Chemistry

Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of acetic acid in the vinegar if no other acid was present?

(A) 1.60 M
(B) 0.800 M
(C) 0.600 M
(D) 0.450 M
(E) 0.200 M

I know the answer is B, but how do you get this? These types of problems completely lose me. Is a rice table necessary?

Vinegar is acetic acid, which I will write as CH3COOH. It's the right end H that reacts with the NaOH.
CH3COOH + NaOH ==> CH3COONa + HOH.

SO, mols NaOH = L x M = 0.0267 L x 0.600 M = ?? mols NaOH.
The whole idea of titration is that the mols of the titrant will tell us the amount of the unknown material in the sample.
So we have ?? mols NaOH.
You can see we replaced the 1 H on CH3COOH with OH of the NaOH, so we have 1 mol CH3COOH per 1 mol NaOH. That means we must ALSO have ?? mols CH3COOH.
Since mols = M x L, you know mols CH3COOH, you know L CH3COOH from the problem, solve for M, the molarity.
It can't get much simpler than that; you may be making it too hard. # mols = M x L is all you need to know.

Okay. I got it now. Thank you so much! :)

  1. 👍
  2. 👎
  3. 👁
  1. 0.800 M

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. chemistry

    10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M and 16.58 ml is required to reach equivalence point what is the molarity of the acetic acid b. if the density of vinegar is 1.006 g/cm3

  2. chemistry

    1. When performing this experiment, a student mistakenly used impure KHP to standardize the NaOH solution. If the impurity is neither acidic nor basic, will the percent by mass of acetic acid in the vinegar solution determined by

  3. Chemistry

    Calculate the molar concentration of acetic acid (CH3COOH) in a 5.00-mL sample of vinegar (density is 1.00 g/mL) if it is titrated with 25.00 mL of NaOH.

  4. chem.

    A 10.0 mL of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5062M NaOH, and 16.58mL is required to reach the equivalence point. A. What is the molarity of the acetic acid? B. If the density of the vinegar

  1. chemistry

    Lab: Determining Ka of Acetic Acid Purpose: The purpose of this experiment is to determine the molar concentration of a sample of acetic acid and to calculate its Ka. Materials: phenolphthalein 125 mL Erlenmeyer flask 25 mL pipet

  2. Chemistry

    Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 4.69 g sample of vinegar was neutralized by 32.97 mL of 0.100 M NaOH. What is the percent by weight of acetic acid in the vinegar?

  3. Chemistry

    A 5.00 mL aliquot of vinegar (acetic acid, FW = 63 g/mol) was diluted and titrated with 0.1104 M NaOH requiring 32.88 mL. If the vinegar has a density of 1.055 g/mL, calculate its acidity as % acetic acid

  4. chem-acid-base titrations

    Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00mL sample of vinegar is titrated with .5052 M NaOH, and 16.88 mL are required to neutralize the acetic acid that is present. a.write a balanced equation for this

  1. Chemistry

    A solution buffered at a pH of 5.00 is needed in an experiment. Can we use acetic acid and sodium acetate to make it? If so, how many moles of NaC2H3O2 must be added to a 1.0L of a solution that contains 1.0 mol HC2H3O2 to prepare

  2. Chemistry

    If you require 30.01 mL of 0.1798 M NaOH solution to titrate 10.0 mL of HC2H3O2 solution, what is the molar concentration of acetic acid in the vinegar?

  3. Chemistry

    A certain vinegar is 6.02% acetic acid (HC2H3O2) by mass. How many grams of HC2H3O2 are contained in a 355-mL bottle of vinegar? Assume a density of 1.01g/mL.

  4. chemistry

    what is the theoretical value for the M of acetic acid in vinegar if the vinegar is 5% acetic acid by a % mass value. Additional info: the experimental M of the acetic acid in vinegar is 0.80M Thank you!!

You can view more similar questions or ask a new question.