'What is the pH of 0.083 M C5H5N if the Kb value for C5H5N is 2.9 x 10^-9?'
I wrote the equation as C5H5N + H2O --> C5H6N^+ + OH^-.
Then, I set up Kb = ([C5H6N^+][OH^-])/[C5H5N]
Next I substituted the numbers in: (2.9 x 10^-9) = (x^2)/0.083 , which makes x about equal to 1.55 x 10^-5.
pOH = - log[OH^-], so pOH is about 4.809, putting pH around 9.191. Is this the right approach?
I'm taking this online and teaching myself everything, and it's immensely helpful to be able to check my answers like this, so thank you very very much!
That looks ok to me. I would make two suggestions.
1. I would write the product in the first equation you have as C5H5NH^+ to make it obvious that the H has been added. At first I thought you had made a typo.
2. I think it is always a good idea to write the K expression in the denominator, in this case, as 0.083-x. The reasoning is that if only the 0.083 is there you may forget in cases where the x cannot be neglected and not take that into account. It's always easy to neglect the x if that is warranted but if it isn't there it's easy to forget it too.
Yes, your approach is correct! You correctly set up the equation for the reaction between C5H5N and water to determine the concentration of OH- ions. Then using the Kb expression, you substituted the given values and solved for x.
To find the pOH, you used the definition pOH = -log[OH-]. You obtained a value of 4.809 for pOH, which is correct.
Finally, to find the pH, you used the fact that pH + pOH = 14 (at 25 degrees Celsius). So, pH = 14 - pOH, which gives you a pH value of approximately 9.191.
Your approach is accurate, and you found the correct pH value for the given concentration of C5H5N and the Kb value. Great job! Keep up the good work in teaching yourself and exploring chemistry concepts.