Question

HAc (aq) + H20 (l) <-- --> Ac- (aq) + H30+ (aq)a. Write the equilibrium expression. Are we using Ka or Kb?b. Label the acid, base, conjugates, and/or salt.c. Which bases are competing for a proton?d. If the K value is 0.06 which base is stronger?

Answer 1

`\begin{gathered} a)\text{ }([H_3O^+][Ac^-])/([HA_c]) \\ b)\text{ HAc - acid} \\ H_2O-base \\ Ac^--conjugate\text{ base} \\ H_3O^+-conjugate\text{ acid} \\ c)\text{ H}_2O \\ d)\text{ Ac}^{-\text{ }}is\text{ stronger since k > 0} \end{gathered}`

Explanations:

Given the equilibrium expression shown below;

`HAc(aq)+H_2O(l)\rightleftarrows Ac^-(aq)+H_3O^+(aq)`

a) The equilibrium expression for the reaction is given as:

`\begin{gathered} k_a=([H_3O^(^+)][Ac^-])/([H_2O][HAc]) \\ Since\text{ }[H_2O]-1,\text{ hence;} \\ k_a=([H_3O^+][Ac^-])/([HA_c]) \end{gathered}`

The equilibrium constant we are using will be Ka.

b) The compound that is given off a proton is an acid while the compound accepting a proton is a base. From the given equation HAc acts as an acid while H2O acts as a base. Ac- is the conjugate base while H3O+ is the conjugate acid.

c) The base that is competing for a proton is H2O (water)

d) Since a lower value of k shows a stronger base, hence the conjugate base hence H2O will be a stronger base than that of Ac. Recall that if k < 1, the reverse reaction is favoured and the forward reaction does not proceed to a great extent showing that H2O will be the stronger base.