posted by .

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??

  • chemistry -

    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.

  • chemistry -

    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^++

    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.

  • chemistry -

    Into the cell assembled, Cu(s)l CuSO4, 0.01M ll CuSO4, 1.0M l Cu(s)...was added 10mL of concentrated 6M NH3

  • chemistry -

    Is there a voltage listed for the assembly?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Honors Chemistry

    "How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2?
  2. chemistry

    for the complex ion equilibria between AgCl and NH3, why does NH3 replace Cl- and how do we know that we have to add 2 infront of NH3?
  3. chemistry

    for the complex ion equilibria between AgCl and NH3, why does NH3 replace Cl- and how do we know that we have to add 2 infront of NH3?
  4. High School Chemistry

    10mL of 0.10M HCl is given. What is the pH?
  5. college chem

    The cation M2+ reacts with NH3 to form a series of complex ions as follows: M2+ + NH3 M(NH3)2+ K1 = 102 M(NH3)2+ + NH3 M(NH3)2 2+ K2 = 103 M(NH3)2 2+ + NH3 M(NH3)3 2+ K3 = 102 A 1.0 × 10–3 mol sample of M(NO3)2 is added to 1.0 L …
  6. chemistry

    what is the Molarity (M) of a 0.87m aqueous solution of ammonia, NH3?
  7. Chemistry

    Consider this equilibrium N2(g) + H2(g) <==> NH3(g) +94 kJ The equilibrium law exoression for the balanced chemical equationwould be A. [N2][H2]/[NH3] B. [NH3]/[H2][N2] C. [NH3]2/[H2][N2] D. [NH3]2/[H2]3[N2] E. 2[NH3]2/3[H2]3[N2]
  8. Chemistry

    in the presence of NH3, Cu+2 forms the complex ion [Cu(NH3)4]+2. If the equilibrium concentrations of Cu+2 and [Cu(NH3)4]+2 are 1.8x10^-17M and 1.0x10^-3M respectively, in a 1.5 M NH3 solution, calculate the value for the overall formation …
  9. chemistry

    50.0 mL of a 0.00200 M Cd^2+ solution is titrated with 0.00400 M EDTA in the presence of 0.0750 M NH3 buffered at pH = 9.00. Determine pCd^2+ when 28.5 mL of EDTA solution have been added. Cd^2+ + Y^4- ?
  10. chem

    Calculate the equilibrium concentration of NH3,Cu^2+,[ Cu(NH3)]^2+ ,[Cu(NH3)2]^2+,[Cu(NH3)3]^2+,and[Cu(NH3)4]^2+ in a solution made by mixing 500.0 ml of 3.00 M NH3 with 500.0 ml of 2.00*10^-3 M Cu(NO3)2. The sequential equilibria …

More Similar Questions