A 25.0 mL sample of 0.100 M pyridine (Kb for pyridine is 1.7 ✕ 10-9) is titrated with 0.100 M HCl solution. Calculate the pH after the addition of the following amounts of HCl with sig figs. a) 24.5 mL b) 25.0 mL c) 26.0 mL d) 28.0 mL e) 30.0 mL

Answer 1

You have five problems here. You must recognize where you are on the titration curve to each of these. Pyridine is C5H5N.
.............C5H5N + HCl ==> C5H5N*HCl
millimols C5H5N initially = mL x M = 25.0 x 0.1 = 2.50
a. millimols HCl added. 24.5 x 0.1 = 2.45
b. millimols HCl added. 25.0 x 0.1 = 2.50
c. millimols HCl added. 26.0 x 0.1 = 2.60
d. millimols HCl added. 28.0 x 0.1 = 2.80
e. millimols HCl added. 30.0 x 0.1 = 3.00
The equivalence point is at b. So a is before the eq. pt. and c,d,e are after.

a. Before the eq. pt. you will still have some pyridine not neutralized and you will have formed some of the salt. You have a buffered solution.Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
.............C5H5N + HCl ==> C5H5N*HCl
I................25.0.......0..............0
add......................24.5.......................
C.............-24.5...-24.5.........24.5
E..............0.5...........0..........24.5
pH = pKa + log (base)/(acid)
pH = pKa + log (0.5/24.5)

b. The equivalence point. The pH is determined by the hydrolysis of the salt which is as follows: The total volume now is 25.0 + 25.0 = 50.0 so the concn of the C5H5N*HCl to start is M = mmols/mL = 2.50/50 = 0.05 M
.................C5H5N*HCl + HOH ==> C5H5N + H3O^+ + Cl^-
I..................0.05.................................0..............0.............0
C..................-x....................................x..............x
E.............0.05-x...................................x..............x
Ka for C5H5N*HCl =Kw/Kb for C6H6N) = (x)(x)/(0.05-x) = (1E-14/1.79E-9)
Solve for x = (H3O^+) and convert that to pH.

c,d,e. All are AFTER the eq. pt so you will have neutralized ALL of the pyridine and you will have an excess of the acid . I will do the c and leave d and e to you.

.............C5H5N + HCl ==> C5H5N*HCl
I..............2.50..........0..............0.
add.........................2.60...........................
C..........-2.50........-2.50.............2.50
E.............0.............0.1.............2.50
The M of the excess HCl. You have 0.1 mmols HCl and you have a volume of 25.0 + 26.0 = 51 mL so M = 0.1/51 = ?. Since that is straight HCl, that gives you the (H^+) and convert that to pH.
d and e are done the same way with the only difference being the total volume which for 28 mL added titrant is 25.0 + 28 and the other is 25.0 + 30.0.
Post your work if you get stuck on any of these. A blackboard is useful for showing this because you can see the titration curve and see where you are with each of these additions.

Answer 2

a.OOOPS. I substituted volumes and not mmols for the a part. I should have done this. Before the eq. pt. you will still have some pyridine not neutralized and you will have formed some of the salt. You have a buffered solution.Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
.............C5H5N + HCl ==> C5H5N*HCl
I................2.50.......0.............0..............0
add......................2.45.......................
C.............-2.45...-2.45.........2.45
E..............0.05...........0..........2.45
pH = pKa + log (base)/(acid)
pH = pKa + log (0.05/2.45)
I don't think that changes the answer any since both numerator and denominator were too high by a factor of 10 so that cancels and the answer stays the same; however, the actual numbers were wrong and I didn't like that. Sorry about the mixup