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The degree of ionizationof a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid. |
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Answer» Solution :Calculation of `[H^+]` of 0.1 M bromoacetic acid `(CH_2BrCOOH)`: Suppose the ionization degree of this acid = `ALPHA` `THEREFORE` Ionised acid = `Calpha = 0.1 alpha` So, at equilibrium acid = `(0.1-0.1alpha)=0.1 (1-alpha)` But VALUE of `alpha` is very LESS so, `0.1 (1-alpha) =0.1 (1) = 0.1` At equilibrium , `[H^+] = [CH_2BrCOO^-]=C alpha= 0.1 alpha` The ionic equilibrium is under : `{:(,CH_2BrCOOH_((aq)) + H_2O HARR, H_3O_((aq))^(+) + , CH_2BrCOO_((aq))^(-)),("Initial M:",0.1 , 0.0 M, 0.0 M),("Change in equilibrium :",-0.1 alpha ,+0.1 alpha,+0.1 alpha),("Concentration at M equilibrium",overset(0.1 (1-alpha)M)(approx0.1),0.1alpha,0.1 alpha):}` `K_a=([H_3O^+][CH_2BrCOO^-])/([CH_2BrCOOH])=0.132` `therefore ((0.1alpha)(0.1alpha))/0.1 = 0.132` `therefore alpha^2= 0.132/0.1 = 0.0132` `therefore alpha` = 0.1149 So, `[H^+]= Calpha = 0.1xx0.1149`=0.01149 M Calculation of pH: pH=-log `[H^+]`=-log (0.01149)=1.9397 = 1.94 Calculation of `pK_a` : `pK_a=-log K_a=-log (0.132)-0.8794` |
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