Question

A solution is made of 1.1 x 10^-3M in Zn(NO₃)² and 0.150 M in NH₃.

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

Kf (Zn(NH₃)₄²⁺) = 2.8 x 10⁹

Answer 1

8.74×10^-10M

Explanation

Consider the complete formation of Zn(NH3)4²⁺

Zn²⁺(aq) + 4NH3(aq) ------> Zn(NH3)4²⁺(aq)

stoichiometrically , 1mole of Zn²⁺ reacts with 4moles of NH3 to give 1mole of Zn(NH3)4²⁺

0.0011M of Zn²⁺ reacts with 0.0044M of NH3 to give 0.0011M of Zn(NH3)4²⁺

After reaction

Remaining concentration of NH3 = 0.1500M - 0.0044M = 0.1456M

Now, consider the dissociation equillibrium of Zn(NH3)4²⁺

Zn(NH3​​​​​​)4²⁺(aq) <---------> Zn²⁺(aq) + 4NH3(aq)

K_d = [Zn²⁺][NH3]⁴/[Zn(NH3)4²⁺]

K_d = 1/K_f

K_d = 1/ 2.8 ×10⁹ = 3.57×10^-10

Initial concentrations

[Zn(NH3)4²⁺] = 0.0011

[Zn²⁺] = 0

[NH3] = 0.1456

Change in concentration

[Zn(NH3)4²⁺] = -x

[Zn²⁺] = +x

[NH3] = +4x

Equillibrium concentration

[Zn(NH3)4²⁺] = 0.0011 - x

[Zn²⁺] = x

[NH3] =0.1456 + 4x

so,

x(0.1456 + 4x)⁴/(0.0011 - x) = 3.57 ×10^-10

we can assume 0.1456 + 4x = 0.1456 and 0.0011-x = 0.0011

x(0.1456)⁴/(0.0011) = 3.57 ×10^-10

x0.4086 = 3.57 ×10^-10

x = 8.74 ×10^-10

Therefore,

at equillibrium ,

[Zn²⁺] = 8.74 ×10^-10M