Hi sorry this seems like a lot. This data is from my solution standardization- t

Question

Hi sorry this seems like a lot. This data is from my solution standardization- titration vinegar lab. Based off this information I need to calculate the

– Moles of oxalic acid in 50-mL Erlenmeyer Flask 0.2484 *0.01L = 0.002484 mols

– Moles of sodium hydroxide (Oxalic Acid Titration) 0.4855 * 0.0035L = 0.00619925*2 (2:1 ratio) = 0.003395 mols oh NaOH

? HINT: Use the stoichiometric ratio!

– Molarity of sodium hydroxide

Trail one: 0.033985/0.011L = 3.090

Trail two: 0.033985/ 0.0106L = 3.206

Trial three: 0.033985/ 0.0085L = 3.998

– Average Molarity of sodium hydroxide 3.090+ 3.206 + 3.998 = 10.294/3 = 3.431

– Moles of sodium hydroxide (Vinegar Titration) 0.4855 * 0.0352L = 0.01699 moles

? HINT: Molarity = Moles per Liter – Moles of acetic acid in vinegar

? HINT: Use the stoichiometric ratio!

I'm STUCK moving foward because I cant determine the molarity of acetic acid (MW = 60.05g)

– Molarity of acetic acid in vinegar

– Average Molarity of acetic acid in vinegar

– g acetic acid/L vinegar solution

– G acetic acid/g vinegar solution

– Wt/wt%

– % Error

VI Datu Law and processed Date de Standortizaten + NaOH perhin if he titva of vines vinegar experimentº* fer?n CH He titvah ) Table 1: 70.2 484 LITrial e Tiad pwo Tnal mvu Volume of Oxalic Acid (mt) 1o jo 10 * Mclos F cxolic Acid (more sle-00243 lo. COO3 + 0.WITHH Initial volume of Noot (mi) 22:9 | 33. 9 21.5 Final Voleme | Nno H (ml) 33. 9 44.5 30 Delivered vas me of Naot // 10. b 8.5 * Moles C+ NaOH liruhus) 66S3to.DOSIS 0.00YTE Hz ** Malarity of NaOH CmJlever U. voto. Se r Nevase molenty NaOH(a)e-UEloo -HSV UFF Table 2: Law and processed clotn fin ta dettuminenhum 91 vineyar ek ee 1 r Por hun 4 NN oprahm V Trial me Thul two IT vid truser Volume 4 Acehc Herd Ciml) 135 135 3S » Modes of AcehC Ariel Cumcu J) Iniha Volume 04 NAOH 24.5 Final Volume * NaOH HO w Y Arepared and proper atom of "C (IUPAC). Sources: IUPAC periodic table 8 January 2016; Pure and Appea umuy 90. r. 22 2 /1 ed Vol ? ? ? // 15.S Avevage malenih NaOH (M) * MO lung d penic Acid (m) hve verse Moleum by 4 Meh| MIA CM)

Explanation / Answer

Molarity of acetic acid = moles of NaOH/volume of acetic acid

Trial 1 : molarity acetic acid = 3.431 M x 0.016 L/0.035 L = 1.568 M

Trial 2 : molarity acetic acid = 3.431 M x 0.0155 L/0.035 L = 1.519 M

Trial 1 : molarity acetic acid = 3.431 M x 0.015 L/0.035 L = 1.470 M

Average molarity of acetic acid = 1.520 M

grams acetic acid per L = 1.520 M x 0.035 L x 60.05 g/mol/0.035 L = 91.276 g/L

density of vinegar is not mentioned above so wt of vinegar solution cannot be determined for w/w% calculation

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