Question

The ionization constant of phenol is 1.0 × 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?

Answer 1

Ionization of phenol:

	C6H5OH	+	H2O	↔	C6H5O-	+	H3O+
Initial conc	0.05				0		0
At equilibrium	0.05 - x				x		x

`"K"_"a" = (["C"_6"H"_5"O"^(-)]["H"_3"O"^(+)])/["C"_6"H"_5"OH"]`

`"K"_"a" = (x xx x)/(0.05 - x)`

As the value of the ionization constant is very less, x  will be very small. Thus we can ignore x in the denominator

`:. x = sqrt(1xx10^(-10) xx 0.05)`

`= sqrt(5xx10^(-12))`

`= 2.2 xx 10^(-6) "M" = ["H"_3"O"^(+)]`

Since `["H"_3"O"^+] = ["C"_6"H"_5"O"^(-)]`

`["C"+6"H"_5"O"^(-)] = 2.2 xx 10^(-6) "M"`

Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa.

\[\ce{C_6H_5ONa -> C_6H_5O- + Na+}\]

Conc                                   0.01

Also

	C6H5ONa	H2O	↔	C6H5O-	+	H3O+
Conc.	0.05 - 0.05 α			0.05 α		0.05 α

`["C"_6"H"_5"OH"] = 0.05 - 0.05 alpha` ; 0.05 "M"`

`["C"_6"H"_5"O"^(-)] = 0.01 + 0.05  alpha  0.01  "M"`

`["H"_3"O"^+] = 0.05 alpha`

`"K"_"a" = (["C"_6"H"_5"O"^(-)]["H"_3"O"^(+)])/["C"_6"H"_5"OH"]`

`"K"_"a" = ((0.01)(0.05alpha))/0.05`

`1.0 xx 10^(-10) = 0.01  alpha`

`alpha = 1 xx 10^(-8)`