InterviewSolution
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InterviewSolution
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`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))` Calculate the `[(Ind^(Theta))/(Hin)]`A. `1.413 xx 10^(-5)`B. `1.413 xx 10^(-4)`C. `3.128xx10^(-5)`D. `3.128xx10^(-14)` |
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Answer» Correct Answer - A i. `{:(,H_(3)PO_(4)overset(K_(1)=10^(-3))hArr,H^(o+)+,H_(2)PO_(4)^(Theta),),("Initial",C,0,0,),("At equilibrium",C(1-alpha),Calpha,Calpha,):}` `1 xx 10^(-3) = (Calpha^(2))/(1-alpha) =(0.05xxalpha^(2))/(1-alpha)` (Neglecting `alpha` wrt 1) Using direct formula for `pH` of a weak acid `(pK_(a_(1)) = 3, C = 0.05)` `pH = (1)/(2) (pK_(a_(1)) - log C) = (1)/(2) (3-log 0.05)` `= (1)/(2) (3-0.7+2) = 2.15` ii. For an acidic indicator. `HIn overset(K_(Ind))hArr H^(o+) + Ind^(Theta)`. `:. pH = pK_(Ind) + "log" ([Ind^(Theta)])/([HIn])` `2.15 = 7 + "log" ([Ind^(Theta)])/([HIn]) (K_(Ind) = 10^(-7))` `"log"([Ind^(Theta)])/([HIn]) = 2.15 - 7 = - 4.85` `([Ind^(Theta)])/([HIn]) = "Antilog" (-4.85)` `= "Antilog" (-4-0.85 +1-1)` `= "Antilog" (bar(5).15) = 1.413 xx 10^(-5)` |
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