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

- decrease the number of plants and animals living in the water.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )