Need help with my post lab 1. Using the data collected for Part A on your data s

Question

Need help with my post lab

1. Using the data collected for Part A on your data sheet, calculate the concentration of your standard sodium hydroxide solution. Show the calculations for trial #1 and the average to receive full credit. (Only include the three closest values in average calculation, or exclude outliers.)

2. Using the data collected for Part B on your data sheet, calculate the molar mass of the unknown acid. To receive credit you must include the unknown number, and show the

calculations for trial #1 and the average for Part B. in average calculation, or exclude outliers.)

3. Using the average value of the data collected for Part C on your data sheet and assuming the density of the vinegar is 1.00g/mL,

a.Calculate the molar concentration of acetic acid (CH3COOH) in the vinegar sample using the average volume of NaOH.

b.Calculate the mass percent of acetic acid in the vinegar sample.

Partner Part A: Standardization of NaOH solution po Trial #1 | | Trial #2 Trial #3 | Trial #4 inal buret reading, mL 49 . Initial buret reading, mLpo0sof Volume NaOH used, mL Part B: Determination of the molar mass of an unknown acid Unknown sample Trial #1 Trial #2 Trial #3 Trial #4 number: Mass unknown acid, g Final buret reading, mL 55 15025 1i0.50 Initial buret reading, mL 1mon Volume NaOH used, mL 2 aa Part C: Analysis of Acetic Acid in Vinegar Trial #1 | Trial #2 | Trial #3 | Trial #4 Volume of vinegar, mL |

Explanation / Answer

Part A)

Standardization of NaOH

Trial#1

moles KHP = 0.508 g/104.23 g/mol = 2.5 mmol

moles NaOH = 2.5 mmol

molarity NaOH = 2.5 mmol/20 ml = 0.125 M

similarly,

Trial#2, molarity NaOH = 0.130 M

Trial#3, molarity NaOH = 0.122 M

Trial#4, molarity NaOH = 0.122 M

Average molarity NaOH = 0.125 M

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Part B)

molar mass of unknown acid

Trial#1,

moles NaOH = 0.125 M x 0.0324 L = 4.05 x 10^-3 moles

moles acic = 4.05 x 10^-3 moles

molar mass acid = 0.254 g/4.05 x 10^-3 moles = 62.72 g/mol

Similarly,

Trial#3, molar mass = 63.69 g/mol

Trial#4, molar mass = 64.36 g/mol

Average molar mass os unknown acid = 63.591 g/mol

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Part C)

analysis of acetic acid in vinegar

average volume NaOH = 36.2 ml = 0.0362 L

moles NaOH = 0.125 M x 0.0362 L = 4.525 x 10^-3 moles

moles acetic acid = 4.525 x 10^-3 moles

molarity acetic acid in vinegar = 4.525 x 10^-3 moles/0.005 L = 0.905 M

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3.

a. molar concentration acetic acid in vinegar = 0.905 M

b. mass % = 4.525 x 10^-3 moles x 60.05 g x 100/mol/5 ml = 5.43%

4.

a. If some amount of KHP was spilled over, the moles of KHP present in lowr, so volume NaOH used will also be lower. Since moles is same, molarity of NaOH would be higher than actual value.

b. If some solid remained undissolved, less volume of NaOh would be used. Moles NaOH would be lower and therefore, molar mass of acid would be higher than actual value.

c. If liquid is blown off, more NaOH is used, higher moles NaOH, higher moles acetic acid. Higher molarity of acetic acid calculated in vinegar.

d. If air bubble is present we would add less NaOH than what we read on burette. So higher moles of NaOH, higher moles of acetic acid calculated. higher molarity of acetic acid will be obtained than actual value.

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