Question

Acid-Base Titration % A. Concentration of Acetic Acid in Vinegar nei ica's 1. Brand clice Volume 5.0 mL (% on label) 2. Molarity (M) of NaOH Oil M M Trial 1 Trial 2 Trial 3 3. Initial NaOH level in buret 1.8 500 4. Final NaOH level in buret 148.3 5. Volume (mL) of NaOH used 6. Average volume (mL) 7. Average volume in liters (L) mole NaOH 8. Moles of NaOH used in titration (Show calculations.) 9. Moles of HC,H,O, neutralized by NaOH. 10. Molarity of HC,H,O, (Show calculations.) mole HC,H,02 M HC2H,02 g HC H202 11. Grams of HC,H,O, (Show calculations.) % HC2H3O2 12. Percent (m/v) HC,H,O, in vinegar (Show calculations.)

Answer 1

Please find the attached sheet.

2011 46.45 mL Volume of NaOH used - 51.2 47.4 average mL of Maon - 48.35 mL ein Ltr 0.04835 Ltr 8- moles of Naon = Mx Vol(L) = 0.1 X 0.04835 0.004835 moles 9. As this is his reaction; thus reaction; thus moles of HGH₂O Meutralized = 0.004835 moles 10. Molarity of Acetic Acid - m no of moles 0.004835 = 0.967 Volume in L 0.005 of acid no locity = 0.967 M 11. Grams of acid = moles x m.wt. 0.004835 X 60.052 = 0.29035 g m of acid in uinegor = 12 % 0,29035 Xloo = 5.807 g./ mL l.