =- log [H_(3)O^(+)] =- log (3.37 xx 10^(-12))` `=12- log 3.37 =12 -0.53 =11.47` |
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