Question

A chemist titrates 220.0 mL of a 0.4168 M pyridine CHN solution with 0.5811 MHBr solution at 25 C. Calculate the pH at equivalence. The pK of pyridine is 8.77. Round your answer to 2 decimal places Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added pH ? Ar

Answer 1

use:
pKb = -log Kb
8.77= -log Kb
Kb = 1.698*10^-9
find the volume of HBr used to reach equivalence point
M(C5H5N)*V(C5H5N) =M(HBr)*V(HBr)
0.4168 M *220.0 mL = 0.5811M *V(HBr)
V(HBr) = 157.7973 mL
Given:
M(HBr) = 0.5811 M
V(HBr) = 157.7973 mL
M(C5H5N) = 0.4168 M
V(C5H5N) = 220 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.5811 M * 157.7973 mL = 91.696 mmol

mol(C5H5N) = M(C5H5N) * V(C5H5N)
mol(C5H5N) = 0.4168 M * 220 mL = 91.696 mmol



We have:
mol(HBr) = 91.696 mmol
mol(C5H5N) = 91.696 mmol

91.696 mmol of both will react to form C5H5NH+ and H2O
C5H5NH+ here is strong acid
C5H5NH+ formed = 91.696 mmol
Volume of Solution = 157.7973 + 220 = 377.7973 mL
Ka of C5H5NH+ = Kw/Kb = 1.0E-14/1.698243652461746E-9 = 5.888*10^-6
concentration ofC5H5NH+,c = 91.696 mmol/377.7973 mL = 0.2427 M


C5H5NH+      + H2O ----->     C5H5N   +   H+
0.2427                    0         0
0.2427-x                  x         x


Ka = [H+][C5H5N]/[C5H5NH+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.888*10^-6)*0.2427) = 1.195*10^-3

since c is much greater than x, our assumption is correct
so, x = 1.195*10^-3 M



[H+] = x = 1.195*10^-3 M

use:
pH = -log [H+]
= -log (1.195*10^-3)
= 2.9225
Answer: 2.92