A weak base with a concentration of 1.3 mol/L has a percent ionization of 0.72%. What is the Kb of this weak base?
Write the ionization for the weak base. I would call it BOH.
Write the Kb expression.
Set up an ICE chart if that will help.
I would call BOH 1.3 M, then concn B^+ = 1.3 x 0.0072.
The unionized BOH is 1.3 x 0.9928,
Substitute these into Kb expression and solve for Kb. post your work if you need further assistance.
To find the Kb (base dissociation constant) of a weak base, we need to use the given information about the base's concentration and percent ionization.
The percent ionization is the ratio of the concentration of ionized base to the initial concentration of the weak base, expressed as a percentage.
In this case, we are given that the percent ionization is 0.72% or 0.0072 (0.72/100 = 0.0072).
Now, let's break down the steps to find the Kb:
Step 1: Calculate the concentration of ionized base.
The concentration of ionized base can be calculated using the percent ionization and the initial concentration of the weak base.
Concentration of ionized base = percent ionization × initial concentration of the weak base
Concentration of ionized base = 0.0072 × 1.3 mol/L
Concentration of ionized base = 0.00936 mol/L
Step 2: Calculate the concentration of non-ionized base.
The concentration of non-ionized base can be obtained by subtracting the concentration of ionized base from the initial concentration of the weak base.
Concentration of non-ionized base= Initial concentration of the weak base - Concentration of ionized base
Concentration of non-ionized base = 1.3 mol/L - 0.00936 mol/L
Concentration of non-ionized base = 1.29064 mol/L
Step 3: Calculate the equilibrium concentration of the weak base and its conjugate acid.
At equilibrium, the concentration of the weak base will be equal to the concentration of non-ionized base since all of it will have been converted to the ionized and non-ionized forms.
Therefore, the equilibrium concentration of the weak base = 1.29064 mol/L.
Step 4: Write the equilibrium expression for the dissociation of the weak base.
The dissociation of a weak base can be represented by the equation:
B ⇌ BH+ + OH-
The equilibrium expression for this reaction is:
Kb = [BH+][OH-] / [B]
Step 5: Substitute the known values into the equilibrium expression and solve for Kb.
Kb = ([BH+][OH-]) / [B]
Kb = ([B] × [OH-]) / [B]
Kb = [OH-]
In this case, the concentration of OH- is equal to the concentration of the weak base since one molecule of OH- is produced for each molecule of the weak base that dissociates.
Therefore, the Kb of this weak base is equal to 1.29064 mol/L.