After the solution reaches equilibrium, what concentration of Zn2+(aq) remains?

NetherCraft 0

A solution is made that is 1.3×10−3M in Zn(NO3)2 and 0.160M in NH3.

After the solution reaches equilibrium, what concentration of Zn2+(aq) remains?

(I need someone who KNOWS how to do this. The last yahoo answers was wrong.)

2 Answers

  • Zn2+ + 4NH3 =======> Zn(NH3)4^2+

    0.0013 + 4(0.0013) =====> 0.0013

    NH3 remaining = 0.160 – 0.0052 = 0.1548 M

    Kf = [Zn(NH3)4^2+]/[Zn2+][NH3]^4 = 4 x 10^8

    Substituting in the above relationship : [0.0013]/[Zn2+][0.1548]^4 and solving for Zn2+

    [Zn2+] = 5.69 x 10^-9

  • the answer should be 8.7 x 10^-10 M. I had this problem in my textbook which has the answers in the back.

    Zn (aq) + 4NH3 = Zn(NH3)4 2+

    kf Zn(NH3)4 2+ = 2.8 x 10^-9

    2.8 x 10^-9 = (1.1 x 10^-3) / [ (0.1456)^4 x (x)]

    x = 8.7 x 10^-10

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