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.
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.
To calculate the concentration of the acetic acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can see that the ratio between moles of CH3COOH and moles of NaOH is 1:1. This means that if 1 mole of acetic acid reacts with 1 mole of sodium hydroxide, they are completely neutralized.
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = concentration of NaOH × volume of NaOH solution
= 1.02 M × 19.60 mL
= 20.03 mmol (since 1 mL = 1 mmol for 1.02 M NaOH)
Next, since the stoichiometry of the reaction is 1:1, the number of moles of NaOH used is also equal to the number of moles of acetic acid in the vinegar.
moles of acetic acid = moles of NaOH = 20.03 mmol
Now, we can calculate the concentration of the acetic acid solution:
concentration of acetic acid = moles of acetic acid / volume of acetic acid solution
Volume of acetic acid solution = 25.00 mL = 25.00 mmol (since 1 mL = 1 mmol for acetic acid)
concentration of acetic acid = 20.03 mmol / 25.00 mmol
= 0.8012 M
Therefore, the concentration of the acetic acid solution is approximately 0.8012 M.