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a solution of household vinegar is to be analyzed. A pipet is used to measure out 10 ml of the vinegar which is placed in a 250ml volumetric flask. Distilled water is added until the total volume of solution is 250 ml portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide. The neutralization reaction is as follows: HC2H3O2 (aq) + OH- (aq) --> C2H3O2- (aq) + H2O (l)It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar. Calculate the molarity of the diluted vinegar.

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Answer:

0.0344 M

Explanation:

  • HC₂H₃O₂ (aq) + OH⁻ (aq) --> C₂H₃O₂⁻ (aq) + H₂O (l)

Because one mol of vinegar (acetic acid) reacts with one mol of NaOH, we can use the formula

  • C₁V₁=C₂V₂

Where C₁ and V₁ refer to the concentration and volume of vinegar, and C₂ and V₂ to those of NaOH. We're given V₁, C₂ and V₂; so we solve for C₁:

  • C₁ = C₂V₂ / V₁
  • C₁ = 0.0500 M * 16.7 mL / 25.0 mL
  • C₁ = 0.0344 M

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