Question

n dite by dispensing NaOH in in aliquots of -1ml ally and 1-2 drops once you observe a vanishing observe a vanishing pink color very Make sure to take burett to take burette readings correctly to two places of decimal (XX. ). Very important to learn it will be asked in final exam Accuracy is important Compare both titrations - if the difference in molarity of HCl of the two trials is greater than 5%, perform a third trial for the same. auton. Do not insert pipet into NaOH or HCI bottle: Always POUR OUTI Calculations: Volume of NaOH dispensed (final-initial) mL (XX. ): NaOH[M] : Given on the bottle. Balanced equation for neutralization with NaOH: HCl(aq) + NaOH(aq) - NaCl(aq) + H2O(0) Stoichiometry dictates 1: 1 mol ratio for HCL: NaOH at the end of titration: mol (HCI) = mol NaOH (mol = molarity X volume) Molarity of Base x volume of base = Molarity of HCl x volume of HCI

Answer 1

mdlaity dNaoHin-0.1486 M voume NaoH M voluume Hel 950m mdatyHel fal molaaty of Hel can be calaulatd by Vm mdlaity Hdma)= VimI31Ml XO1486M 327ML C.19LB1M Va 250mL Teial 2 vdume NaOH V20-8 m remaining values ahe m a same molarity Hcl Ma) Vimi- 30-8 maLX01486 M0 18301m 25-0 MAL Teial 3 volume a NaoH Remaining valus m, V re tee Bam molaaity d HelM)Vims 323na X0-1486M 32 8 mL = 0-19496 M 25 ml During dilution Number d molus 4 Compound dlocs ndt change Titeant is a solution coh Couentration is Kmouwu Avalgu Gampond ohse Covantiatio to bu Calaulatid so NaoH is tietsant and Hdis Analyle sdlution deceoses and molarty Indicator is added Analyt Hore inditater is addid to Hd NacHo ) +Hcl Nao H Caq) Caa) Ht t N+0t N + +Hao Ht Cap) Nat Tonic equaton H20