Question

Calculate pH for a strong acid/strong base titration. Determine the pH during the titration of 31.0 mL of 0.342 M HNO3 by 0.342 M NaOH at the following points: (a) Before the addition of any NaOH (b) After the addition of 15.5 mL of NaOH (c) At the equivalence point (d) After adding 40.0 mL of NaOH

Answer 1

she neutralisation Rotatisahe action of NGOH Halos lasst alaohay 10 mole - stolli omenic moles Nanos coast than to mule l'onde l'omole d Addition of 1. ömL of nooh 10.6mmoly Acortion of . 106 mmules is's mL of nigoh S3 mmoles Abted the reaction 53 mmeole o mmoler 5:30 moles S3 minutes c) 10.6 mmoles 106 mmoles Addition of 31.0mL ob 1GOH (equivalence point) Abted the deation o mmoley immules 10.6 106 to 6 mmuley 13.68 mmole - - d) Addition of your ob naon Abreu reaction ommer - lobi 106 3.08 mmoles @ Bebore addition ob NGOH, CH4]: 0:3424 pH. - log CH4] : 6.46 6 Abten the addition of 155mL of nault : Cht]= [14+- 31852 - 0.2284 pH, -log[ht] = -log[ 0.224) = 0.64 465

c) At the point; [41]= [oi] = 10tm ( from water in equivalence pH=7 40 m2 ob aholt, Afted the addition of COH-] = 3.08 = 0.04374 3) ports -log Con]: -1991 0.043) = 1:36 - PH - 14- 2015 (41 34 : -6