millimoles of pyridine = 25 x 0.22 = 5.5 kb= 1.7x10^-9 pKb = -logKb = -log (1.7x
ID: 699859 • Letter: M
Question
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
Explanation / Answer
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
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