Final answer:
HCN will ionize the least in a 0.10 M NaOH solution due to the presence of excess hydroxide ions, which shift the equilibrium and suppress the ionization of HCN.
Step-by-step explanation:
To determine in which of the listed solutions HCN will ionize less than it does in pure water, we need to consider the properties of the other components in solution. Ionization will be affected by the relative strengths of the acids, bases, and the resulting equilibrium in the presence of added ions.
HCN is a weak acid with a conjugate base CN-, which is a weak base. In a solution with strong acids or bases, the ionization of HCN can be suppressed due to the common ion effect or by shifting the equilibrium.
0.10 M KNO3: Potassium nitrate is a salt derived from a strong acid (nitric acid) and a strong base (potassium hydroxide). It does not alter the ionization of HCN significantly as it dissociates into K+ and NO3- ions, neither of which reacts significantly with HCN or its ions.
0.10 M NaOH: Sodium hydroxide is a strong base that will fully dissociate in solution. The presence of excess OH- ions will suppress the ionization of HCN by shifting the equilibrium according to Le Châtelier's principle.
0.10 M NaCl: Sodium chloride is a neutral salt from a strong acid (hydrochloric acid) and a strong base (sodium hydroxide). Like KNO3, it dissociates into Na+ and Cl- ions, which do not affect the ionization of HCN significantly.
0.10 M NaCN: Sodium cyanide contains the CN- ion which is the conjugate base of HCN. The presence of CN- will reduce the ionization of HCN through the common ion effect where the dissociation is suppressed to maintain the equilibrium.
Among the given solutions, HCN will ionize the least in the 0.10 M NaOH solution due to the presence of excess hydroxide ions (OH-). The lower ionization in the presence of a strong base is a result of the equilibrium shifting to the left to maintain balance, favoring the reactants and reducing the concentration of ionized HCN.