A 10.0 ml sample of vinegar which is an aqueous solution of acetic acid, requires 16.5ml of a 0.500M NaOH solution to reach the endpoint in a titration (where the moles of acid are equal to the moles of base). What is the molarity of the acetic acid solution?
.303m
To determine the molarity of the acetic acid solution, we can use the concept of stoichiometry in a titration.
Here's how you can calculate the molarity of the acetic acid solution:
1. Write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
2. Determine the stoichiometry of the equation. From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
3. Calculate the number of moles of NaOH used in the titration:
Moles of NaOH = Molarity of NaOH × Volume of NaOH used (in liters)
Moles of NaOH = 0.500 M × 0.0165 L = 0.00825 moles
4. Since the reaction between acetic acid and NaOH has a 1:1 ratio, the number of moles of acetic acid is also 0.00825 moles.
5. Calculate the molarity of the acetic acid solution:
Molarity of acetic acid = Moles of acetic acid / Volume of acetic acid (in liters)
Volume of acetic acid = 10.0 mL = 0.0100 L
Molarity of acetic acid = 0.00825 moles / 0.0100 L = 0.825 M
Therefore, the molarity of the acetic acid solution is 0.825 M.