Treat this as just another ammonia, NH3, problem. Remember, NH3 + HOH ==> NH4^+ + OH^- and you know how to solve those problems. This is the same thing.
See here for the structure.
https://en.wikipedia.org/wiki/Pyridine
Let's call pyridine a simple PN. Then
......PN + HOH ==> PNH^+ + OH^-
I.....0.1............0.......0
C.....-x.............x.......x
E....0.1-x...........x.......x
Kb pyr = (PNH^+)(OH-)/(PN)
Substitute the E line and solve for x = (OH^-), then convert to pH.
A 25.0 mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.
Calculate the pH at 0 mL of added acid.
2 answers
hi I had a similar question too this on an assignment and when I calculated it my pOH was 4.89 and my pH was 9.11 which doesn't seem right and I don't know what I did wrong