How do I calculate Q?

Question

When I add 1.50 mL of 0.69 M NH3 to 1.0 L of 1E-3 M FeSO4, what will Q be?

I determined when NH3 reacts with H2O, it forms 3.52E-3 M OH.
Was I suppose to do this?
How do I calculate Q from here?

I've tried two different ways and have gotten
2.64E-9 and 1E-6. Both are wrong. Please help me!

Answer 1

Let us first calculate the initial concentration after mixing but before reaction:
NH3: 0.0015*0.69/(1.0 + 0.0015) = 1.035e-3 (M)

FeSO4: 1E-3 /(1.0 + 0.0015) = 1E-3 (M)
Second, let us write out the reaction:
NH3 + H2O <==> NH4+ + OH-, pKa = 9.25 and pKb = 4.75
Fe(2+) + 2OH- <==> Fe(OH)2, pKsp = 15.097
I do not exactly know what Q you are talking about: the first, the second, or the overall reaction:
2NH3 + 2H2O + Fe(2+) <==> Fe(OH)2(s) + 2NH4(+)
Remembering that solid Fe(OH)2 or water H2O should not appear in the representation of Q. For the overall reaction:
Q = [NH4(+)]^2/{[NH3]^2*[Fe(2+)]}
= [NH4(+)]^2*[OH-]^2/{[NH3]^2*[Fe(2+)]*[OH-]^2}
p(Q) = -log(Q)
= 2*pKb - pKsp = -5.60
Q = 10^5.60