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The ionization constant of phenol is 1.0xx10^(-10) .What is the concentration of phenolate ion is 0.05 M solution of phenol ? What will beits degree of ionization if the solution is also 0.01 M in sodium phenolate ? |
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Answer» Solution :The CONCENTRATION of Phenolate ion `(C_6H_5OH)` in 0.05 M Phenol `(C_6H_5O^-)` is =`ALPHA` `{:(,C_6H_5OH_((aq)) + H_2O hArr,C_6H_5O_((aq))^(-) + , H_3O_((aq))^(+)),("Initial :", 0.05 M, -,-),("Change in equilibrium :", -alpha,+alpha,+alpha),("At equili. M:" ,(0.05-alpha),alpha,alpha) :}` `K_a=([C_6H_5O^-][H_3O^+])/([C_6H_5OH])=1.0xx10^(-10)` `therefore alpha^2/(0.05-alpha)=1.0xx10^(-10)` `therefore alpha^2/0.05 = 1.0xx10^(-10)` (`because alpha lt lt` 0.05 , So take 0.05 - `alpha`=0.05) `therefore alpha^2= 0.05xx10^(-10) = 5.0xx10^(-12)` `therefore alpha=sqrt(5.0xx10^(-12))=2.236xx10^(-6) M = [C_6H_5O^-]` If 0.01 M sodium phenolate degree of ionization COMPLETE ionization of sodium phenolate is under. `{:(C_6H_5ONa to , C_6H_5O^(-)+ , Na^+),(0.01 M,0.01 M,0.01M):}` If dissociation degree of phenol is equal to x . So, `therefore` Dissociated phenol = `C xx x`= (0.05 )x `therefore` At equilibrium [Phenol]=(0.05-0.05x) but x is very less, So , [Phenol] `approx` 0.05=C At equilibrium `[H_3O^+]` in [Phenol] =`Cxxx`= 0.05 x (Phenolet ion at equilibrium)=(Phenolet formed from phenol)+(Phenolet of sodium phenolet ) `[C_6H_5O^-]=(C alpha+0.01)` `=(0.05 alpha+0.01)` `approx` 0.01 (`because alpha` is very less) This values put in table : `{:(,C_6H_5OH_((aq))+H_2O_((l)) hArr, C_6H_5O_((aq))^(-)+,H_3O_((aq))^(+)),("At equili.",0.05 M, 0.01 M, x(0.05)):}` `K_a=([C_6H_5O^-][H_3O^+])/([C_6H_5OH])=1.0xx10^(-10)` `therefore K_a=((0.01)(0.05x))/((0.05))=1.0xx10^(-10)` `therefore x=(1.0xx10^(-10))/0.01 = 1.0xx10^(-8)` |
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