Answer
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.
