Question

aniline, c6h5nh2, is a weak base. its base-dissociation constant at 25°c is 7.5 ✕ 10−10. calculate the poh of a 0.467 m aniline solution in water at 25°c. round your answer to two decimal places.

Answer 1

To calculate the pOH of a 0.467 M aniline solution, use the base-dissociation constant (Kb) of aniline and apply the equilibrium constant equation. Solve for the concentration of OH- and then determine pOH by taking the negative logarithm of the OH- concentration.

Step-by-step explanation:

To calculate the pOH of a 0.467 M aniline solution, we need to use the base-dissociation constant (Kb) of aniline. The Kb value given is 7.5 ✕ 10−10, which represents the equilibrium constant for the reaction:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

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

The equilibrium constant can be used to find the concentration of OH- ion, which is related to pOH. Using the given Kb value and the concentration of aniline, we can proceed with the calculation:

Let x be the concentration of OH-

Kb = (x)(x) / (0.467 - x)

Since Kb is very small and x is much smaller than (0.467 - x), we can assume that x ≈ Kb

Kb = (x)(x) / 0.467

Solving for x gives us x = √(Kb * 0.467)

Finally, pOH = -log10(x)