Question

HCl (aq) + H2O (l) ⇌ H3O^+ (aq) + Cl^- (aq)
In 1.0 M HCl (aq), is nearly 100 percent dissociated, as represented by the equation above. Which of the following best helps to explain why, in 0.01 M HCN (aq), less than 1 percent of HCN is dissociated?

A. Compared to the HCl (aq) solution, the concentration of the HCN (aq) solution is much too dilute to achieve 100 percent dissociation.

B. The CN^- ion is not very soluble in water, and a solid precipitate would form if more of the HCN dissociated..

C. The equilibrium constant for the dissociation of HCN (aq) is much smaller than that for the dissociation of HCl (aq).

D. HCN (aq) reacts with water to form a basic solution, and the high concentration of OH^- (aq) interferes with the dissociation process.

Answer 1

There is lesser dissociation of hydrogen cyanide as compared to HCl, as  equilibrium constant for the dissociation of hydrogen cyanide has been smaller than HCl. Hence, option C is correct.

Hydrogen cyanide has been a weak acid that has been dissociated into the constituent ions. The dissociation of the compound in water has been corresponds to the equilibrium constant.

Equilibrium constant for hydrogen cyanide and HCl

The equilibrium constant has been defined as the ratio of the concentration of solute and solvent ions at equilibrium.

The sample with lesser equilibrium constant tends to dissociate at lesser concentration as well.

The equilibrium constant for the dissociation of hydrogen cyanide has been smaller than HCl. Thus, there will be lesser dissociation of hydrogen cyanide as compared to HCl. Hence, option C is correct.

Learn more about equilibrium constant, here:

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