Question

Silver chloride and silver iodide form equilibria when added to water.
AgCl(s) = Ag+(aq) + Cl –(aq) Kc = K1
AgI(s) = Ag+(aq) + I–(aq) Kc = K2
Each equilibrium position lies well to the left.
Silver iodide will not dissolve in aqueous ammonia. Silver chloride will dissolve in aqueous ammonia. Another equilibrium is formed.
Ag+(aq) + 2NH3(aq) = Ag(NH3)2+(aq) Kc = K3
The position of this equilibrium lies to the right.
What is the order of magnitude for these three equilibrium constants?
A. K1 > K2 > K3
B. K2 > K1 > K3
C. K3 > K1 > K2
D. K3 > K2 > K1

PLEASE SOMEONE EXPLAIN HOW THE ANSWER IS C..........

Answer 1

Answer: C

Explanation:

When the equilibrium lies to the right, the concentration of products will increase. This causes the value of KC to increase. So, K3 should be the largest among the three. AgCl dissolves in aqueous ammonia. Ag+ would react with ammonia to form soluble complex ion. This causes the concentration of Ag+ in first equation to decrease. Once the concentration of Ag+ decreases, the equilibrium shift to the right to oppose the change. So, the concentration of products would increases as there is Cl- ions as well. Therefore, K1 is larger than K2.

Answer 2

Answer:c

Explanation:equilibrium to right and left