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
The equilibrium concentration of free silver ion in the solution = 1 * 10⁻⁸ M
Given data :
0.398 mol of AgNO₃ added initially at equilibrium
2.73 mol NH₃ added initially at equilibrium
Temperature = 25°C
Volume of solution = 1.00 L
Determine the equilibrium conc of [Ag+]
Chemical reaction : Ag⁺ + 2NH₃ ⇄ Ag(NH₃)₂⁺ ------ ( 1 )
kf = 1.5 * 10⁷ = [Ag(NH₃)₂⁺] / [NH₃]² [Ag⁺] ------- ( 2 )
From equation ( 1 )
∴ NH₃ = 2.73mol - 2X
Ag⁺ = 0.398mol - X
back to equation ( 2 )
1.50 * 10⁷ = (X) / ( 2.73 - X )² ( 0.398 - X ) ----- ( 3 )
Resolving equation ( 3 )
X = 0.39799999
∴ The equilibrium conc of [ Ag⁺ ] = 0.398 - 0.39799999
= 1 * 10⁻⁸ M
Hence we can conclude that The equilibrium concentration of free silver ion in the solution = 1 * 10⁻⁸ M.
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