The acidic substance in vinegar is acetic acid, HC2H3O2 When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH, 25.56 mL of the NaOH solution were required to reach the equivalence point. What is the mass percent of HC2H3O2in the vinegar?

7.68%

Explanation:

Let's consider the neutralization reaction between acetic acid and sodium hydroxide.

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

25.56 mL of 0.200 M NaOH reacted. The reacting moles of NaOH are:

0.02556 L × 0.200 mol/L = 5.11 × 10⁻³ mol

The molar ratio of HC₂H₃O₂ to NaOH is 1:1. The reacting moles of HC₂H₃O₂ are 5.11 × 10⁻³ moles.

The molar mass of acetic acid is 60.05 g/mol. The mass corresponding to 5.11 × 10⁻³ moles is:

5.11 × 10⁻³ mol ×  60.05 g/mol = 0.307 g

0.307 g of acetic acid are in 4.00 g of vinegar. The mass percent of HC₂H₃O₂ is:

0.307 g/4.00 g × 100% = 7.68%

it's c. elements are placed in rows based on how may valencies electrons have.

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