calculating the molarity of acetic acid in a vinegar sample given that 5.00 of vinegar required 43.50 mL of 0.1050 M NaOH to just reach the methyl red endpoint.
Is that 5.00 mL )0.00500L)vinegar?
mols NaOH = M x L = ?
Then M = mols/L soln
To calculate the molarity of acetic acid in the vinegar sample, we can use the concept of stoichiometry. Here are the steps to find the answer:
1. First, let's write the balanced equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
2. According to the balanced equation, the stoichiometric ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1. This means that for every mole of acetic acid, we need one mole of sodium hydroxide to react completely and vice versa.
3. Convert the volume of NaOH used to liters:
43.50 mL = 0.04350 L
4. Use the given concentration of NaOH and its volume to calculate the amount of NaOH (in moles) used:
Moles of NaOH = Concentration × Volume
= 0.1050 M × 0.04350 L
5. Now, since the stoichiometry between acetic acid and NaOH is 1:1, the moles of NaOH used will also be equal to the moles of acetic acid in the vinegar sample.
6. Convert the mass of vinegar used (5.00 g) to moles of acetic acid:
Moles of acetic acid = Mass ÷ Molar mass
= 5.00 g ÷ 60.052 g/mol (molar mass of acetic acid)
7. Finally, calculate the molarity of acetic acid in the vinegar sample using the formula:
Molarity = Moles ÷ Volume
By following these steps, you can determine the molarity of acetic acid in the vinegar sample.