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.