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Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74 . How is thedegree of dissociation affected when its solutionalso contains(a)0.01 M , (b) 0.1 M in HCl ? |
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Answer» SOLUTION :(i)Calculation of `K_a` on base of `pK_a` : `pK_a=-log (K_a)=-4.74 = bar5.26` So, `K_a=1.8197xx10^(-5)` `THEREFORE K_a= 10.0^(-4.74)` `=1.8197xx10^(-5)approx 1.82xx10^(-5)` (ii)Calculation of `K_a` on base concentration of `H^+` : `{:(,CH_3COOH_((aq))+H_2O_((l)) hArr, CH_3COO_((aq))^(-)+ , H_3O_((l))^(+)),("Inital M :" , 0.05 M, 0.0 M , 0.0 M),("At equilibrium M ", underset(approx 0.05 M) (0.05-x),CALPHA=xM, x M=Calpha):}` `therefore K_a=([CH_3COO^-][H^+])/([CH_3COOH])= 1.82xx10^(-5)` `therefore ((x)(x))/0.05 = 1.82xx10^(-5)` `therefore x^2=0.05 xx 1.82 xx 10^(-5) = 0.91 xx 10^(-6)` `therefore x=sqrt(0.91xx10^(-6))= 0.95396xx 10^(-3) M = [H^+]` (iii) Calculation of degree of ionization : At equilibrium `[H^+]= alpha C = 0.05 alpha =x` `therefore [H^+]=0.9539xx10^(-3)M =Calpha` `therefore alpha=([H^+])/C=(0.9539xx10^(-3))/0.05` `=1.9078xx10^(-2)` =0.019078 `approx` 0.0190 = Dissociation degree (iv) Degree of ionization in 0.01 M HCl : `H^+` is formed from HCl So, initial `[H^+]` = 0.01 M `{:(,CH_3COOH_((l)) hArr, H_((aq))^(+)+,CH_3COO_((aq))^(-)),("Initial M:",0.05,0.01 M,0.0),("Change M:",-Calpha,+Calpha,+Calpha),("At equilibrium M",(0.05-0.05alpha),0.01+Calpha,+Calpha),(,=0.05(1-alpha),(0.01+0.05alpha),0.05 alpha),(,approx 0.05,approx 0.01 M,0.05 alpha):}` `K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.82xx10^(-5)` `therefore ((0.05alpha)(0.01))/(0.05)=1.82xx10^(-5)` `therefore alpha=1.82xx10^(-3)`=0.00182 (v)Calculation of degree of ionization `alpha` of 0.01 M HCl : `((0.005alpha)(0.1))/(0.05)=1.82xx10^(-5) therefore alpha = 1.82xx10^(-4)` |
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