Calculate the pH for each of the following solutions:

Question

(a) 0.10 M NH3
(b) 0.050 M C5H5N (pyridine)

Answer 1

NH3 + HOH ==> NH4 + OH^-

Kb = (NH4^+)(OH^-)/(NH3)
Prepare an ICE chart, plug into the Ka expression, and solve for (OH^-). Convert to pOH, then to pH.
The pyridine is worked the same way.

C5H5N + HOH ==> C5H5NH^+ + OH^-

Post your work if you get stuck.