Map A Sapling Learning Ilculate the pH of the solution after the addition of the
ID: 508075 • Letter: M
Question
Map A Sapling Learning Ilculate the pH of the solution after the addition of the following amounts of 0.0647 M HNO3 to a 60.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04. d) 66.1 mL of HNO3. a) 0.00 mL of HNO3 Number Number pH 6.7571 pH 0.4575 b) 5.36 mL of HNO e) Volume of HNO3 equal to the equivalence point Number Number pH 9.118 pH 4.7496 c) Volume of HNO3 equal to half the equivalence point volume 74.4 mL of HNO3 Number Number pH 8.04 pH Incorrect Previous Try Again Next I Exit ExplanationExplanation / Answer
aziridine is a weak base with pKb = 14-8.04 = 5.96
[base] =0.075 M and volume of base = 60mL
[acid] =0.0647 M
a) before the titration
pH = 14 - pOH
= 14 - 1/2[pkb -log C]
= 14 - 1/2[5.96 - log 0.075]
= 10.4575
b) 5.36mL of 0.0647M HNO3
B + H+ --------------> BH+
60x0.075 =4.5 0 0 initial mmoles
- 5.36x0.0647=0.3467 - change
4.153 0 0.3467 fter mmoles
The solution is a buffer whose pH is given by Hendersen equation as
pH = 14 -{pkb +log [conjugate acid]/[base]}
= 14 -{5.96 + log 0.3467/4.153}
= 9.118
c) At half equivalenc epoint
pOH = pKb and pH = 14-pKb
pH = 14 -5.96
= 8.04
d) 66.1mL of HNO3
B + H+ --------------> BH+
60x0.075 =4.5 0 0 initial mmoles
- 66.1x0.0647= 4.277 - change
0.223 0 4.277 after mmoles
Thus the pH of this buffer
pH = 14 -{pkb + log [conjugate acdi]/[base]}
= 14 -{5.96 + log 4.277/0.223]
= 6.7571
e) At equivalence
60x0.075 = VmL x 0.0647
Thus volume of HNo3 at equivalence = 69.55mL
B + H+ --------------> BH+
60x0.075 =4.5 0 0 initial mmoles
- 69.55x0.0647=4.5 - change
0 0 4.5 after mmoles
thus the solution has a salt of weak base and strong acid
concentration of salt = mmoles / volme
= 4.5/129.55 = 0.03473
whose pH is givenby
pH = 1/2[pKw -pKb - logC]
= 1/2 [14 -5.96 -log 0.03473]
= 4.7496
e) Vof HNo3 =74.4 mL
excess acid = 74.4 -69.55= 4.85 mL
Thus [H+] in solution = 4.85x0.0647/134.4
= 0.002334 M
thus pH = -log 0.002334
= 2.63
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