Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.140 M pyridine, C5H5N(aq) with 0.140 M HBr(aq): (a) before addition of any HBr (b) after addition of 12.5 mL of HBr (c) after addition of 23.0 mL of HBr (d) after addition of 25.0 mL of HBr (e) after addition of 31.0 mL of HBr

Answer 1

millimoles of pyridine = 25 x 0.140 = 3.5

kb= 1.7x10^-9

pKb = -logKb = -log (1.7x10^-9) = 8.77

a) before the addition of any HBr

pOH = 1/2 [pKb -logC] = 1/2 [8.77 -log0.140] = 4.81

pH + pOH = 14

pH = 9.19

b) after the addition of 12.5 mL HBr

it is half equivalence point

here salt millimoles = base millimoles

so pOH = pKb

    pOH = 8.77

pH +pOH =14

pH = 5.23

c) after the addition of 23 mL HBr

millimoles of acid = 23 x 0.140 = 3.22

C6H5N + HBr ----------------------> C6H5NH+Br-

3.5          3.22                              0

0.28        0                                  3.22

pOH = pKb + log (3.22 /0.28)

pOH = 9.83

pH = 4.17

d) after the addition of 25 mL HBr

it is equivalence point only salt is formed

salt concentration = millimoles / total volume = 3.5 / (25+25) = 0.07 M

salt is from strong acid weak base so pH <7

pH = 7 -1/2 [pKb + logC]

pH = 7 - 1/2 [8.77 + log 0.07]

pH = 3.19

e) after the addition of 31 mL HBr

strong acid remained in the solution

pH = 1.82