calculate molarity of acetic acid in vinegar Volume of Vinegar used in each tria

Question

calculate molarity of acetic acid in vinegar Volume of Vinegar used in each trial = 5.00 mL Molarity of NaOH used in each trial = 0.200 M Volume of NaOH used in trail#1 (Final - Initial burnt reading) = 21.5 mL Volume of NaOH used in trail#2 (Final - Initial buret reading) = 21.6 mL Average Volume of NaOH used = (#3 + #4)/2 = 21.55 mL Calculate Molarity of Acetic acid in vinegar = 300.25 M (molarity of acetic acid times volume of acid) = (molarity of base times volume of base) Calculate the molecular weight of Acetic acid using a Periodic table = 60.05 g/mol Grams of acetic acid in one liter of Vinegar = Molarity times Molecular weight = step-6 times step-7 =_________ g Grams of acetic acid in 100 mL of Vinegar = step-8 divide 10 = ___________ g Calculate % of acetic acid in vinegar: (Assuming density of vinegar is 1.00 g/ml so 100 mL of Vinegar = 100 g Vinegar) % of acetic acid in vinegar = (step-9 g/100 g) times 100 = __________ % (numerically #9 and #10 are same}

Explanation / Answer

NaOH + Acetic Aci d= H2O + NaAc

so...

mmol of base used = MV = 0.2*(21.5) = 4.3 mmol of base

so

4.3 mmol of acid is present

since we used V = 5 mL

then

[HAcid] = mmol/mL = 4.3/5 = 0.86M

molarity of acid = 0.86 M

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.