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Could you help me to discuss this data and what conclusion can we make? Thank you

GLYCINE TITRATION CURVE Correio point of pH = 25 (30-16) - 14 mL pila (17.6) = 1l ml p#30 (1-3) - 8 ml PH+3,3 (6-1) = 5 mL Co
Determining pkai (HCl titration) At pH = 2.5, 30 mL of 0.1 N HCI was used for the glycine titration and 16 mL was used for wa
Formula: pH = pka1 + log (conjugate base]/ [acid] Conjugate base = 0.6 meq Acid = 1.4 meq 2.5 = pka1 + log [0.6]/[1.4] 2.5 =
Determining pka2 (NaOH titration) At pH = 9.4, 18 mL of 0.1 N NaOH was used for the glycine titration and 1 mL was used for t
Formula: pH = pka2 + log (conjugate base]/[acid] Conjugate base = 1.7 meq Acid = 0.3 meq 9.4 = pka2 + log (1.7]/[0.3] 9.4 = p
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Contlusien is ytine 11 Alou Yhe glycene s an amin aud and heuce f exts exicls af Zurtor ion. As belod HN (Zwitr Pon)NOw tn Yhe drs t case, yeu tikaled the HL and af.5 to 3.3. LOCHeases y uith Myerafane Ait Lout pH yhe aly cine faum Catton abelween 7he COOO 1s0tkeetie point fy T

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