Determination of Kf of Cu(NH3)4 (aq) with given 10mL of 6M NH3 mix with 40mL of 1M Cu.
This is what I have so far:
nCu= .040L x 1= .040mol
nNH3= .010L x 6= .060mol
Cu^2+ + 4NH3 -> Cu(NH3)4
.040 .06 0
-.040 4(-.040) +.040
0 -.1 .040
[Cu(NH3)]= .040/(.010+.040)=.80M
for Kf=[Cu(NH3)]/[Cu][NH3]^4
= .80/(1*6^4)=6.17E-04
I'm not so quite sure with my answer. Help please??
4 answers
a Kf wasn't given too? Was told to use ice table to find concentration of Cu(NH3)4 then I'll be able to find Kf.
There MUST be something missing here but I will go as far as I can.
initial (Cu^2+) = 0.8 M as you have it.
initial (NH3) = 0.06 mols/0.05L = 1.2M
............Cu^2+ + 4NH3 ==> Cu(NH3)4^++
I...........0.8......1.2........0
C...........-.3.....-1.2.......+1.2
E...........0.5.......0........1.2
Note that all of the Cu CAN'T be used, as you've done, because that would take 0.8 x 4 = 3.2M NH3 and you don't have that much. So the limiting reagent is NH3. But this is where we get into trouble. If we try to plug into Kf expression we can't use zero for NH3 OR if we use it as you did and use all of the Cu, the equilibrium Cu is zero and we can't use that either. Kf for the complex is about 1.1E13 according to a web site I used.
initial (Cu^2+) = 0.8 M as you have it.
initial (NH3) = 0.06 mols/0.05L = 1.2M
............Cu^2+ + 4NH3 ==> Cu(NH3)4^++
I...........0.8......1.2........0
C...........-.3.....-1.2.......+1.2
E...........0.5.......0........1.2
Note that all of the Cu CAN'T be used, as you've done, because that would take 0.8 x 4 = 3.2M NH3 and you don't have that much. So the limiting reagent is NH3. But this is where we get into trouble. If we try to plug into Kf expression we can't use zero for NH3 OR if we use it as you did and use all of the Cu, the equilibrium Cu is zero and we can't use that either. Kf for the complex is about 1.1E13 according to a web site I used.
Into the cell assembled, Cu(s)l CuSO4, 0.01M ll CuSO4, 1.0M l Cu(s)...was added 10mL of concentrated 6M NH3
Is there a voltage listed for the assembly?