The pH of a 0.10 M salt solution is found to be 8.10. Determine the Kb and the degree of ionization of the basic ion.
A^- + HOH ==> HA + OH^-
Kb = (HA)(OH^-)/(A^-)
You have pH which gives you OH and HA and you have A^-. Calculate Kb
To find the Kb and the degree of ionization of the basic ion, we need to first identify the basic ion present in the salt solution. Since the pH is greater than 7, the solution is basic, and the basic ion is the one responsible for this alkaline nature.
Given that the salt solution is 0.10 M, we can assume it fully dissociates in water, so the concentration of the basic ion is also 0.10 M.
Step 1: Determine the pOH of the solution.
Since pH + pOH = 14, we can calculate the pOH as follows:
pOH = 14 - pH
pOH = 14 - 8.10
pOH = 5.90
Step 2: Calculate the concentration of OH- ions in the solution.
We can use the equation pOH = -log[OH-] to find the concentration of OH- ions:
[OH-] = 10^(-pOH)
[OH-] = 10^(-5.90)
[OH-] = 1.02 x 10^(-6) M
Step 3: Calculate the Kb for the basic ion.
Kb represents the equilibrium constant for the ionization of the basic ion. It can be calculated using the equation:
Kw = Ka * Kb
where Kw is the ion product constant for water (1.0 x 10^(-14) at 25°C), and Ka is the ionization constant for the conjugate acid of the basic ion.
Assuming that the conjugate acid of the basic ion is fully ionized in water, its concentration can be assumed as [H+] = 1.0 x 10^(-7) M (since pH = 7). Therefore, we can calculate Ka for the conjugate acid:
Ka = [H+][OH-] / [Salt]
Ka = (1.0 x 10^(-7))(1.02 x 10^(-6)) / 0.10
Now, rearranging the equation Kw = Ka * Kb, we can solve for Kb:
Kb = Kw / Ka
Kb = (1.0 x 10^(-14)) / [(1.0 x 10^(-7))(1.02 x 10^(-6)) / 0.10]
Calculating Kb using the given values:
Kb = 0.098 M
Step 4: Calculate the degree of ionization of the basic ion.
The degree of ionization, denoted by the Greek letter alpha (α), represents the fraction of the initial concentration of the basic ion that actually ionizes.
To find α, we need to compare the concentration of the ionized species ([OH-]) to the initial concentration of the basic ion ([Salt]):
α = [OH-] / [Salt]
α = (1.02 x 10^(-6)) / 0.10
Calculating α using the given values:
α = 1.02 x 10^(-5)
So, the Kb of the basic ion is 0.098 M and the degree of ionization is 1.02 x 10^(-5).
To determine the Kb and the degree of ionization of the basic ion, we need to use the equation for the ionization of a weak base in water:
B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)
First, we need to calculate the concentration of hydroxide ions (OH-) in the solution based on the pH.
pH = -log[H+]
To find OH- concentration, we can use the fact that:
[H+] * [OH-] = 1.0 x 10^-14 (from the ion product of water, Kw)
[H+] = 10^(-pH)
Now we can calculate [OH-] using the equation:
[OH-] = 1.0 x 10^-14 / [H+]
[OH-] = 1.0 x 10^-14 / (10^(-pH))
[OH-] = 1.0 x 10^-14 / (10^(-8.10))
[OH-] = 7.94 x 10^-7 M
Now that we have the concentration of hydroxide ions, we can calculate the concentration of the basic ion (BH+) by subtracting the hydroxide ion concentration from the initial concentration of the salt solution:
[BH+] = initial concentration of salt - [OH-]
[BH+] = 0.10 M - 7.94 x 10^-7 M
[BH+] ≈ 0.10 M (since the [OH-] concentration is very small compared to the initial concentration)
Now we can calculate the degree of ionization (α) using the equation:
α = [OH-] / initial concentration of salt
α = 7.94 x 10^-7 M / 0.10 M
α ≈ 7.94 x 10^-6
Finally, we can calculate the Kb (base dissociation constant) using the equation:
Kb = [BH+] * [OH-] / [B]
Kb ≈ (7.94 x 10^-7 M) * (7.94 x 10^-7 M) / (0.10 M)
Kb ≈ 6.30 x 10^-14
Therefore, the Kb of the basic ion is approximately 6.30 x 10^-14 and the degree of ionization is approximately 7.94 x 10^-6.