alculate the pH for each of the following cases in the titra

alculate the pH for each of the following cases in the titration of 25.0 mL of 0.190 M pyridine, CsHsN(aq) th 0.190 M HBr(aq) Number (a) before addition of any HBr Number (b) after addilion of 12.5 mL of HBr D Number (c) after addition of 16.0 mL of HBr L Number (d) after addition of 25.0 mL of HBr lip Number (e) after addition of 36.0 mL of HBrL O Previous Give Up & View Solution check Answer 0 Next Exit Hin about ua

Solution

millimoles of pyridine = 25 x 0.190 = 4.75

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 -log0.190] = 4.75

pH + pOH = 14

pH = 9.25

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 16 mL HBr

millimoles of acid = 16 x 0.190 = 3.40

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

4.75          3.40                              0

1.71        0                                  3.40

pOH = pKb + log (3.40 /1.71)

pOH = 9.07

pH = 4.93

d) after the addition of 25 mL HBr

it is equivalence point only salt is formed

salt concentration = millimoles / total volume = 4.75 / (25+25) = 0.095 M

salt is from strong acid weak base so pH <7

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

pH = 7 - 1/2 [8.77 + log 0.095]

pH = 3.13

e) after the addition of 36 mL HBr

strong acid remained in the solution

[H+] = 36 x 0.190- 4.75 / (25+36) = 0.0343 M

pH = -log(0.0343)

pH = 1.46