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Chem Titration help!!?
The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion: CH3COOH(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq). If 2.16 mL of vinegar needs 42.5 mL of 0.105 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?
Ok, can someone teach me how to set this us using stoichiometery?
Thanks
1 Answer
- 10 years agoFavorite Answer
Ooh, titrations... fun stuff
Start by looking at the equation to determine how many moles of your titrant (NaOH solution) you need to react with one mole of your unknown (Vinegar)... in this case it is 1:1, so for every mole of NaOH added, 1 mole of CH3COOH is completely reacted.
Next, calculate the total moles of titrant that you added
if you added 42.5 mL of titrant at .105 M then using stoichiometry
42.5 mL x (1 L / 1000 mL) x (.105 Mol NaOH / L) = .0044625 Moles NaOH used
So because 1 mole of NaOH is needed to react 1 mole of CH3COOH... .0044625 moles of NaOH ill be needed to react .0044625 moles of CH3COOH.
knowing this we can calculate the molarity of the CH3COOH solution
(.0044625 mol CH3COOH) x (1 / 2.16 ml) x (1000 ml / 1 L) = 9.68 mol CH3COOH / L
Next convert the liters to quarts
(9.86 mol CH3COOH / L) x (.946 L / qt) = 9.33 mol CH3COOH / qt
Then convert moles CH3COOH to grams by multiplying by its atomic mass (60 g/mol)
(9.33 mol CH3COOH / qt ) x (60 g / mol) = 559.8g
