The pKb of pyridine is 8.75. What is the pH of a 0.205M solution of pyridine?

To find the pH of a 0.205 M solution of pyridine, we first need to determine the concentration of hydroxide ions (OH-) in the solution.

The pKb of pyridine is given as 8.75, which is the negative logarithm (base 10) of its equilibrium constant for the reaction with water:

C5H5N + H2O ⇌ C5H5NH+ + OH-

In this reaction, pyridine (C5H5N) acts as a base and donates a proton to water, forming the pyridinium ion (C5H5NH+). The concentration of hydroxide ions (OH-) in the solution is equal to the concentration of the pyridinium ion (C5H5NH+).

Since pKb = -log(Kb), we can find the value of Kb (the equilibrium constant) by taking the inverse logarithm:

Kb = 10^(-pKb)

Kb = 10^(-8.75)

Now, we can set up an ice table to calculate the concentration of C5H5NH+ (which is equal to [OH-]).

Initial:
[C5H5N] = 0.205 M
[C5H5NH+] = 0 M
[OH-] = 0 M

Change:
[C5H5N] decreases by x
[C5H5NH+] increases by x
[OH-] increases by x

Equilibrium:
[C5H5N] = 0.205 - x
[C5H5NH+] = x
[OH-] = x

Using the Kb value, we can write the equilibrium expression for the reaction:

Kb = [C5H5NH+][OH-] / [C5H5N]

Since [C5H5NH+] = x and [OH-] = x, we can substitute these values into the equilibrium expression:

10^(-8.75) = x * x / (0.205 - x)

Simplifying the equation, we get:

10^(-8.75) = x^2 / (0.205 - x)

To solve this equation, we can make an approximation and assume that x is much smaller than 0.205. This makes the denominator essentially equal to 0.205.

10^(-8.75) = x^2 / 0.205

Now, we solve for x:

x^2 = 0.205 * 10^(-8.75)

x ≈ √(0.205 * 10^(-8.75))

At this point, we can use a calculator to find the approximate value of x. After calculating x, we can substitute it back into the equation to find the concentration of hydroxide ions ([OH-]) in the solution.

Finally, we can use the concentration of hydroxide ions to calculate the pOH and then convert it to pH using the equation:

pOH = -log10[OH-]
pH = 14 - pOH

So, by following these steps, you will be able to find the pH of the 0.205 M solution of pyridine.