Answer:
The Ka for phenol (C6H5OH) is given as 1.00×10^−10. The dissociation reaction for phenol can be written as:
C6H5OH (aq) ⇌ C6H5O− (aq) + H+ (aq)
The Ka expression for this reaction can be written as:
Ka = [C6H5O−] [H+] / [C6H5OH]
Since the phenol is a weak acid, its conjugate base, C6H5O− is a weak base. The Kb expression for C6H5O− can be written as:
Kb = [C6H5OH] [OH−] / [C6H5O−]
where OH− is the hydroxide ion concentration.
Now, we can use the relationship Kw = Ka x Kb, where Kw is the ionization constant of water, which is 1.00 x 10^-14 at 25°C, to solve for Kb:
Kw = Ka x Kb
1.00 x 10^-14 = 1.00 x 10^-10 x Kb
Kb = (1.00 x 10^-14) / (1.00 x 10^-10)
Kb = 1.00 x 10^-4
Therefore, the value of Kb for the conjugate base C6H5O− is 1.00 x 10^-4.