Answer:Equilibrium concentration of free silver ion is 3.21×10^6 mol
Explanation: Since we have enough water in the reaction, we have aqueous NH3 reacting with AgNO3 (aq)
Equation of reaction:
AgNO3(aq) + 2NH3(aq) <----> Ag(NH3)2+ at 25°C
Formation of complexes is also a thermal dynamic phenomenon. For the equilibrium,
Ag+ + 2NH3 <----> Ag(NH3)2+
AgNO3(aq) + 2NH3(aq) <----->Ag+ + 2NH3(aq)
•From the question, we have 0.214 mol of AgNO3 reacting with 3.58mol NH3 to produce a solution containing Ag(NH3)2+
•We can relate the number of moles to say that since we have 2 moles of NH3 reacting to form 2 moles of NH3 as product,
•Therefore 3.58mol of NH3 produces 3.58 mol of NH3 as well
AgNO3+ 2NH3 <----> Ag+ + 2NH3
0.214mol 3.58mol 3.58mol
We can calculate the equilibrium concentration of free Ag+
Kf= [Ag+] [NH3]^2 / [AgNO3] [NH3]^2
Since we have [NH3]^2 as numerator and denominator we can easily cancel out both, leaving
Kf= [Ag+] / [AgNO3]
cross multiply
[Ag+] = Kf × [AgNO3]
Kf is given as 1.50× 10^7 at 25°C
Substitute into equation we have
[Ag+]= 1.50×10^7 × 0.214mol
[Ag+]= 3.21×10^6mol
Equilibrium concentration of free silver ion is 3.21×10^6 mol
Based on the data provided, the equilibrium concentration of free silver ion [ Ag⁺ ] in the solution is 1.43 × 10^⁻9 M
The equilibrium concentration of a substance is the concentration of that substance when it has achieved equilibrium.
From the data provided, equilibrium concentration of free silver ion can be calculated as follows:
0.214 mol of AgNO₃ was added
2.73 mol NH₃ was also added
Temperature = 25°C
Volume of solution = 1.00 L
Since Kf is very high, reactants get converted to product before equilibrium is attained.
The equilibrium concentration of [Ag+] is determined from the equation for the reaction at equilibrium:
Initially:
[Ag⁺] = 0.214 - 0.214 = 0
[Ag(NH₃)₂⁺] = 0 + 0.214 = 0.214
[NH₃] = 3.58 mol - 2 × 0.214 = 3.152
At equilibrium:
[Ag⁺] = X
[Ag(NH₃)₂⁺] = 0.214 - X
[NH₃] = 3.152 + X
kf = 1.5 * 10⁷ = [Ag(NH₃)₂⁺] / [NH₃]² [Ag⁺]
1.50 * 10⁷ = (0.214 - X) / ( 3.152 + X)² × (X)
assuming X to be very small:
Then;
1.50 * 10⁷ = (0.214) / (3.152)² × (X)
solving for X;
X = 7.09 × 10⁻⁷ M
Therefore, the equilibrium concentration of [ Ag⁺ ] is 1.43 × 10^-9 M
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