Question

Given that Fe2+ forms a complex ion with CN-, should Fe(NO3)2 be more or less soluble in a solution of 0.5M NaCN than in pure water?

Group of answer choices

A.more soluble; the presence of CN- will shift the solvation of Fe(NO3)2 to the right

B.more soluble; the presence of CN- will shift the solvation of Fe(NO3)2 to the left

C.less soluble; the presence of CN- will shift the solvation of Fe(NO3)2 to the left

D.less soluble; the presence of CN- will shift the solvation of Fe(NO3)2 to the right

Answer 1

The dissolution equilibrium of Fe(NO3)₂ is given below:

Fe(NO3)₂ is said to be more soluble if more Fe(NO3)₂ molecules break into Fe²⁺ and NO₃^- ions. That is, Fe(NO3)₂ is said to be more soluble if the equilibrium shifts right.

In water, no stable complex is formed between Fe²⁺ and NO₃^- ions and water.

In NaCN, Fe²⁺ forms a complex ion with CN^-. This will decrease the concentration of Fe²⁺ ions. According to Le-Chatlier's principle, this will result in the equilibrium shifting right, resulting in more solubility.

Therefore the correct answer is the first option, that is:

A. more soluble; the presence of CN^- will shift the solvation of Fe(NO₃)₂ to the right