Question

0.398 mol AgNO3 and 2.73 mol NH3 were added to enough water to give 1.00 L of solution at 25 °C. Calculate the equilibrium concentration of free silver ion , [Ag+], in this solution. The formation constant for Ag(NH3)2+ (Kf) = 1.50x107 at 25 °C.

Answer 1

Answer:

[Ag⁺] = 1x10⁻⁸M

Explanation:

In the reaction:

Ag⁺ + 2NH₃ ⇄ Ag(NH₃)₂⁺

Kf = 1.50x10⁷ =  [Ag(NH₃)₂⁺] / [NH₃]²[Ag⁺]

If in the beginning you add 0.398 moles of Ag⁺ and 2.73 moles of NH₃. Concentrations in equilibrium are:

[Ag(NH₃)₂⁺]: X

[NH₃]: 2.73mol - 2X

[Ag⁺]: 0.398mol - X

Replacing:

1.50x10⁷ =  [X] / [2.73-X]²[0.398-X]

1.50x10⁷ =  [X] / (7.4529-5.46X+X²)(0.398-X)

1.50x10⁷ =  [X] / 2.9662542-9.62598X+5.858X²-X³

-1.50x10⁷X³+8.787x10⁷X²-1.443897x10⁸X + 4.4493813x10⁷ = X

-1.50x10⁷X³+8.787x10⁷X²-1.443897x10⁸X + 4.4493813x10⁷ = 0

Solving for X:

X = 0.39799999

Thus, equilibrium concentration of Ag⁺, [Ag⁺] is 0.398mol - X. Replacing:

[Ag⁺] = 0.398mol - 0.39799999 = 1x10⁻⁸M