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18 Dec 2019

A 28.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 24.5 mL of 0.450 M NaOH to reach the endpoint in a titration.

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

What is the molarity of the acetic acid solution?

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Reid Wolff
Reid WolffLv2
31 Dec 2019

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Related questions

A 25.0-mL sample of vinegar (which contains the weak acid acetic acid, CH 3 CO 2 H) requires 28.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point. What is the mass of acetic acid (molar mass = 60.05 g/mol), in grams, in the vinegar sample, and what is the concentration of acetic acid in the vinegar? CH 3 CO 2 H(aq) + NaOH(aq) → NaCH 3 CO 2 (aq) + H 2 O(l)

 
 

1. A 10.0 mL sample of vinegar is titrated with 0.100 M NaOH. If it requires 45.00 mL of titrant to reach the endpoint, what is the molarity of acetic acid in the vinegar?

2. (a) If a titration of a different 10.0 mL sample requires 0.00500 moles of base, what mass of acetic acid is in the solution?

2. (b) Assuming the solution has a density of 1.0 g/mL, what is the mass % of acetic acid in the solution?

1. In a titration between HCl and NaOH, if the volume of theNaOH used is larger than the volume of HCl used, would thiscalculate to a base concentration larger than or smaller than theacid concentration? Briefly explain.

2. In a titration, 13.51 mL of NaOH(aq) are titrated using 12.85mL of 0.1104 M HCl(aq). Calculate the molarity of NaOH(aq). (Noticethat there is a (volume acid)/(volume base) ratio embedded in thecalculation) !

3. A 10.00 mL volume of vinegar (concentrated acetic acidsolution) is diluted in a 100.0 mL volumetric flask to make adilute acetic acid solution. The NaOH in #2 is used to titrate thisdilute acetic acid. 14.53 mL of this diluted acetic acid solutionare neutralized by 12.51 mL NaOH of the molarity in #2. !

Calculate:

a. Volume Base/Volume Acid Ratio in the titration (4 s.f.)

b. Molarity of the diluted acetic acid using the ratio in a. (Ma= Mb*Ratio)

c. Molarity of the Acetic acid in the concentrated acetic acidsolution (vinegar) (Mc = Md*Vd/Vc where Vd = 100.0 mL and Vc =10.00 mL)

d. Mass of acetic acid per liter of the concentrated acetic acidsolution (vinegar) {the molar mass of acetic acid is 60.052 g/mol}(grams/L = mol/L * g/mol)

e. The percent by mass of acetic acid in the concentrated aceticacid (vinegar) The density of the vinegar is 1.011 g vinegarsolution/mL vinegar solution. (The percent Acetic Acid in Vinegaris usually 4% to 8% by mass)

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