Let's call pyridinium chloride just BNHCl so it's the BNH^+ that is hydrolyzed.
...........BNH^+ + H2O ==> BN + H3O^+
I..........0.2..............0.....0
C...........-x..............x.....x
E.........0.2-x.............x.....x
Ka for BNH^+ = (Kw/Kb for pyridine) = (x)(x)/(0.2-x)
Solve for x = (H3O^+) and convert to pH. Looking at the answers, you may need to solve the quadratic but I should point out that the answers are given to 4 significant figures and Kb is known only to 2 s.f. so all of that precision in the answer choices is not warranted.
What is the pH of a 0.2 M solution of pyridinium
chloride (C5H5NHCl)? Kb for pyridine
(C5H5N) is 1.5 × 10−9
.
1. 8.824
2. 2.788
3. 4.761
4. 3.068
5. 3.308
6. 2.698
7. 2.938
8. 5.176
9. 9.239
1 answer
3 answers