-te. *meyer ffa. 15 mL of dr. 1. the Instri UNKNOWN # B. Titration of Acetic Aci

Question

-te. *meyer ffa. 15 mL of dr. 1. the Instri UNKNOWN # B. Titration of Acetic Acid in Vinegar Average molarity of NaOH (see Procedure A) 0.963M 10- O CO mL ml volume of vinegar 2 .6 12.90 mL mL final buret reading 0.80 ? ? ? ? initial buret reading ml 0.00 ml ml 31, 60 ml - 230 CU ml volume of NaOHG Show the calculation for the molarity of acetic acid for trial 1 (see Example Exercise 20.2). CYCLE memical Vaste elwal Na re , Imd #lets 2 = 0.00 620 7000 HL Storien 7mm/ GH ažic acid 0.00 62.) mol Aral eociel (090 ML 10.0 mL ile on * TL * = 0,62 m m / acet cacia, 32.60 Al Salian x0,219 l Max m / 4C2302 / 32.00 alSallia7 x 0.210 l af x/a/42, 0, / 0.0020 al. ACEH,302, 1000 mL 1980 - Salidz 10.0 mL solon mel Na OA L - 0 32,00 ML lohien x 0.345 ml NaOH ma H CH 3 7 / 0 .OnO 4 nel HCAH 50 a x 7800 mL 1000 ml Solution 1 ml NaOH / 10.0 mL so/ * - = 0.0 ||oy mol ocolicocid 0.621 m 0,70LM - 10_M Molarity of HC-H302 ) 0.807 M Average molarity of HC,H302 Show the calculation for the percent concentration of acetic acid for trial 1. (Assume the density of the vinegar solution is 1.01 g/mL.) Mass/mass percent HC2H302 * Average mass/mass percent HC2H502

Explanation / Answer

Acetic acid in vinegar

Trial 1,

moles NaOH used = 0.253 M x 32 ml = 8.096 mmol

moles acetic acid reacted = 8.096 mmol

molarity acetic acid = 8.096 mmol/10 ml = 0.8096 M

mass acetic acid in vinegar = 8.096 x 10^-3 moles x 60.05 g/mol = 0.486 g

mass vinegar solution = 10 ml x 1.01 g/mol = 10.1 g

mass% acetic acid in vineagr = 0.486 x 100/10.1 = 4.812%

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