Need help!!! A solution is made from 1.4×10−3 M in Zn(NO3)2 and 0.180 M in NH3.

Question

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

Kf = [Zn(NH3)4]2+)=2.8×109.

Answer 1

Zn(NO₃)₂ + 4NH₃ ==> [Zn(NH₃)₄]²⁺ and since NO3- is a spectator, we can re-write this as...

Zn²⁺(aq) + 4NH₃ <==> [Zn(NH₃)₄]²⁺

1.4x10-3....0.180..............0..........Initial

-x.............-4x..................+x..........Change

1.4x10^-3-x..0.18-4x............x...........Equilibrium

K = 2.8x10⁹ = (x) / (1.4x10^-3-x)(0.18-4x)⁴

Solve for x and then subtract that value from 1.4x10^-3 M to get equilibrium [Zn²⁺]

It will obviously been very low because the Kf is so large, and because the original [Zn²⁺] is low to begin with.