Question

9. A 4-gram sample of vinegar containing acetic acid CH3C 0.2 M NaOH. After 15 mL of NaOH was added, the stoichiometric p reached as evidenced by the change in color of the indicator. Calculate percent of acetic acid in the vinegar. Molar mass of acetic acid=60.05 ining acetic acid CH3COOH was titrated with s added, the stoichiometric point was or of the indicator. Calculate the mass

Answer 1

g. Acetic acid cetic acid (CH₂COOH) was tétrated with 0.2M Naot solution. 0,2 M NaOH solution = 0.2 moles Naot is 1000 ml solution JODO 8. 15 mL of NaOH solution will have 0.2 x 15 moles of oHo cons = 3 x 103 moles of oth cons. Principle of Acid-Base neutralization reaction us, eque molar oHo ions will neutralize equimolar amount of to cons. ". Ho cous concentration - Concentration of Acetic acid 8. 40 cous present = 3x10 s moles: i. Aletic acid present in the Vinigar sample=3x10 moke & Molecular weight of acetic acid = 60.05 g/mol. 8. Acetic acid present= ( 8 x 10 mole x 60.05 g/mole) = 0.180 g This 4 gram vinigar sample contains 0.180 gram acetic mase percent of acetec acid present famowita famount of acetic acid " Tamount of vinigar - (x 100 % - 4.5% (Ams) - acid. e x 100);