The ionization constant of phenol is `1.0 xx 10^(-10)`. What is the concentration of phenolate ion in `0.05` M solution of phenol? What will be its degree of ionization if the solution is also `0.01M` in sodium phenolate?

Explanation

` {:( ,C_6H_5,hArr, C_6H_5O^(-) ,+,H^(+)),("Initial conc." ,0.05M,, -,,-),( " Conc. at equilibrium ", 0.05-x,,x,,x):}`
` therefore K_a= ([C_6H_5O^(-)][H^(+) ])/([C_6H_5OH]) = (x.x)/(0.05-x) =1.0 xx 10 ^(-10) ( " given ") `
or ` " " (x^(2))/(0.05-x)= 1.0 xx 10 ^(-10) `
Since , phenol is not much dissociated , ` 0.05 -x ~~ 0.05`
Hence
` " " (x^(2))/(0.05) = 1.0 xx 10 ^(-10) `
or ` x= (0.05 xx 1.0 xx 10^(-10)) ^(1//2) = 2.24 xx 10 ^(-6) `
Hence `[C_6H_5O^(-)]` in solution ` = x = 2.24 xx 10 ^(-6) M.`
In the presence of 0.01 M ` C_6H_5ONa: `
Suppose is the amount of phenol dissociated . Therefore at equilibrium
` [C_6H_5OH ]=0.05 -y, ,[C_6H_5O^(-)]=1.0 +y and [H^(+) ]=y`
` therefore " " K_a( (0.01+y)(y))/((0.05- y))= 1.0 xx 10^(-10) (" given ") `
Since , in the presence of sodium phenolate , the dissociation of phenol will get further suppressed , we can take
` 0.01 +y~~ 0.01 " " and " " 0.05 - y ~~ 0.05`
Hence
` " " ( 0.01xx y )/(0.05) = 1.0 xx 10 ^(-10) `
or ` " " y = (1.0 xx 10 ^(-10)xx 0.05)/(0.01) = 5.0 xx 10^(-10)`
Hence , degree of dissociation of phenol
` " " = alpha =(" Number of moles dissociated")/(" Total no. of moles initially taken ") `
` " " ( 5.0 xx 10 ^(-10))/(0.05) = 1.0 xx 10 ^(-8) `