Question

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.140 M pyridine, C H N(aq) with 0.140 M HBr(aq). A. before addition of any HBr 9.198 B. after addition of 12.5 mL of HBr 5.25 m D. after addition of 25.0 mL of HBr 5.23 Ø C. after addition of 16.0 mL of HBr 5.00 E. after addition of 32.0 mL of HBr

Answer 1

Q. No ΜΑΙ Chs Nht + Be CS HAN THBS - 25X0.14 25x0.14 - 25Xo-14 25X0-14.m Hudsolysis ests ont +40 IS XO:14 n GHN thot a 25X0.14 a kw I 25x0.14 2 22 km - ko 95 0.19 25X0.14 ? > U = 0.63 X 10 lut 71 __"am - pH = -logo.63X163 pH = -log 0.65 - PI e 3.27 0-02 - (-2 xo-У 0:07 X10 7.7X10-9 n o od ko foronxious fo.4x600)

С- пnt - Дg — | (HN + ИЗ 20ҳр-1, б.x 32 2. г ч . ч& То ч.ч. — 3. до И Вg iѕ ѕts in soi, & ft - иде, is / come ч, ч% — 2 9 + 32 - НИВ) ---380-ot- +) - о, от 7 р. р" е „л(ht) 1 = — Клто-от- (р = |та —
above ans is that when the 25ml of HBr add and below is that when the 32ml of HBr add.