2. In a typical titration, someone needs to "standardize" the solution of the titrant-i.e., to determine its molarity accurately and precisely. In this experiment, we don't have a way for you to standardize the vinegar solution. As such, since the vinegar is sold as "5% acidity" (5% by mass acetic acid, HC;H3O2), we will assume a value of 5.00% of acetic acid, even though it is unclear how close companies need to be to 5% in order to use that value on the label. What is the molarity of a "5%" acetic acid solution? This problem will take you through that calculation. The "molarity" of a solute A in a solution is defined to be the "number of moles of A dissolved per liter of solution". As such, one needs two “quantities" for a given solution in order to calculate molarity: “Moles of solute" and "Liters of solution". Division of the first by the second will yield molarity: molarity = # mol solute # liters of solution Since molarity is an intensive property (it doesn't depend on the sample size), we can pick any sized solution we want for calculational purposes! Let's choose a sample of exactly 100. g of vinegar and calculate its molarity. (a) How many grams of HC2H3O2 are in a sample of 100. g of vinegar if we assume vinegar is 5.00% by mass HC2H3O2? NOTE: this is not a trick question! 5.00% mass here means "5.00 grams of HC2H3O2 per 100. grams of solution! (b) How many moles of HC2H3O2 are in the sample? (The molar mass of HC2H3O2 = 60.05 g/mol) 9 Prelab, Analysis of Milk of Magnesia We now have "half" of what we need to get molarity-we have "moles of solute"! Now we need the volume of the solution. (c) What is the volume, in mL, of a sample of 100. g vinegar? Assume dvinegar = 1.01 g/mL NOTE: The density is close to that of water because most of the matter in a solution of vinegar is, in fact, water! (d) What is the volume of the sample in L? (1000 mL = 1 L) (e) What is the molarity of the sample in mol/L (or M)?
2. In a typical titration, someone needs to "standardize" the solution of the titrant-i.e., to determine its molarity accurately and precisely. In this experiment, we don't have a way for you to standardize the vinegar solution. As such, since the vinegar is sold as "5% acidity" (5% by mass acetic acid, HC;H3O2), we will assume a value of 5.00% of acetic acid, even though it is unclear how close companies need to be to 5% in order to use that value on the label. What is the molarity of a "5%" acetic acid solution? This problem will take you through that calculation. The "molarity" of a solute A in a solution is defined to be the "number of moles of A dissolved per liter of solution". As such, one needs two “quantities" for a given solution in order to calculate molarity: “Moles of solute" and "Liters of solution". Division of the first by the second will yield molarity: molarity = # mol solute # liters of solution Since molarity is an intensive property (it doesn't depend on the sample size), we can pick any sized solution we want for calculational purposes! Let's choose a sample of exactly 100. g of vinegar and calculate its molarity. (a) How many grams of HC2H3O2 are in a sample of 100. g of vinegar if we assume vinegar is 5.00% by mass HC2H3O2? NOTE: this is not a trick question! 5.00% mass here means "5.00 grams of HC2H3O2 per 100. grams of solution! (b) How many moles of HC2H3O2 are in the sample? (The molar mass of HC2H3O2 = 60.05 g/mol) 9 Prelab, Analysis of Milk of Magnesia We now have "half" of what we need to get molarity-we have "moles of solute"! Now we need the volume of the solution. (c) What is the volume, in mL, of a sample of 100. g vinegar? Assume dvinegar = 1.01 g/mL NOTE: The density is close to that of water because most of the matter in a solution of vinegar is, in fact, water! (d) What is the volume of the sample in L? (1000 mL = 1 L) (e) What is the molarity of the sample in mol/L (or M)?
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 57P
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Transcribed Image Text:2. In a typical titration, someone needs to "standardize" the solution of the titrant-i.e., to determine its molarity accurately
and precisely. In this experiment, we don't have a way for you to standardize the vinegar solution. As such, since the
vinegar is sold as "5% acidity" (5% by mass acetic acid, HC2H3O2), we will assume a value of 5.00% of acetic acid, even
though it is unclear how close companies need to be to 5% in order to use that value on the label. What is the molarity of
a "5%" acetic acid solution? This problem will take you through that calculation.
The "molarity'" of a solute A in a solution is defined to be the “number of moles of A dissolved per liter of solution". As
such, one needs two "quantities" for a given solution in order to calculate molarity: "Moles of solute" and "Liters of
solution". Division of the first by the second will yield molarity: molarity =
# mol solute
# liters of solution
Since molarity is an intensive property (it doesn't depend on the sample size), we can pick any sized solution we want for
calculational purposes! Let's choose a sample of exactly 100. g of vinegar and calculate its molarity.
(a) How many grams of HC2H3O2 are in a sample of 100. g of vinegar if we assume vinegar is 5.00% by mass HC2H3O2?
NOTE: this is not a trick question! 5.00% mass here means "5.00 grams of HC2H3O2 per 100. grams of solution!
(b) How many moles of HC2H3O2 are in the sample? (The molar mass of HC2H3O2 = 60.05 g/mol)
Prelab, Analysis of Milk of Magnesia
We now have “half" of what we need to get molarity-we have “moles of solute"! Now we need the volume of the solution.
(c) What is the volume, in mL, of a sample of 100. g vinegar? Assume dvinegar = 1.01 g/mL
NOTE: The density is close to that of water because most of the matter in a solution of vinegar is, in fact, water!
(d) What is the volume of the sample in L? (1000 mL = 1 L)
(e) What is the molarity of the sample in mol/L (or M)?
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