Question

- Calculating the pH at equivalence of a titration A chemist titrates 180.0 ml of a 0.1215 M nitrous acid (HNO) solution with 0.5478 M KOH solution at 25 "C. Calculate the pH at equivalence. The pk of nitrous acid is 3.35. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. PHU X 5 ?

Answer 1

volume of HNO = 180m2 = 0-16 Lt Come at HNO2 = 0-1215 m. Cone of kom = 0.5476M - p19 = 3.35 & 4.46*10-4 .=Kq Reni HNO + koH KNO + H₂O vol af kon required for eg point MiVi =MV2 0.18 X0-1215 - 0.5498 X V2 V2 = 0.039 Ir = 39m) KNO2 + H₂0 .. HNO2 + kon Inetically moles (0-18X0.1215) CO-S4 Pb yo osa) 0.02187 0.02187 After Rxn o o mole 0.021 67 mole = 0.219 44 Total volyme = 0.18+ 0.039 mole. 0:02167 = 0.098m. VOL NOW 0-219 KNO₂ will react with the water and will get convested to and.

NO + H₂O 2 — Hng foto Core 0.096 Finally 0.096-2 Kar Kb = Kwon Kas 4,468104 Kb = 181044 = Kb = 4146x104 0.22 X10-10 - Now is= THN0275 06 07 = xid . 0098-) Pare, 7 22 0-22 X1070 - 0-096-22 2+0.2281070 x = 0.022x10-10 =0 from quadratic eghe 0-22 X1070 + ) (0 :22x1070)2- uplx (-0022x1070) 2X1 me = 1048X10-6 X = TOHOT = 1- 48x10-6 Se Ph = 14-POH = lhem-log(010). Elu + 109 (1.4881009 pH = 8.17