Question

a)The equilibrium constant, Ka, for the reaction given below is is 6.0 x 10-3. Calculate the pH of a 0.798 M solution of Fe(H20)63+.

Fe(H20)63+(aq) + H2O(l) <--> Fe(H20)5OH2+(aq) + H3O+(aq)

b)Calculate the pH necessary for 97.4 % of the iron(III) to be in the form Fe(H20)₆³⁺

Answer 1

It should be written as [Fe(H2O)6]3+, and its name is the Iron(III) hexahydrated complex ion. It is a Fe3+ ion that has a complex of 6 H2O groups around it. Since it has a Ka of 8.9*10^-4, we can infer that it is acidic, donating a proton (hydrogen ion) to basic species. Therefore it has a dissociation equation as follows:

[Fe(OH)6]3+ (aq) + H2O (l) <--> [Fe(OH)5OH]2+ (aq) + H3O+ (aq)

Now the Ka expression is the concentration of products over reactants:

Ka = [[Fe(OH)5OH]2+][H3O+]/[[Fe(OH)6]3+]

The complex ion only dissociates in a 1:1 mol ratio; therefore, each product is created as much as the amount that the complex ion dissociates from its original concentration(3.0 - x). Plug in everything and solve.

6.0 x 10^-3. = (x)(x)/(0.798 - x)

we can assume that the dissociation is so insignificant that it doesn't reduce the formal concentration by much( say if x = 4.5*10^-5, it's 5 orders of magnitude smaller than 0.798, so 0.798 - 4.5*10^-5 = 0.798, so we say it's insignificant and okay to ignore, as long as there is a less than 5 percent dissociation of the acid, which you can compute later.

So now the equation, since we ignore x in 0.798- x because it will be small (based on the size of the Ka to the concentration; 4 magnitudes lower, 0.798 - 6.0 x 10^-3 =0.79) compared the formal concentration of the ion. It becomes:

6.0 x 10^-3 = x^2/ 0.798

x = 0.0870 M

Now see if it's okay to ignore x: (x/Concentration) * 100%. If smaller than 5, okay to ignore. If not, not okay.

(0.0870/ 0.798) * 100% = 10.90 %, it's okay to ignore x since the ion only dissociates 10.90 %, of its original concentration!

To find pH, use the formula: pH = -log([H3O+]). The ion only gives 1 mol per dissociation, so therefore x = [H3O+] = 0.0870 M. Put this concentration into the pH equation.

pH = -log(0.0870) = 1.060. The pH of a 3.0 M aqueous Fe 3+ solution is 1.060