Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0576 M HNO2 to a 50.0 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 pH 0.00 mL HNO,? What is the pH of the solution after the addition of pH 7.41 mL HNO? What is the pH of the solution after the addition of a volume pH of HNO equal to half the equivalence point volume? What is the pH of the solution after the addition of pH 61.8 mL HNO? What is the pH of the solution after the addition of a volume pH of HNOg equal to the equivalence point?

Answer 1

Aziridine (7 & HNO3 reaction: + HNO F X + Nog 6) Aziridinium Aziridin ion (WA) Here, Angiridium is conjugate bare (GB) of Aziridinium ion which are an. weak acid (WA First of all we have to determine volume of Ano (ve) needed to reach equivalence point. At eauivalence poust all Azjri divm (CB) converted to Azirridinium ion (WA) According to stoichiometric reaction at earävalence point Initial number of & Number of moles of moes of Aziridium HNO3 added MARX VA . MAN & VANO sinamolarity Ivavolume To reach eamur dence ponnit vol of HNo needed. ve = VAND- - MARX VAB MANUB 0.0750 px soon 0.0576MY - 65.1 mm .: When 65. anon No Guill be added ther 100% reaction will occun.

# When 0.00 ml HNon added Then solution contain only Aziridine arhich in a weak bare! pka of azieri dinivm - 8.04. PKA pk, 19 - 8.04 + pkb = 14 pk6=14-8:04 = 596 pkip of apiridine. 5.96 Pota de 8 kb - aloge [c= concentration of weak bare = 5x5.26 - Ex log(0-075) = 0:07 Som) = 3.54 i pittpot14. & PH214-polt=14-354= 10.46. Pt 10.46 hen 7.41 med Horog added 6ilme Hno needed for 100% reaction. Ta a alu .: 741 100x74) in t # 3 65. 17. u = 11.38%. At this condition in reaction mixtine, [WA] = 11:38%. [CB-C -128). - 8862%. wow, from Henderson. Hansel balch eanation al get [CR] .... (ii) pHopkat log - [wal PH - 8:04 of log 1686%) 10 higgy.) - 8.93 & Phraga

It half equivalence point [CB) = 50%. & [Was= 50% using equation (ii) pt= 8:04 + log, (50%) =804 + 8 = 8.04. PH28:04] * When 61m HNon added, then reaction occured = 61'6 *100%. = 94.99% : (WA) = 94.97% [CB)= (180 - 9493%. = 5:07%. veing eanation (in) pH = 8:04 +log (24.937.) for - 6.77 (pH = 6.77 4 at earnivalence all converted to W.A guitial Moles of CB tolnolenot B r Vee- volume of I conjugate bare VCB & ve & sooor MCBX VB Ver equivalence Vestve Volume of HNd3 65.1 -0.0750x56 Tsot 65.12 00926 M im solution was present. c. it will release puton. Pha apka, Log (wa I was concentration of w,a at = 2 804 / log (0.00) emma = 476 PM - 976 T.