Question

Please show all work.

Be sure to answer all parts. Find the pH of the two equivalence points and the volume (mL) of 0.0496 M KOH needed to reach them in the titration of 17.3 mL of 0.130 M H2CO3. Second equivalent point First equivalent point mL KOH mL KOH pH= pH =

Answer 1

Answen Accordng tothe given problem, to delemine u pt al fsi,Second Equivalence point Gitven Rvation o 3m o.cuqbH koH voume m) 6 calculated suquisud volume; H3 XD130M o. ouabM 2.2u mk O Oua6 KDH us. 343mL KoH Vi ax 3xo.130M C,oua6M KoH Va 90.65mi KoH Fiostuivalence Point O130M x 3m (3mL415. 3u3 ml)

P०q९:ot ३. २u१ M 62.6u3 4Co5) : ০০35৭1H balaneed uafion is, co + +la ০ -> +2C03 + oH taen, Kw Ka ০০3S৭) Hors kw: (০ -l५ ५5x6 ০০3S१। ४৮ ४ ০ ০85१। ५sx। : ३-२२2४ট x 0'০35৭। । (१५००3नxu6 = ।०१3४t6 . HCog] : 7 : 1+०१३४।63

pH t4-10g1-043xuo3)) page:03 ua.61 pH . vst equivalence pofnt, p 039 second Equivalence potnt:- O.3XH 3 ml 3mlt 90. 68 5m Ccog O02082H Kwka 2 Ckau Xioxo.02082 X - ३.१०५ xu63

page:ou PH (-log (2.104x163)) u- 2 6 pt 33 Setond Equivalenca point , pH- l1-33|