# Chapter 13 - Exercises - Page 575b: 36

a) $HCN(aq) + H_2O(l) \lt -- \gt CN^-(aq) + H_3O^+(aq)$ $K_a = \frac{[H_3O^+][CN^-]}{[HCN]}$ b) $HOC_6H_5(aq) + H_2O(l) \lt -- \gt OC_6{H_5}^-(aq) + H_3O^+(aq)$ $K_a = \frac{[H_3O^+][OC_6{H_5}^-]}{[HOC_6H_5]}$ c) $C_6H_5N{H_3}^+(aq) + H_2O(l) \lt -- \gt C_6H_5NH_2(aq) + H_3O^+(aq)$ $K_a = \frac{[H_3O^+][C_6H_5NH_2]}{[C_6H_5N{H_3}^+]}$

#### Work Step by Step

a) 1. Write the ionization chemical equation: - Since $HCN$ is an acid, write the reaction where it donates a proton to a water molecule: $HCN(aq) + H_2O(l) \lt -- \gt CN^-(aq) + H_3O^+(aq)$ 2. Now, write the $K_a$ expression: - The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants: $K_a = \frac{[Products]}{[Reactants]}$ $K_a = \frac{[H_3O^+][CN^-]}{[HCN]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution. b) 3. Write the ionization chemical equation: - Since $HOC_6H_5$ is an acid, write the reaction where it donates a proton to a water molecule: $HOC_6H_5(aq) + H_2O(l) \lt -- \gt OC_6{H_5}^-(aq) + H_3O^+(aq)$ 4. Now, write the $K_a$ expression: - The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants: $K_a = \frac{[Products]}{[Reactants]}$ $K_a = \frac{[H_3O^+][OC_6{H_5}^-]}{[HOC_6H_5]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution. c) 5. Write the ionization chemical equation: - Since $C_6H_5N{H_3}^+$ is an acid, write the reaction where it donates a proton to a water molecule: $C_6H_5N{H_3}^+(aq) + H_2O(l) \lt -- \gt C_6H_5NH_2(aq) + H_3O^+(aq)$ 6. Now, write the $K_a$ expression: - The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants: $K_a = \frac{[Products]}{[Reactants]}$ $K_a = \frac{[H_3O^+][C_6H_5NH_2]}{[C_6H_5N{H_3}^+]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.

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