0.214 mol agno3 and 3.58 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) = 1.50 x 10^7 at 25 °c.

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

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