/18/2018 11:55 PM 0/5 Print Calculator Periodic Table tion 4 of 4 eneral Chemist
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/18/2018 11:55 PM 0/5 Print Calculator Periodic Table tion 4 of 4 eneral Chemistry 4th Edition University Science Books predorted by Sapling Leaning Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.240 M pyridine, CHsNaq) with 0.240 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBr Number (c) ater addition of 23.0 mL of HBr Number (d) after addition of 25.0 mL of HBr Number (e) after addition of 29.0 mL of HBr I O Previous e Check Answer Next Exit Hin about us careers privacy poscyterms of use contact usExplanation / Answer
millimoles of pyridine = 25 x 0.240 = 6
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 -log0.240] = 4.70
pH + pOH = 14
pH = 9.30
b) after the addition of 12.5 mL HBr
it is half equivalence point . 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 0.24 = 5.52
C6H5N + HBr ----------------------> C6H5NH+Br-
6 5.52 0
0.48 0 5.52
pH = pKa + log (0.48 /5.52)
pH = 4.17
d) after the addition of 25 mL HBr
millimoles of HBr = 25 x0.24 = 6
it is equivalence point only salt is formed
salt millimoles = 6
salt concentration = millimoles / total volume = 6 / (25+25) = 0.06 M
salt is from strong acid weak base so pH <7
pH = 7 -1/2 [pKb + logC]
pH = 7 - 1/2 [8.77 + log 0.06]
pH = 3.23
e) after the addition of 29 mL HBr
millimoles of HBr = 29 x 0.24 = 6.96
pH = 1.75
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