Question

ratra 1z 1 t B: Detes Name Date inoe Acid-Base Titration Data and Calculations Part A: Standardization of NaOH solution, Data: Trial 1 Trial 3 Trial 2 Initial level, HCI buret 26-60 mL Final level, HCI buret 33.85 mL 43.15 2u.95 mL mL Initial level, NaOH buret 00mL Final level, NaoH buret 23.45 mL 44.35 mL 24.25 mL Calculations: Trial 1 Trial 2 Trial 3 Total volume HCI solution 24.45 mL 33-86 mL 43.95 mL 0.1030 Molarity of standardized HCI 2345mL Total volume NaOH solution 34 25 mL M M Molarity of NaOH solution, Average Molarity of NaOH solution, NaOH. bt eled Part B: Determination of the Molar Mass of a Solid Acid Unknown Number 5113to80 Number of moles of ionizable H ions per mole of solid acid (darker) MNaoH (The average molarity of NaOH from Part A.) Data: ueginat 0.559 Trial 2 Trial 1 O. 212 g Mass of sample 0.23子。 Initial volume, NaOH solution 0.00mL 0.00 mL 23.45mL 37.25 mL Final volume. NaOH solution Calculations: Trial 2 Trial 1 33.95 mL 37.25 mL Volume of NaOH solution used mol Moles of NaOH (OH) used mol Moles of H* in sample mol mol Moles of solid acid in sample mol mol g/mol Molar mass of solid acid g/mol Average Molar Mass g/mol

Answer 1

wodalion 1ct $C) (Naon)-274m = 0, 1050M a4.46ml $ = 0 0917 M 33.85X0.1930 S 0.1018 M 34.25 3rd 3.95 x01 J030 6 1021M ty.35 Average molani hy f Naot so lution =009 8 M(NAOR) 0.09852M Volume ued = 31. 25 ml so utin Naot moles Naoft O.09852 moles &x37.25ml O09852molt37.15m 1600md 3.67X 103 moles. 33.95ml Vo lume of Naot 3.34SX10 moles holes st Na o H 4t in thesamp le b moles sf and 3.345K10 moles. 3.67 X103olep moles of solid acid in Hhe samp le f15X10moles 122 Xfo ules and Molar nash 0 557 0 5579 1Itsxsmel 22XDmul = 499.55Imol = 456.56/m.l 45656 t499 55 mass Avacage molore 2 = 478.0 63/mv1 Best wishes