The equilibrium constant for the dissolution of AgI(s) in water is 1.5×10-16. Ag

Question

The equilibrium constant for the dissolution of AgI(s) in water is 1.5×10-16. AgI(s) Ag +(aq) + I-(aq) A solution is prepared by diluting 2.0 mol AgI, 3.0×10-8 mol Ag+, and 3.0×10-8 mol I- to a volume of 1.0 L. Calculate the reaction quotient and determine whether a reaction occurs. A.Q = K; no reaction occurs. B.Q < K; Ag+ and I- combine to form AgI(s) until equilibrium is established. C.Q > K; AgI(s) dissolves until equilibrium is established. D.Q < K; AgI(s) dissolves until equilibrium is established. E.Q > K; Ag+ and I- combine to form AgI(s) until equilibrium is established.

Explanation / Answer

For the given dissociation of AgI(s) in water,

AgI(s) = Ag+(aq) + I-(aq)

The equilibrium constant Kc = 1.5×10-16

[Ag+] = 3.0×10-8 mol

[I-] = 3.0×10-8 mol

[AgI] = 2 mol

The reaction quotient Qc can be calculated as

Qc = [Ag+] [I-] / [AgI]

Qc = 3.0×10-8 * 3.0×10-8 / 2

Qc = 4.5×10-16

Now comparing between Kc and Qc, we find that

Qc > Kc

When Qc > Kc, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. Hence Ag+ and I- combine to form AgI(s) until equilibrium is established.

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