Chemistry Solubility Math Question 4

Question

If 401.0 mg of AgBr is completely dissolved in 118.0 mL of 0.07 M NH₃, what is the equilibrium concentration of Ag⁺(aq) in the final solution? K_f[Ag(NH₃)₂⁺] = 1.7E7

Answer 1

Complex ion formation:

Ag⁺(aq) + 2NH₃(aq) <==> [Ag(NH₃)₂]⁺(aq)

Kf = 1.7x10⁷ = [Ag(NH₃)₂]⁺ / [Ag⁺][NH3]²

moles AgBr: 401.0 mg x 1 g/1000 mg x 1 mole/188 g = 0.00213 moles AgBr = 0.00213 moles Ag⁺

[Ag+] = 0.00213 moles/0.118 L = 0.0181 M Ag+

[NH3] = 0.07 M

Ag⁺(aq) + 2NH₃(aq) <==> [Ag(NH₃)₂]⁺(aq)

0.0181......0.07.................0...........Initial

-0.0181...-0.0362.........+0.0181....Change

0...........0.0338..........0.0181.........Equilibrium

1.7x10⁷ = (0.0181) / (x)(0.0338)²

x = 9.3x10^-7 M = [Ag⁺]