In an experiment we used 10 ml of vinegar, water solution of acetic acid, CH3COO

Question

In an experiment we used 10 ml of vinegar, water solution of acetic acid, CH3COOH. The mass of vinegar was 9.292.We added 20 ml of distilled water in the flask.Next we poured NaOH solution into the beaker so that it was 1/3 full. Then we began titrating. Put two drops of of phenolthalein indicator solution in the flask. The initial base volume in ml is 24 and the final base volume in ml is 34. The difference between the initial and final volume readings is due to the volume of base solution used: 10 ml or 0.01 in lit.

Calculate Concentration of base (in moles / liter)=   

Moles of base = V x M=

   Moles of acetic acid = moles of base =

   Grams of acetic acid = moles x MW

% Acetic acid = grams of acid x 100 / grams of vinegar =

Explanation / Answer

10 mL of vinegar solution (acetic acid solution) = 9.292 g

Commercially available vinegar is 5% (w/w).

i.e., 10g of solution contains 0.5g of acetic acid.

Thus, 9.292 g of solution contains = 0.4646 g of acetic acid = present in 10 mL of solution

Thus the concentration of vinegar, M = (0.4646/60)x(1000/10)= 0.774 Molar.

Moles of base = 10mL of 0.774 M of acetic acid used = 7.74 milli moles = moles of acetic acid

% mass acetic acid = 5 %.

Conc of base = 7.74 millimolesx1000/24mL = 0.3225 Molar

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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