Answer and Explanation:
(a) The H3O+ concentration in a 0.66 M NH4+(aq) solution can be determined using the equilibrium equation of the reaction between NH4+ and water. In this reaction, NH4+ acts as an acid and donates a proton (H+) to water to form NH3 and H3O+.
The balanced equation for the reaction is:
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
The Kb value of NH3 is given as 1.8 * 10^-5.
Since NH4+ and NH3 have a 1:1 stoichiometric ratio in the balanced equation, the concentration of NH3 formed will also be 0.66 M.
Using the equation for Kb:
Kb = [NH3][H3O+] / [NH4+]
We can rearrange the equation to solve for [H3O+]:
[H3O+] = (Kb * [NH4+]) / [NH3]
Substituting the given values:
[H3O+] = (1.8 * 10^-5 * 0.66) / 0.66
[H3O+] = 1.8 * 10^-5
Therefore, the H3O+ concentration in the 0.66 M NH4+(aq) solution is 1.8 * 10^-5 M.
(b) The percent dissociation for this reaction can be calculated by comparing the concentration of NH3 formed with the initial concentration of NH4+. Since NH4+ and NH3 have a 1:1 stoichiometric ratio, the percent dissociation can be expressed as:
% dissociation = ([NH3] / [NH4+]) * 100
Substituting the given values:
% dissociation = (0.66 / 0.66) * 100
% dissociation = 100
Therefore, the percent dissociation for this reaction is 100%. This means that all of the NH4+ in the solution has dissociated to form NH3 and H3O+.