Data Table: Known % acetic acid in vinegar: assume 5.0% Known Molarity of NaOH s

ID: 559504 • Letter: D

Question

Data Table: Known % acetic acid in vinegar: assume 5.0% Known Molarity of NaOH solution (read from container): AVERAGE NaOH volume(Vo) in mL Trial 3 2.00 ml. Trial 4 2.00 ml.? Trial I Volume of vinegar (must be 2.00 ml.7 Trial 2 2.00 mL. ? precisely 2.00 mL each time): The initial veleme for Tril 2a Initial volume reading of NaOH in the buret:V mL. Final volume of NaOH in the buret: ½ Subtract the initial volume reading from the final volume to determine the amount of NaOH delivered with each titration SHOW ALL WORK FOR EACH OF THE FOLLOWING CALCULATIONS 1. Determine the amount in moles of NaOH used to titrate (neutralize) the acid by using the known molarity (M) of the NaOH and the average volume of NaOH used. Remember to convert mL to L (1000mL-IL). Molarity (M)- moles of solute Moles of NaOH: 2. At the point of neutralization (equivalence pointhe amourat acid (noles)-the amount of base (moles Use the moles calculated above to determine how many moles of acetic acid (CH COOH) were present in the Erlenmeyer flask (2.00ml.) when the equivalence point was reached Moles of acetic acid (CHOOOH).- _at equivalence point. 3. Determine the number of grams of acetic acid (CH,COOH) by converting the mole amount to grams (use the molar mass as the conversion factor) Grams of acetic acid: V02

Explanation / Answer

The acetic acid content of a vinegar may be determined by titrating a vinegar sample with a solution of sodium hydroxide of known molar concentration (molarity).

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

(acid) + (base) — > (salt) + (water)

Trial 1

Trial 2

Trial 3

Trial 4

Volume of vinegar must be used 2 ml per time

2 ml

Initial Na0H reading in burette no-1 , V1 ml

Final Na0H reading in burette no-1

2.7

Amount of naoh needed for neutralize acetic acid

2.7

Average naoh volume in ml, V1 ml

2.7 ml

Average vinegar 5% in acetic acid

We know that,

3. M CH 3 COOH + VCH 3 COOH = M NaOH VNaOH

MCH3 COOH= M NaOH VNaOH/ VCH3COOH

0.5 x 2.7 /2=1.35/2 = 0.675 mol/L

Strength of acetic acid=0.675 x 60 =40.5 g/L
percent ammunt of vnegar in solution is

40.5 gm in 1000 ml

X gm in 100 ml

% of acetic acid in vinegar is 4.05 %. i.e 4.05 gm in 100 ml solvent

milimoles of naoh

40 gm Na0H in 1 lit =1M
20 gm Na0H in 1 lit=0.5 M

Molarity of Na0H=moles of solute/moles of solution

moles of solute =Molarity of Na0H x moles of solution

moles of solute=0.5 M/1000 ml=0.0005 Moles of Na0H