millimoles of pyridine 25 x 022 55 kb 17x109 pKb logKb l

millimoles of pyridine = 25 x 0.22 = 5.5

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 -log0.22] = 4.71

pH + pOH = 14

pH = 9.29

b) after the addition of 12.5 mL HBr

millimoles of HBr = 12.5 x0.22 = 2.75

it is half equivalence point . so

pOH = pKb

pOH = 8.77

pH +pOH =14

pH = 5.23

c) after the addition of 24 mL HBr

millimoles of acid = 24 x 0.22 = 5.28

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

5.5           5.28                                0

0.22            0                                   5.28

pOH = pKb + log (5.28 /0.22)

pOH = 10.15

pH = 3.85

d) after the addition of 25 mL HBr

millimoles of HBr = 25 x0.22 = 5.5

it is equivalence point only salt is formed

salt millimoles = 5.5

salt concentration = millimoles / total volume = 5.5 / (25+25) = 0.11 M

salt is from strong acid weak base so pH <7

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

pH = 7 - 1/2 [8.77 + log 0.11]

pH = 3.09

e) after the addition of 36 mL HBr

millimoles of HBr = 36 x 0.22 = 7.92

[H+] = 2.42 / (25+36) = 0.0397 M

pH = -log(0.0397)

pH = 1.40

Solution

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.220 M pyridine, CsHsN(aq) with 0.220 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBr Number (c) after addition of 24.0 mL of HBr Number (d) after addition of 25.0 mL of HBr Number (e) after addition of 36.0 mL of HBr L