Question

The equilibrium constant, Ka, for the reaction below is 6.0*10^-3. Fe(H2O)6 ^3+ (aq) + H2O(l) <--> Fe(H2O)5 (OH)^2+(aq) + H3O^+ (aq) a)Calculate the pH of the 0.10M solution of Fe(H2O)6^3+. b)Will a 1.0 M solution of iron (II) nitrate have a higher or lower pH than a 1.0 M solution of iron (III) nitrate?

Answer 1

          [Fe(H2O)6]^3+ (aq) + H2O(l) --> [Fe(H2O)5(OH)]^2+ (aq) + H3O^+ (aq)
initially        0.1                                  0                   0
at equi      (0.1 - x)                                                      x                     x
K = ([Fe(H2O)5(OH)]^2+)*(H3O+)/[Fe(H2O)6]^3+
6 * 10^-3 = x*x/(0.1-x)
X = 0.0216
[H3O+] = 0.0216
pH = -log(0.0216) = 1.6655