Asked by Allie
A sample of 25.00 mL of vinegar is titrated with a standard 1.02 M NaOH solution. It was found that a volume of 19.60 mL of the standard NaOH solution is used to completely neutralize the acetic acid in the solution. Calculate the concentration of the acetic acid solution.
Answers
Answered by
DrBob222
Acetic acid is CH3COOH. The H on the right end is the acid hydrogen. The H atoms on the left end are not acidic.
Equation:
CH3COOH + NaOH ==> CH3COONa + HOH
mols NaOH used = M x L
mols CH3COOH in the vinegar = mols NaOH (1:1 ratio from the equation)
M vinegar = mols/L. You know mols from above and you know volume. Calculate molarity. Post your work if you get stuck.
Equation:
CH3COOH + NaOH ==> CH3COONa + HOH
mols NaOH used = M x L
mols CH3COOH in the vinegar = mols NaOH (1:1 ratio from the equation)
M vinegar = mols/L. You know mols from above and you know volume. Calculate molarity. Post your work if you get stuck.
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