InterviewSolution
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InterviewSolution
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0.6 mL of acetic acid (CH_(3)COOH)having a density of 1.06 g "mL"^(-1)is dissolved in 1 litre of water . The depression in freezing point observedfor this strength of the acidwas 0.0205^(@)C , Calculate the van't Hoff factor and the dissociation constant of the acidK_(f) for water= 1.86 K kg" mol"^(-1) |
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Answer» Solution :Calculation of van't HOFF factor (i) 0.6 mL of acid means solute `(w_(2)) = 0.6 xx 1.6 ` 1 Litre of water means solvent `(w_(1)) = 1000 g , (DELTA T_(f))_("observed") = 0.0205 ^(@) C` ` :. `Observed molar mass , `(M_(2)) _("observed") =(1000 xx K_(f) xx w_(2))/(w_(1) xx Delta T _(f))= (1000 " g mol"^(-1) xx 1.86" K kgmol"^(-1) xx 0.636g )/(1000g xx 0.0205 K)` ` = 57.7 g " mol"^(-1)` Calculatedmolar mass of `CH_(3) COOH = "60 g mol"^(-1)` ` :.`van't Hoff factor (i) = `("Calculated molar mass")/("Observed molar mass ") = (60"g mol"^(-1))/(57.7"g mol"^(-1))= 1.04` Calculation of degreeof dissociation `(alpha)`from van't Hoff factor (i) If `alpha` is the degreeof dissociation of ACETIC acid, then ` {:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial moles",1,,,,),("Moles at eqm.",1-alpha,,alpha,,alpha",Total" = 1+alpha):}` `:. i (1+alpha)/1 = 1 + alphaor alpha = 1 - i = 1.04 - 1.0 = 0.04` Calculation of dissociation constant , If we startwith C mol `L^(-1)`of acetic acid, then ` {:(,CH_(3) COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial conc.",C " mol "L^(-1),,,,),("Conc.at eqm.",C(1-alpha),,Calpha,,Calpha):}` Dissociation constant`(K_(a)) = ([CH_(3)COO^(-1)][H^(+)])/([CH_(3)COOH])=(C alpha. C alpha )/(C(1-alpha)) = (Calpha^(2))/(1-alpha) ` ButC =0.636 g `L^(-1) = (0.636 g L^(-1))/(60 g " mol"^(-1)) = 0.0106 "molL"^(-1) :. K _(a) = ((0.0106)(0.04)^(2))/(1-0.04) = 1.76 xx 10^(-5)` |
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