& 3/17/2017 11:00 PM A 84.5/100 3/15/2017 10:38 PM Gradebook Print Calculator Pe
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& 3/17/2017 11:00 PM A 84.5/100 3/15/2017 10:38 PM Gradebook Print Calculator Periodic Table Question 10 of 16 Map University Science Books General Chemistry 4th Edition presented by Sapling Leaming calculate the pH for each of the following cases in the titration of 25.0mLor 0.220 M pyridine, c5H5N(aq) with 0.220 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBr O Number (c) after addition of 16.0 ml of HBr L Number (d) after addition of 25 o mL of HBr Number (e) after addition of 30.0 mL of HBr You ans atteExplanation / Answer
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
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 0.22 = 3.52
C6H5N + HBr ----------------------> C6H5NH+Br-
5.5 3.52 0
1.98 0 3.52
pOH = pKb + log (3.52 /1.98)
pOH = 9.02
pH = 4.98
d) after the addition of 25 mL HBr
it is equivalence point only salt is formed
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 30 mL HBr
strong acid remained in the solution
[H+] = 30 x 0.22 - 5.5 / (25+30) = 0.02 M
pH = -log(0.02)
pH = 1.70
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