University of l CHEM 1415 Spl Oklahoma Overall Score Gr 4/7/2017 11:55 PM 4.315
ID: 508034 • Letter: U
Question
University of l CHEM 1415 Spl Oklahoma Overall Score Gr 4/7/2017 11:55 PM 4.315 4/6/2017 og 28 PM Print Ca Periodic Table estion 3 of 14 Sapling Learning Calculate the pH of the solution after the addition of the fo amounts of o.o620 M HNO3 to a 60.0 mL solution of 0.0750 M azinidine. The pka of aziridinium is 8.04 d) 67.6 mL of HNO3 a) 0.00 mL of HNO3 pH pH 10.46 e) Volume of HNO3 equal to b) 5.89 ml of HNO the equivalence point pH- pH c) Volume of HNOs equal to half 77.3 mL of HNO the equivalence point volume pH O Previous Check Answer Next AExitExplanation / Answer
pKa = 8.04
pKb = 14 - 8.04 = 5.96
millimoles of azridinium = 60 x 0.0750 = 4.5
b) after the addition of 5.89 mL HNO3
millimoles of HNO3 = 5.89 x0.0620 = 0.365
B + HNO3 -----------------> BH+
4.5 0.365 0
4.14 0 0.365
pOH = pKb + log [salt / base]
= 5.96 + log [0.365 / 4.14]
= 4.90
pH = 9.10
c) At equivalence point :
pOH = pKb
pOH = 5.96
pH = 8.04
d) after the addition of 67.6 mL HNO3
millimoles of acid = 67.6 x 0.0620 = 4.19
B + HNO3 ----------------------> BH+
4.5 4.19 0
0.3088 0 4.19
pOH = pKb + log 4.19 /0.3088)
pOH = 7.09
pH = 6.91
d) At equivalence point
millimoles of HNO3 = millimoles of base
0.0620 x V = 4.5
V = 72.58 mL
it is equivalence point only salt is formed
salt millimoles = 4.5
salt concentration = millimoles / total volume = 4.5 / (60 + 72.58) = 0.0339 M
salt is from strong acid weak base so pH <7
pH = 7 -1/2 [pKb + logC]
pH = 7 - 1/2 [5.96 + log 0.0339]
pH = 4.76
f)
millimoles of HNO3 = 77.3 x 0.0620 =4.7926
[H+] = 4.7926 - 4.5 / 60 + 77.3 = 2.13 x 10^-3
pH = 2.67
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