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?
could someone help me with this question or how to get started?
Call the weak base BOH.
BOH <==> B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH)
If 0.72% ionized, that means(H^+) must be 1.3 M x 0.0072 = ??
Of course (OH^-) must be the same.
If BOH is 0.72% ionized, it must be 100% - 0.72% unionized.
Substitute into Kb expression and solve for Kb. Post your work if you get stuck.
Sure! To find the Kb value of a weak base, we need to use the equation for percent ionization and the initial concentration of the weak base.
The percent ionization of a weak base is defined as the ratio of the concentration of the dissociated base to the initial concentration of the weak base, multiplied by 100. The formula is:
% Ionization = (Concentration of dissociated base / Initial concentration of weak base) * 100
In this case, we are given the percent ionization (0.72%) and the initial concentration of the weak base (1.3 mol/L). Let's denote the concentration of the dissociated base as x mol/L.
So, we can rewrite the formula as:
0.72% = (x / 1.3) * 100
Now, we can solve for x by rearranging the equation:
0.72% = (x / 1.3) * 100
0.72 / 100 = x / 1.3
0.0072 = x / 1.3
Multiplying both sides by 1.3, we get:
0.00936 = x
Therefore, the concentration of the dissociated base is 0.00936 mol/L.
Next, we can determine the concentration of the undissociated base by subtracting the concentration of the dissociated base from the initial concentration of the weak base:
Concentration of undissociated base = Initial concentration of weak base - Concentration of dissociated base
Concentration of undissociated base = 1.3 - 0.00936
Concentration of undissociated base = 1.29064 mol/L
Finally, we need to use the equilibrium constant expression for the dissociation of the weak base (B) into the conjugate acid (BH+), given by:
Kb = (Concentration of BH+ * Concentration of B-) / Concentration of B
Since we know the concentrations of the undissociated base (B) and conjugate acid (BH+), we can substitute these values into the equation:
Kb = (Concentration of BH+ * Concentration of B-) / Concentration of B
Kb = (0.00936 * 0.00936) / 1.29064
Kb = 6.7488 * 10^-5
The Kb value of this weak base is approximately 6.7488 x 10^-5.