Consider the reaction Hg²⁺(aq)+ Hg(l) Hg₂²⁺(aq). Find the equilibrium constant for this reaction at 25°C.

Question

Quizplus · Accepted Answer

The equilibrium constant expression can be written as:

Kc = [HgS1U1B12S1U1B0S1U1P12][S1S1P0] / [HgS1U1P12][S1S1P0]

Since Hg(l) is a pure liquid, its concentration is essentially constant and does not affect the equilibrium constant.

Therefore, we can simplify the equilibrium constant expression to:

Kc = [HgS1U1B12S1U1B0S1U1P12] / [HgS1U1P12]

Using the solubility product constant (Ksp) for HgS, we can write:

[HgS1U1B12S1U1B0S1U1P12] = Ksp[HgS1U1P12]

Substituting into the equilibrium constant expression, we get:

Kc = Ksp

At 25°C, the solubility product constant for HgS is approximately 1.6 x 10^-50. Therefore, Kc ≈ 2 x 10^-50, which corresponds to choice E.

Consider the Reaction Hg2+(aq)+ Hg(l) Hg22+(aq) ×\times× 101 B)6 ×\times× 1059 C)4