Question

For a solution equimolar in HCN and NaCN, which statement is false?

A. This is an example of the common-ion effect.

B. Addition of NaOH will increase [CN^�] and decrease [HCN].

C. Addition of more NaCN will shift the acid-dissociation equilibrium of HCN to the left.

D. [H⁺] is equal to K_a.

E. [H⁺] is larger than it would be if only the HCN were in solution.

Answer 1

the answer is E.

HCN(aq) <=> H^+(aq) + CN^-(aq)

This is the equilibrium. Basically LeChatelier states that if a stress is placed on a system at equilibrium then the system will act in such a way as to relieve that stress. This means that the system will adjust to keep the equilibrium value constant.

D) Addition of more NaCN will shift the acid dissociation equilibrium of HCN to the left.
i.e., forcing this reaction to occur HCN --> CN^- + H^+
If you add CN^- to the equilibrium the [CN^-] will increase. Then the system will shift its equilibrium so that [CN^-] will decrease and [HCN] will increase. In order for that to happen the direction of shift in the equilibrium must be to the left toward the formation of more HCN.
NOTE: [CN^-] means concentration of CN^- ion. [HCN] means concentration of HCN.

E) Addition of NaOH will increase [CN-] and decrease [HCN].

If you add NaOH to the equilibrium mixture then the following reaction will occur.

H^+(aq) + OH^- --> H2O

This means that the equilibrium must shift toward the right to produce more H^+. When it does that the [HCN] must decrease. Since the HCN is producing more H^+ to replace that which was lost, it must also produce more CN^- and increase [CN^-].