Calculate the ph of an aqueous solution at 25°c that is 0.31 m in phenol (c6h5oh). (ka for phenol = 1.3 × 10−10.)

Given:

Concentration of phenol = 0.31 M

Ka (acid dissociation constant) = 1.3 * 10⁻¹⁰

To determine:

The pH of the solution

Explanation:

            C6H5OH ↔ C6H5O⁻ + H⁺

Initial       0.31 M          0             0

change   -x                 +x             +x

Equlib        0.31-x         x             x

Ka = [C6H5O⁻][H⁺]/[C5H5OH]

1.31*10⁻¹⁰ = x²/0.31-x

x =[H+]= 6.37*10⁻⁶M

pH = -log[H+] = -log(6.37*10⁻⁶) = 5.19

Ans: pH of the solution is 5.19

pH of the solution is 5.19