Find pOH with Concentration and Kb

Question

Determine the pOH of a 0.227 M C₅H₅N solution at 25°C.  

The K_b of C5H5N is 1.7 × 10^-9

^{If you have hints how to remember these more easily, that would be appreciated.}

^{Using pH to find the concentration or vice versa are starting to get more confusing, instead of easier.}

^{Thank you!}

Answer 1

There really aren't any hints to solve the current type of problems, it is simply a matter of being able to write the chemical equation and then using the Ka or Kb expression and solving it. Perhaps your problem is writing the original equation. For a weak base such C₅H₅N or any base, we write the hydrolysis as follows:

C₅H₅N + H₂O <==> C₅H₅NH⁺ + OH^- .. hydrolysis of C₅H₅N

You can always just substitute B as the base, and write the hydrolysis as...

B + H₂O <==> B⁺ + OH^- .. hydrolysis of a generic base

Now, to find the pOH, we must first find the [OH^-] and then simply take the negative log to get pOH...

Kb = [C₅H₅NH⁺][OH^-] / [C₅H₅N]

1.7x10^-9 = (x)(x) / 0.227 - x and assume x is small relative to 0.227 and ignore it

1.7x10^-9 = (x)² / 0.227

x² = 3.86x10^-10

x = [OH^-] = 1.96x10^-5 M

pOH = -log 1.96x10^-5

pOH = 4.71

If this doesn't make sense, or there is a part you don't understand, leave a "comment".

If you want hints for using pH to find the concentration, then the hint is to use the pH or pOH to calculate the [H⁺] or the [OH^-], and then use that value in the Ka or Kb expression. An example of this type question is the one you posted asking about the [NH₃] given a pH = 11. Please refer to the solution for that problem.