Question

# Calculate the pH of a 0.334 M aqueous solution of phenol (a weak acid) (C6H5OH, Ka...

Calculate the pH of a 0.334 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.

 pH = [C6H5OH ]equilibrium = M [C6H5O- ]equilibrium = M

C6H5OH acid dissociates as

C6H5OH + H2O C6H5O- + H3O+

Ka = [C6H5O- ][H3O+] / [C6H5OH]

but  [C6H5O- ] = [H3O+] = x

Ka = [x][x] / [C6H5OH]

Ka of phenol = 1.0 X10-10

Substitute the value in equation

1.0 X10-10 = [x]2/ 0.334

[x]2 = 1.0 X10-10 X  0.334 = 3.34 X 10-11

[x] = 5.779 X 10-6

[C6H5O- ] = [H3O+] = x= 5.779 X 10-6 M

[C6H5O- ] = 5.779 X 10-6 M

equlibrium concentration of C6H5OH = intial concentration of C6H5OH - equilibrium cooncentrtion of C6H5O-

[C6H5OH] equilibrium = 0.334 - 5.779 X 10-6 = 0.33399 M

[C6H5OH] equilibrium = 0.33399 M

[H3O+] = 5.779 X 10-6 M

pH = - log[H3O+]

pH = - log (5.779 X 10-6)

pH = 5.24

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