Calculate the pH of a 0.393 M aqueous solution of phenol (a weak acid) (C6H5OH, Ka = 1.0×10-10) and the equilibrium concentrations of the weak acid and its conjugate base.

See explanation below

Explanation:

First to all, we need to write the general reaction for this.

The phenol is C₆H₅OH:

C₆H₅OH + H₂O > C₆H₅O⁻ + H₃O⁺

Now, to get the equilibrium concentrations, we need to write the ICE chart for this reaction:

       C₆H₅OH + H₂O > C₆H₅O⁻ + H₃O⁺

i)        0.393                               0            0

c)           -x                                 +x          +x

e)      0.393-x                              x            x

Now, let's write the equilibrium expression:

Ka = [C₆H₅O⁻] [H₃O⁺] / [C₆H₅OH]

Replacing the above values:

1x10⁻¹⁰ = x² / 0.393-x

However Ka is a very small value, so we can assume that the substration between 0.393-x would be very small, so we can round this to 0.393 only:

1x10⁻¹⁰ = x²/0.393

Solving for x:

x = √1x10⁻¹⁰ * 0.393

x = 6.27x10⁻⁶ M

This would be the equilibrium concentration of the conjugate base, the weak acid would be practically the same.

[C₆H₅OH] = 0.393 - 6.27x10⁻⁶ = 0.39399373 M

Finally for the pH:

pH = -log(6.27x10⁻⁶)

pH = 5.2

chemistry sucks a**

explanation: