Aniline, C6H5NH2, is a weak base. If the hydroxide ion concentration of a 0.201 M solution is 9.2 10-6 M, what is the ionization constant for the base?

Let's call aniline BNH2

...........BNH2 + HOH ==> BNH3^+ + OH^-
I..........0.201...........0........0
C...........-x.............x........x
E.........0.201-x..........x........x

Kb = (BNH3^+)(OH^-)/(BNH2)
You know x. Substitute into Kb expression and solve for Kb.

To find the ionization constant for the base, you need to calculate the equilibrium constant for the dissociation of aniline. The equation for the dissociation of aniline (C6H5NH2) in water can be written as:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

The ionization constant, also known as the base dissociation constant (Kb), can be calculated using the equilibrium concentrations of the reactants and products. The equation for Kb is:

Kb = [C6H5NH3+][OH-] / [C6H5NH2]

Given: [OH-] = 9.2 × 10^(-6) M and initial concentration of aniline [C6H5NH2] = 0.201 M.

First, let's determine the change in concentrations of the species in the equation above by assuming x as the change or molar concentration of [C6H5NH3+]. The change in [OH-] concentration would be equal to x.

[C6H5NH2] = 0.201 - x
[C6H5NH3+] = x
[OH-] = x

Now, substitute these values into the Kb equation:

Kb = ([C6H5NH3+][OH-]) / [C6H5NH2]
Kb = (x * x) / (0.201 - x)

Given that Kb is the ionization constant for the base, this implies that it is a small value, indicating that the reaction has limited dissociation.

Since the value of [OH-] is very small compared to the initial concentration of the base [C6H5NH2], we can assume that 0.201 - x ≈ 0.201.

Therefore, the equation simplifies to:

Kb ≈ (x * x) / 0.201

To solve for x, we can use the quadratic formula. However, we can also make an approximation since Kb is a small value. We can assume that the value of x is close to 0, so the x^2 term in the equation is negligible.

Kb ≈ (0 * 0) / 0.201
Kb ≈ 0

Hence, the ionization constant for aniline (C6H5NH2) is approximately 0.