Question

In: Chemistry

A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of...

A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of vinegar. If you titrate a 10.00 mL sample of undiluted vinegar, what volume (in mL) of 0.1014 M NaOH should you expect to need to use to reach the endpoint of the titration?

Solutions

Expert Solution

molarity    = W*1000/G.M.Wt * volume in L

                = 5.38*1000/60*100

                 = 0.896M

CH3COOH + NaOH------------> CH3COONa + H2O

1 mole             1 mole

CH3COOH                                       NaOH

M1 = 0.896M                                    M2 = 0.1014M

V1 = 10ml                                          V2 =

n1 = 1                                                 n2 = 1

                       M1V1/n1   =    M2V2/n2

                          V2            =   M1V1n2/M2n1

                                          = 0.896*10*1/0.1014   = 88.36ml of NaOH >>>>answer

queen_honey_blossom answered 2 years ago
Next > < Previous

Related Solutions

A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of...
A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of vinegar. You pipet 10.00 mL of that vinegar into a 100 mL volumetric flask, add water to the mark, and mix. You then pipet 10.00 mL of the diluted vinegar into a flask and titrate with 0.1014 M NaOH. What volume (in mL) of 0.1014 M NaOH should you expect to need to use to reach the endpoint of the titration?
A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of...
A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of vinegar. If you titrate a 10.00 mL sample of undiluted vinegar, what volume (in mL) of 0.1014 M NaOH should you expect to need to use to reach the endpoint of the titration?
given a 5.0 g sample of a vinegar having a mass percentage of acetic acid equal...
given a 5.0 g sample of a vinegar having a mass percentage of acetic acid equal to 9.1%(m/m), calculate what volume of NaOH solution of molarity 0.82M will be necessary to threats this vinegar sample.
The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium...
The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium hydroxide in the following fashion: HC2H3O2(aq) + NaOH(aq)-> H2O(l) + NaC2H3O2(aq). If 2.52 mL of vinegar requires 34.9 mL of 0.1031 M (molarity) NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? (1 L = 1.0567 qt)
A 15.3 mL sample of vinegar, containing acetic acid, was titrated using 0.809 M NaOH solution....
A 15.3 mL sample of vinegar, containing acetic acid, was titrated using 0.809 M NaOH solution. The titration required 24.06 mL of the base. Assuming the density of the vinegar is 1.01 g/mL, what was the percent (by mass) of acetic acid in the vinegar?
What is the pH of a 100 mL solution of 100 mM acetic acid at pH...
What is the pH of a 100 mL solution of 100 mM acetic acid at pH 3.2 following addition of 5 mL of 1 M NaOH? The pKa of acetic acid is 4.70. A) 3.20. B) 4.65. C) 4.70. D) 4.75. E) 9.60.
Titration experiment: Determination of the acid content of vinegar. 25.00 mL of vinegar with a 5%...
Titration experiment: Determination of the acid content of vinegar. 25.00 mL of vinegar with a 5% concentration of acetic acid is combined with distilled water to prepare a dilute vinegar solution of 250.00 mL. The burette is filled with a standardised NaOH solution with a molarity of 0.09979 M. 25.00 mL of diluted vinegar is pipetted into a conical flask, and phelphthalein indicator is added to it. The volume of NaOH that needed to be added in order for the...
1. Vinegar consists of acetic acid, CH3COOH, and water. Write out the equilibrium expression for acetic...
1. Vinegar consists of acetic acid, CH3COOH, and water. Write out the equilibrium expression for acetic acid reacting with water. 2. Citric acid, H3C6H5O7, contributes to the acidity of many foods. Write out the equilibrium expression for citric acid reacting with water to lose one acidic proton.
The solubility of benzoic acid in water is 6.80g per 100 mL at 100 ℃, and 0.34g per 100 mL at 25 ℃.
The solubility of benzoic acid in water is 6.80g per 100 mL at 100 ℃, and 0.34g per 100 mL at 25 ℃.(a) Calculate the min. volume of water needed to dissolve 1.00g of benzoic acid at 100 ℃. (b) Calculate the maximum theoretical percent recovery from the recrystallization of 1.00g benzoic acid from 15 mL of water assuming the solution is filtered at 25 ℃?
Calculate the weight % of acetic acid in the vinegar. Calculate your percent error. For this...
Calculate the weight % of acetic acid in the vinegar. Calculate your percent error. For this calculation, assume that the density of vinegar is 1.01g/mL. 0.8488 Molarity of acetic acid in my experiment. I'm a little confused-if you solve it can you be sure to specify what the conversion factors are in the calculations. Thank you!
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT