If 2.50 g of CuSO4 are dissolved in 9.4 102 mL of 0.34 M NH3, what are the concentrations of Cu(NH3)42+, NH3 and Cu2+ at equilibrium?

Question

i tried doing Cu2+ first

and i did

Kf=5e13= salt/cu2+ nh3+^4
and i did

molesCuSO4/.94 L = .01666 M salt

so then i did

.01666M/[cu][.34-(4*.01666)^4]

to find cu+2

and i got like 5 different answers each time i dot it

is this even right

please help me
im running out of submissions

Answer 1

I think I would approach the problem a little differently.
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Write the expression for Kf which is products/reactants. I don't know what "salt" means in your equation but the numerator should be [Cu(NH3)4^+2] and the denominator should be (Cu^+2)(NH3)^4.
I would find the moles CuSO4 which is 2.5/159.61 = a very small number.
moles NH3 = 9.4 x 10^2 x 0.34 = approximately 320.
With such a small number of moles of Cu^+2 and a large excess of NH3,(and a huge Kf), we figure (complex ion) = moles Cu we started with. That divided by 9.4 x 10^2 should give the (complex). Since we used almost none of the NH3, then final (NH3) should be what we started with. Then you can plug in (complex) and (NH3) into Kf and calculate final (Cu^+2). Check my thinking.