Question

3. The dibasic compound B (pKbi = 4.00, pK2 = 8.00) was titrated with 1.00 M HCl. The initial solution ofB was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: V,- 0, 1, and 10 mL. (40 points)

Answer 1

phbl a 4.00 KB, 2 104-0010 21-0x104 ² Button I с B+HO o-toom -х x x 0-100 - neglect x a Kac 103. (I do do calculations calculations wthat neglecting & also K2 BitCoir) a x2 C3) o-loo Cost = x u 2 J Loxoxo-Toom 0 -0032 m 100x104 2 x² o-loo-> 2) Xez o x + 10x10x - 0.00ool zo using avadranc eavation solver a get & 2 0.0031m. Con 2) pon z tog Co-oosim) E 2-51 14.00 - 2.51 2 2) pl a 11.49) [on from second rondsation negingible. varl mL & I moles Bonitas Otrom y me 2 mxulos E temal a 0 1000 x 100 ml z 10 mmoles and cudded - 1-oom & ImL 2 Immoie mole

B & H 2 But اسم 10 mmole innale ن لما ammore o immole Now solution conjugate And has But weak so base solution B' and is buffer pona pk bl + log (Acid) C Base) por 24.00+ log [ Immore] Cammore) 23.046 23.05. 1 2 PH2 14-00 -3.05 2. LOMLE+ moles it added a 1-oom x loml 210mmoles Here moles B2 moles sit added – so stequivalence so all o es converted to But But as amphoteric Cacts as and a base) so spill & plaz epean picar tekbi 14.00 ) plaz, 10.00 Ka P pk by a 14-00 1 14.00 -8.00 26.00 2 pH a 10.00 +6.00 E 18-00)