Question

(part a) Determine the percent ionization of the following solutions of phenol at 25°C: (a) 0.751 M % (b) 0.270 M % (c) 1.56 × 10−6 M × 10 % (Enter your answer in scientific notation.)

(part b) The pH of an aqueous acid solution is 6.20 at 25°C. Calculate the K_a for the acid. The initial acid concentration is 0.013 M.× 10 (Enter your answer in scientific notation.)

Answer 1

Solution:

Since, phenol is a weak acid and it dissociated as,
a) C₆H₅OH = H⁺ + C₆H₅O^-
0.751 0 0
(0.751-x) x x
Then, dissociation constant (Ka) of phenol,

Ka = [H⁺] [C₆H₅O^-] / [C₆H₅OH] = 1.29 x 10^-10

(x) (x) / (0.751-x) = 1.29 x 10^-10

Since, phenol is a weak acid, hence, x <<< 0.751 then above equation becomes as,

x²/0.751 = 1.29 x 10^-10
x = 0.984 x 10^-5 M

Hence, x = [H+] = 0.984 x 10^-5 M

% ionization = [H⁺] / [C₆H₅OH] x 100% = (0.984 x 10^-5 M /0.751 M) x 100% = 0.0013 %

b) 0.270 M %

Using dissociation constant equation for 0.270 M %

(x) (x) / (0.270-x) = 1.29 x 10^-10

Since, phenol is a weak acid, hence, x <<< 0.270, then above equation becomes as,

x²/0.270 = 1.29 x 10^-10
x = 4.78 x 10^-5

Hence, x = [H+] = 0.59 x 10^-5 M

% ionization = [H⁺] / [C₆H₅OH] x 100% = (0.59 x 10^-5 M / 0.270 M) x 100% = 0.0022 %

c) 1.56 x 10^-6 M x 10 % = 1. 56 x10^-5 M

Using dissociation constant equation for 1.56 x10^-5 M

(x) (x) / (1.56 x10^-5 M -x) = 1.29 x 10^-10

Since, phenol is a weak acid, hence, x <<< 1.56 x10^-5 M , then above equation becomes as,

x²/1.56 x10^-5 M = 1.29 x 10^-10
x² = 2.0124 x 10^-15 , then, x = 4.49 x 10^-8 M

Hence, x = [H+] = 4.49 x 10^-8 M
% ionization = [H⁺] / [C₆H₅OH] x 100% = (4.49 x 10^-8  M / 1.56 x10^-5 M) x 100% = 0.288 %

Part B)

Suppose acid is HA, it dissociated in aqueous solution as,

Ka = [H⁺] [A^-] / [HA]

pH = -log [H⁺], then [H⁺] = 10^-pH = 1 X 10^-6.20 = 6.31 X 10^-7 M

Since the acid ionizes to produce 1 mol of H⁺ and 1 mol of A-, then the concentration of A^- will be the same and the concentration of HA will be 0.013 - 6.31 X 10^-7 , hence,

Ka = (6.31 X 10^-7) x (6.31 X 10^-7) / (0.013 - 6.31 X 10^-7) = 39.8161 x 10^-14 / 0.013

Ka =3.06 X10^-11