A student titrated a 25.00 mL sample of vinegar containing acetic acid, HC2H3O2,

Question

A student titrated a 25.00 mL sample of vinegar containing acetic acid, HC2H3O2, with a 0.9980M NaOH solution. It required 22.78 mL of the base to reach the equivalence point. Based on this information determine the following. What is the molarity of the OH ? How many moles of OH- are used to reach the equivalence point? How many moles of H+ are present at the equivalence point? What is the molarity of the H+ (MH+)? What is the molarity of the acetic acid (Macid)? What is the % by mass of acetic acid in vinegar?

Explanation / Answer

1) Molarity of OH- = 0.998 M 2) Moles of OH- used = 0.998 x 0.02278 = 0.02273 moles 3) Molarity of H+ at equivalence point = 10^-7 M ..........Volume = 25 + 22.78 = 47.78mL Therefore moles = 10^-7 X 0.04778 = 4.778 X 10^-9 moles 4) 25 X M = 22.78 X 0.998 ==> M = 0.91 M = Molarity of Acetic Acid 5) Let density of Vinegar = 1 g/mL => Mass = 25 g Moles of Acetic Acid = 0.91 x 0.025 = 0.02273 moles => Mass = 0.02273 x 60 = 1.364 g % by mass = 1.36 X100 / 25 = 5.456 %

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