Question

stion 3 of 4 Calculate the pH of the solution after the addition of cach of the given amounts of 0.0683 M HNO to a 80.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL. HNO,? pH What is the pH of the solution after the addition of pH 5.90 mL. HNO,? What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH What is the pH of the solution after the addition of 83.5 mL. HNO,? pH What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? pH What is the pH of the solution after the addition of 91.23 mL HN0,? pH ASUS

Answer 1

-6 Kb1x 10 PD oazinidine 5.96 After additn o00 m HNO3 C2HSN aa) thola (0.075-07 - 6 11x10 ル-Sto.o 2.87 x 1O 3.54 :pH t the ution 14-3 SU = 16. 46 ALter addion of 5.9 .0683 M HNO3 Moleo Asividiniem ion 5.9 x0.0683x10 3 ニ403 x me、 ExaenへAdi ve 403x104 Gx10 4.03 x10 5,ち99 ×163mle Azividne Puet Airidiniem en 8 O44 5 S97103 81.b ALLer addion o HNOS n to l eguis loer pat PH PLA 8.04 ARTer addithon of 83 5 m HNO3 M.e Agividinium on 83 5xO0683 16 mek 5.7x103 mi. Excem Azividive O.3A10mel -3 O.3x10 8.04 17 Coohution 5-子メ163 6.76 Scanned with CamScanner

HNOR emal to eguisalence paint LANiniuiu sul 80 x0.07 7 g0+87 85 036 M 7 Pub 7 5 96 l 0 036 2 4 74 Ater addiH of 91.23 m HND3 Aler auivalore bant 91.23 X6 0683 g0X 6075 nN HMO 17 231 m M HO3 O.231 1 1 35 x 10 M 171.23 -Log 1.3s x 10 Pohe lution 2.87 CS