Question

Determine the percent ionization of the following solutions of phenol at 25°C:

(a) 0.795 M

_____%

(b) 0.280 M

_____%

(c) 1.57×10⁻⁶M

_____×10_____%

(Enter your answer in scientific notation.)

Answer 1

Percent ionization for a weak acid (HA) is calculated by the formula:

Percent ionization=[HA] ionized / [HA] initial × 100%

a) Molarity of phenol = 0.795 M, Ka = 1.3 x 10-10

C6H5OH ----------> H+ + C6H5O-

Initial 0.795 0 0

Equilibrium 0.795 -x x x

Ka = [H+][C6H5O-]/[C6H5OH] = 1.3 x 10-10

(x) (x)/(0.795-x) = 1.3 x 10-10

x2 = 1.3 x 10-5 x 0.795

x = 1.016 x 10-5 M = [H+]

Percent ionization=[HA] ionized / [HA] initial × 100% = (1.016 x 10-5 M / 0.795 M) x 100% = 1.27 x 10-5 x 100% = 0.00127%

a) Molarity of phenol = 0.280 M, Ka = 1.3 x 10-10

C6H5OH ----------> H+ + C6H5O-

Initial 0.280 0 0

Equilibrium 0.280 - x x x

Ka = [H+][C6H5O-]/[C6H5OH] = 1.3 x 10-10

(x) (x)/(0.280-x) = 1.3 x 10-10

x2 = 1.3 x 10-5 x 0.280

x = 0.364 x 10-5 M = [H+]

Percent ionization=[HA] ionized / [HA] initial × 100% = (0.364 x 10-5 M / 0.280 M) x 100% = 1.3 x 10-5 x 100% = 0.0013 %

c) Molarity of phenol = 1.57 x 10-6 M, Ka = 1.3 x 10-10

C6H5OH ----------> H+ + C6H5O-

Initial 1.57 x 10-6 0 0

Equilibrium 1.57 x 10-6 - x     x x

Ka = [H+][C6H5O-]/[C6H5OH] = 1.3 x 10-10

(x) (x)/(1.57 x 10-6 -x) = 1.3 x 10-10

x2 = 1.3 x 10-5 x 1.57 x 10-6

x2 = 20.41 x 10-10 M

x = 4.51 x 10-5 = [H+]

Percent ionization = [HA] ionized / [HA] initial × 100% = (4.51 x 10-5 M /1.57 x 10-6 M) x 100 % = 2.87 x 10 x 100% = 2870 %