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The value of Ka for phenol (a weak acid) is 1.00×10−10. What is the value of Kb for its conjugate base, C6H5O−?

User Losbear
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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.

User Gukoff
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