Question

Calculate the pH and the percent dissociation of the hydrated cation in 0.048 M solutions of the following substances.

a) Calculate the pH of the hydrated cation in Fe(NO3)2 (Ka=3.2×10−10).

b)Calculate the percent dissociation of the hydrated cation in Fe(NO3)2 (Ka=3.2×10−10).

c) Calculate the pH of the hydrated cation in Fe(NO3)3 (Ka=6.3×10−3).

d) Calculate the percent dissociation of the hydrated cation in Fe(NO3)3 (Ka=6.3×10−3).

Answer 1

a)

Fe(H2O)6 + H2O ---------> [Fe(H2O)5]2+ +   H3O+

0.048                                          0                      0

[H+] = sqrt (Ka x C)

        = sqrt (3.2 x 10^-10 x 0.048)

        = 3.92 x 10^-6 M

pH = -log (3.92 x 10^-6)

pH = 5.41

b)

% dissociation = (x / C) x 100

                        = (3.92 x 10^-6 / 0.048) x 100

% dissociation = 0.0082 %

c)

[H+] = 0.0145 M

pH = -log (0.0145)

     = 1.84

pH = 1.84

d)

% dissociation = [0.0145 / 0.048] x 100

                         = 30.2 %