Asked by Derrick
The titration of 10ml sample of vinegar requires 30ml of 0.20M of NaOH solution. Find molarity and mass percent concentration of the acetic acid.
Answers
Answered by
DrBob222
NaOH + CH3COOH ==> H2O + CH3COONa
moles NaOH = M x L = 0.20 x 0.030 L = 0.006
You see 1 mol CH3COOH to 1 mol NaOH; therefore, 0.006 mols NaOH is due to 0.006 mols CH3COOH (vinegar--;i.e., acetic acid).
M acetic acid = moles/L = 0.006/0.010 = 0.60 M.
To do mass percent you need the mass of the 10 mL (you don't have that and no way to get it that I see). You also need the mass of the acetic acid in the sample. That is grams = mols x molar mass = ?
moles NaOH = M x L = 0.20 x 0.030 L = 0.006
You see 1 mol CH3COOH to 1 mol NaOH; therefore, 0.006 mols NaOH is due to 0.006 mols CH3COOH (vinegar--;i.e., acetic acid).
M acetic acid = moles/L = 0.006/0.010 = 0.60 M.
To do mass percent you need the mass of the 10 mL (you don't have that and no way to get it that I see). You also need the mass of the acetic acid in the sample. That is grams = mols x molar mass = ?
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