Final answer:
a. Approximately 0.513 grams of acetic acid are in 8.3 mL of vinegar. b. Approximately 0.00855 moles of acetic acid are in 8.3 mL of vinegar. c. Approximately 0.00855 moles of NaOH would react with the acetic acid in the vinegar sample. d. Approximately 34.2 mL of a 0.25 M NaOH solution would be required to react completely with the acetic acid in 8.3 mL of vinegar.
Step-by-step explanation:
a. To find the grams of acetic acid in 8.3 mL of vinegar, we can set up a proportion using the mass-to-volume percentage: (6.2 grams acetic acid / 100 mL vinegar) = (x grams acetic acid / 8.3 mL vinegar). Cross-multiplying and solving for x gives us approximately 0.513 grams of acetic acid in 8.3 mL of vinegar.
b. To find the moles of acetic acid in 8.3 mL of vinegar, we can use the molar mass of acetic acid (60.06 g/mol). First, convert the grams of acetic acid to moles: 0.513 grams acetic acid * (1 mol / 60.06 g) = 0.00855 moles acetic acid.
c. To find the moles of NaOH required to react completely with the acetic acid, we need to use the balanced equation: CH3COOH + NaOH → CH3COONa + H2O. The mole ratio between acetic acid and NaOH is 1:1, so there would be approximately 0.00855 moles of NaOH required.
d. To find the mL of a 0.25 M NaOH solution required to react completely with the acetic acid, we can use the mole ratio from part c and the molarity equation: (0.00855 moles NaOH) * (1 L / 0.25 moles) * (1000 mL / 1 L) = 34.2 mL NaOH.