Question

You are charged to perform a titration with 20 mL of 1.0 M hydrazine (KB = 1.7x10-6) as analyte and 0.5 M HBr as titrant. What is the initial pH of the solution? pH = At what volume would you reach the stoichiometric point? stoichiometric point = mL At what volume would pH = pka? pH = pka @ ml What is the pH at the stoichiometric point? pH @ stoichiometric point =

Answer 1

It is a weak base vs strong acid titration. pH initially is due to base only and it is found to be 11.11

Volume of stoichiometric point 40 ml

Volume at half. That is pH is equal to pka =20 ml

pH at equivalence point is 4.35

wear base vs sliting Acid a) Initially there is no titrant, hence pre only due do hydranine - 1.om rydigine 101) - Kbx concentration of tase = 11:4x10-6 X 1.0 con'] = x103 - 1.30 X 10% pon = -log (1.3041013) = 3loglu - logis pon = 2.89. Ph=14-poy = 14-9.89 = 11:11 TpH=11:11] Stoichiomelic point is equirolence pout. number of moles of agerazine = number of moles of HB oded moles of hydrazne = 80% = 0.02 moles - lovo moles of HBr = 0.02 moles volume of HBr = cronoles of HBV = 0.02 . Concentration of HBV 0.5 | Volume=40mL =0.04 liter =40mL pH= pia at half equivalence point number of moles of acid colded = number of more of base left) = 0.02 volume of HBr = 0.01 - PH=11.11 ml = 0ool more of = (20 mL) Ap 0-5

- Na Hy + HBr → N 2 HS + Bro pH at equivalence point Total volume = 20mL ttomL = 60mL=0.066 moles of Na Hst formed = 0.02 moles N₂ HS ut thone thon - No Hy + Hool (429) w 0.02 - 0 - (2012 (29) Initially 0.06 44 tu Change- equilibrium 0.02 -1 0.06 Kb 0.333-2 10-14 = x2 ka = [ Na Hu] [1420 - [Ninst Approximation Kw = a** (0.338-2 4 0.333 because a is very small in 1.7x10-6. 0.323 comparison to 5.88x10 9 = x2 0.333 0.333 42 = 1958X10-9 a = 4.42X1025 ( 13 = P2] = 4.42X105 pH = -108 (4.42x1015) = 5 - 10g 4.42 TpH = 4.35) Ap