Question

The acid ionization constant for Fe(H2O)62+(aq) is 3.0×10-6. Calculate the pH of a 0.0248 M solution of Fe(NO3)2. pH =

Answer 1

As above, at the start (before ionization), the concentration of Fe(NO₃)₂, is 0.0248 M an after dissociation Fe⁺² forming complex with H₂O. The concentration of H3O+ is essentially zero (~ 0), assuming that the contribution from the self-ionization of water can be neglected. Concentration (M) Fe⁺² + H₂O <==> Fe(H₂O)₆. The dissociation   [H₃O+ ][Fe+2 ] [Fe(NO₃)₂] = Ka or x2 (0- x) = 3.0 x 10^-6

X²/0.0248 = 3.0 x 10^-6

X² = 3.0 x 10^-6 x 0.0248

X = 27 x 10^-5

P^H = -log(27 x 10^-5)

= 7.15