Determine the pH of a 1 M aniline solution (Kb = 4.2 10-10).

I am unsure how to start, any help?

Write the hydrolysis equation. Call aniline BNH2, then

........BNH2 + H2O ==> BNH3^+ + OH^-
I........1M.............0.......0
C........-x.............x.......x
E........1-x............x.......x

Kb = (x)(x)/(1-x)
Solve for x = OH and convert to pH.

To determine the pH of a 1 M aniline solution, you need to use the given value of Kb (base dissociation constant) for aniline.

Aniline (C6H5NH2) is a weak base, so it undergoes partial dissociation in water according to the following equation:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

The Kb expression for aniline is:

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

Given that the Kb value for aniline is 4.2 × 10^-10, we can assume that the concentration of OH- formed by the dissociation of aniline is very small compared to the original concentration of aniline.

To solve this problem, we'll make the approximation that [OH-] is negligible compared to the initial concentration of the aniline. This allows us to simplify the Kb expression to:

Kb = [C6H5NH3+]/[C6H5NH2]

Since aniline is a weak base, we can assume that the concentration of aniline before dissociation ([C6H5NH2]) is equal to the initial concentration of the solution (1 M). Thus, the equation becomes:

Kb = [C6H5NH3+]/1

Simplifying further gives:

[C6H5NH3+] = Kb

Now, we can use the given Kb value to find the concentration of C6H5NH3+:

[C6H5NH3+] = 4.2 × 10^-10

To determine the pH, we'll need to find the pOH first. The pOH is the negative logarithm (base 10) of the hydroxide ion concentration.

pOH = -log10([OH-])

Since we made the approximation that [OH-] is negligible compared to the initial concentration of aniline, we can assume that [OH-] ≈ 0. Hence, the pOH is approximately equal to 0.

Now, we can use the relationship between pH and pOH:

pH + pOH = 14

Since the pOH is approximately 0, we have:

pH + 0 = 14

Therefore, the pH of the 1 M aniline solution is approximately 14.