Question

Part 2: Standardization of HCL Solution: Normality of base (NaOH) (from label) N=0.1352 Indicator chosen: bromythymol blue Trial 1 Trial 2 0.85 mL Trial 3 (BONUS) 0.90ml Initial reading of acid (HCI) Final reading of acid Volume of acid used 0.75 mL 31.12mL 30.37mL - 28.95 mL 28.10 mL - 29.50ml - 28. 5L Initial reading of base(NaOH) 0.61 ML Final reading of base 26.85mL 25.24 ML -0.45 mL 23.65 mL 23.20mL 0.95 mL 24.35mL 23.40mL Volume of base used Normality of acid (HCI) Mean normality of acid; Rel. Ave. Dev.: Molecular equation for the neutralization reaction: Net ionic equation for the reaction: Calculation for obtaining the normality of the acid: Trial 1:

Answer 1

Normality - Gran Equivalent Mass Naunality - w Dung Niv, = N₂V2 Normality of NaOH = 0. 1352N (NV) NaOH = (NV) HCl) Terial Fient: NaOH = 25.24m2 VHU = ( 0:1 353 x 25 24 ) = 30:37 x NHA NHA = 3 +41 - 01 | 23N 30.37 Trial 2 VNOH=23.20m2 VHU = 28.10m2 (0.1352 x 23.20) = 28.108 NHL NHU= 3.16 = 0.1116N 28.10 Turial 3 NOWH = 23.45 me VHU = 28.50mL (0:1352X23:45) - (28.50xNH) NHU = 3 • 1701 = 0 || 12 N 28.50

Mean Norunality = 0.1123 + 0.1116 +0.1112 2 0.117 N • molecular Equation for the neutralisation et eaction NaOH the NaCl + H₂O • Net Tonic Reaction Ht (aq) + ce (aq) + Natt allaq Nat (aq) + chloq) +H2O common ions will cancel out and we will get Per Ht toq) + ut laq) 7H2O