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ID: 1071301 • Letter: M
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ming.com/ibiscms/modlibis/view.php?id 3016492 12/17/2016 10:55 PM 6/10 12/8/2016 11:49 PM Gradebook Print calculator Periodic Table Question 7 of 13 Sapung Learming Map A Calculate the pH of the solution after the addition of the following amounts of 0.0594 M HNO3 to a 50.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04. a) 0.00 mL of HNO3 d) 59.6 mL of HNO, Number pH b) 6.82 mL of HNO3 e) Volume of HNO3 equal to the equivalence point Number pH Li c) Volume of HNO3 equal to half the equivalence point volume 66.8 ml of HNO3 Number pH pH Previous 8 Give Up & View solution Check Answer Next Exit o to e o D F11 Ther O eTe O Help O Web O TechnExplanation / Answer
aziridinium is weakbase
pkb of aziridine = 14-pka = 14-8.04= 5.96
a) pH of pure aziridine = 14- 1/2(pkb-logC)
= 14 - 1/2(5.96-log0.075)
= 10.46
b) no of mole of HNO3 = M*v = 6.82*0.0594 = 0.405 mmole
no of mole of aziridine = M*V = 50*0.075 = 3.75 mmole
pH of buffer = 14 - (pkb+log(salt(or)acid/base)
= 14 - (5.96+log(0.405/(3.75-0.405)))
= 8.95
c) at half equivalence point
pH = 14 - pkb = 8.04
d) no of mole of HNO3 = M*v = 59.6*0.0594 = 3.54 mmole
no of mole of aziridine = M*V = 50*0.075 = 3.75 mmole
pH of buffer = 14 - (pkb+log(salt(or)acid/base)
= 14 - (5.96+log(3.54/(3.75-3.54)))
= 6.81
e) at equivalence point , no of mole of aziridine = HNO3
SO that,
no of mole of aziridine = M*V = 50*0.075 = 3.75 mmole
no of mole of HNO3 = 3.75 mmole
volume of HNO3 = n/M = 3.75/0.0594 = 63.13 ml
concentration of salt = 3.75/(63.13+50) = 0.03315 M
pH of salt = 7 - 1/2(pkb+logC)
= 7 - 1/2(5.96+log0.03315)
= 4.76
f) excess HNO3 added = 66.8 - 63.13 = 3.67 ml
concentration of excess HNO3 added = 3.67*0.0594/(50+66.8) = 0.0019 M
pH = -log[H+]
= -log0.0019
= 2.72
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