Aniline, C6H5NH2, is a weak base. If the hydroxide ion concentration of a 0.223 M solution is 9.7 10-6 M, what is the ionization constant for the base?

Let's call this BNH2 to save some typing.

..........BNH2 + HOH --> BNH3^+ + OH^-
I........0.223.............0.......0
C.........-x...............x.......x
E.......0.223-x............x.......x
But you know x = 9.7E-6. Evaluate 0.223-x and substitute into the Kb expression and solve for Kb.
Kb = (BNH3^+)(OH^-)/(BNH2)

To determine the ionization constant for the base aniline (C6H5NH2), we need to use the equation for the base dissociation:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

The ionization constant is given by the expression:

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

To calculate the ionization constant (Kb), we need the concentrations of C6H5NH3+ and OH-, as well as the concentration of C6H5NH2.

Given that the hydroxide ion concentration ([OH-]) is 9.7 * 10^-6 M, we can assume that the concentration of C6H5NH3+ is also equal to 9.7 * 10^-6 M (since they are in a 1:1 ratio based on the balanced equation).

Now, we need to determine the concentration of C6H5NH2.

The initial concentration of C6H5NH2 in a 0.223 M solution is 0.223 M, but since it is a weak base, it will not completely dissociate. To find the concentration of C6H5NH2, we must consider the equilibrium expression for the reaction:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

At equilibrium, the concentration of C6H5NH2 is equal to its initial concentration minus the concentration of C6H5NH3+ formed. Since the concentration of C6H5NH3+ is 9.7 * 10^-6 M, the concentration of C6H5NH2 is:

[C6H5NH2] = 0.223 M - 9.7 * 10^-6 M = 0.223 M

Now, we can substitute the values into the Kb expression:

Kb = ([C6H5NH3+][OH-]) / [C6H5NH2]
= (9.7 * 10^-6 M * 9.7 * 10^-6 M) / 0.223 M
≈ 4.21 * 10^-10

Therefore, the ionization constant (Kb) for the base aniline is approximately 4.21 * 10^-10.