Question

35 mL of a 0.0250 M pyridine is titrated with 0.00 mL, 10.00mL 15.00mL 20.00 mL and 25.00mL of a 0.0438 M HCl solution. Calulate the pH of the solution after each addition and sketch the resulting graph

Answer 1

Kb of C5H5N = 1.7 x 10^-9
pKb = - log 1.7 x 10^-9=8.77

moles C5H5N = 0.0350 L x 0.0250M= 0.000875

1
moles HCl added = 0.010 L x 0.04380 M = 0.000438
the reaction is
C5H5N + H+ = C5H5NH+
moles C5H5NH+ formed = 0.000438
moles C5H5N in excess = 0.000875 - 0.0004380=0.000437
total volume = 0.045 L
[C5H5N]= 0.000437 / 0.045 L=0.00971 M
[C5H5NH+] = 0.0004380/ 0.045 L=0.00973 M
pOH = pKb + log [C5H5NH+]/ [C5H5N] ( Handerson-Hasselbalch equation)
pOH = 8.77 + log 0.00973/ 0.00971=8.77089
pH = 14 - pOH = 14 - 8.77089 = 5.2291

2

moles HCl added = 0.015 L x 0.04380 M = 0.000657
the reaction is
C5H5N + H+ = C5H5NH+
moles C5H5NH+ formed = 0.000657
moles C5H5N in excess = 0.000875 - 0.000657=0.000218
total volume = 0.050 L
[C5H5N]= 0.000218 / 0.05 L=0.00436 M
[C5H5NH+] = 0.000657/ 0.05 L=0.01314 M
pOH = pKb + log [C5H5NH+]/ [C5H5N] ( Handerson-Hasselbalch equation)
pOH = 8.77 + log 0.01314/ 0.00436=9.25
pH = 14 - pOH = 14 - 9.25 = 4.75

3

moles HCl added = 0.020 L x 0.04380 M = 0.000876
the reaction is
C5H5N + H+ = C5H5NH+
moles C5H5NH+ formed = 0.000876
moles C5H5N in excess = 0.000875 - 0.000876= - 0.000001

Which means moles of C5H5NH+ formed = moles of C5H5N consumed = 0.000875
total volume = 0.055 L
[C5H5N]= [C5H5NH+]
pOH = pKb + log [C5H5NH+]/ [C5H5N] ( Handerson-Hasselbalch equation)
pOH = 8.77
pH = 14 - pOH = 14 - 8.77 = 5.23

4

moles HCl added = 0.025 L x 0.04380 M = 0.001095
the reaction is
C5H5N + H+ = C5H5NH+
moles C5H5NH+ formed = 0.001095
moles C5H5N in excess = 0.000875 - 0.001095= - 0.00022

Which means moles of C5H5NH+ formed = moles of C5H5N consumed = 0.000875
total volume = 0.06 L
[C5H5N]= [C5H5NH+]
pOH = pKb + log [C5H5NH+]/ [C5H5N] ( Handerson-Hasselbalch equation)
pOH = 8.77
pH = 14 - pOH = 14 - 8.77 = 5.23