InterviewSolution
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InterviewSolution
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500 mL of 0.2M aqurous solution of aceticacid is mixed with 500 mLof 0.2M HClat25^(@)C (i) Calculatethe degreeof dissociation acidin the resulting solutionandpH ofthe solution . (b) If 6 g of NaOH is addedto the abovesolution , determinethe final pH . [ There is nochangein volume on mixing , Ka of aceticacid is1.75 xx 10^(-5) mol L^(-1) |
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Answer» Solution :The volumebeingdoubledby mixingtwo SOLUTIONS, the molarityof eachcomponentwill be halved i.e ` [CH_(3)COOH] = 0.1 M , [HCl] = 0.1 M ` HClbeing a strongacidwill remaincompletelyionised and hence`H^(+)`ion concentrationfurnishedby it will be `0.1 `M . This wouldexert commonion effecton the dissociation ofaceticacid ,a weak acid. `{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),(C,,0,,0),(C(1-alpha),,Calpha,,Calpha +0.1):}` `K_(a)= (Calpha(Calpha+0.01))/(C(1-alpha)) = (Calpha^(2)+0.1alpha)/((1-alpha))` Neglecting `alpha `in comparisonto UNITY and `Calpha^(2)` i.e ` 0.1 alpha^(2)` `K_(a) = 0.1 alpha ` oe `alpha= (K_(a))/(0.1) = (1.75 xx10^(-5))/(0.1) = 1.75xx 10^(-4)` ` [H^(+)] TOTAL = 0.1 +Calpha ,Calpha ` is negligiableas compared to `0.1 ` ` :. [ H^(+)] _(Total) = 0.1 ` ` :.pH = 1 ` (ii) `6g NaOH = 6/40= 0.15 ` mol `0.1 ` moleof NaOH will beconsumedby`0.1 ` mole of HCL . Thus, `0.05` mole of NaOH will react with aceticacid(No. of molof `CH_(3)COOH = 1 xx 0.2 = 0.1 ` ) accordingto the equation `{:(CH_(3)COOH,+,NaOH,to,CH_(3)COONa,+,H_(2)O),(0.1" mol",,0.05 " mol",,0,,0),(0.05 " mol",,0 " mol" ,,0.05 " mol",,0.05 " mol"):}` Thus , solutionof aceticacidand sodium acetate will becomeacidicbuffer . So pH of the buffer will be `pH = pK_(a) + log.(["salt"])/(["acid"])= - log(1.75 xx 10^(-5)) + log1 = 4.75` |
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