DATA SHEET Acid in a vinegar Solution Part B: Determining the Concentration of A

Question

DATA SHEET Acid in a vinegar Solution Part B: Determining the Concentration of Acetic Mass of Cl in the vinegar solution 5L TRAL 1 TRAL 2 TRAL 3 TRIAL 4 TRIAL 5 mom label on borde) Vol. of Vinegar -5 -5 Initial Buret vol. L2 2 2 Final Buret Vol. (va T 2e SLee vol of NaOH used Molarity of CH,COOH Show a sample calculation Average Molarity of CH,COOH (using best two trials) use the average molarity of vinegar to calculate the Mass of CHscooH in the vinegar solution Mass CH3COOH Lab Day M T W R F instructor. 10 Room 103 109 117 125 Lab Time:

Explanation / Answer

Solution:- First of all we will calculate the molarity of NaOH from given second data table. Mass of KHP is given in the table. Convert this into moles and using stoichiometry we will calculate the moles of NaOH. KHP reacts with NaOH in 1:1 mol ratio. Molar mass of KHP is 204.22 g/mol.

Balanced equation is....

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

where KHC8H4O4 is potassium hydrogen phthalate and in short it is written as KHP.

for Trial 1:-

0.566 g KHP x (1mol KHP/204.22 g KHP) x (1mol NaOH/1mol KHP) = 0.00277 mol NaOH

Volume = 12.10 ml = 0.01210 L

molarity = 0.00277 mol/0.01210L = 0.229 M

Trial 2:-

0.526 g KHP x (1mol KHP/204.22 g KHP) x (1mol NaOH/1 mol KHP) = 0.00258 mol NaOH

volume = 15.00 ml = 0.01500 L

molarity = 0.00258 mol/0.01500 L = 0.172 M

Trial 3:-

0.565 g KHP x (1mol KHP/204.22 g KHP) x (1mol NaOH/1mol KHP) = 0.00277 mol NaOH

volume = 9.00 ml = 0.00900 L

molarity = 0.00277 mol /0.00900 L = 0.308 M

Trial 4:-

0.516 g KHP x (1mol KHP/204.22 g KHP) x (1mol NaOH/1 mol KHP) = 0.00253 mol NaOH

volume = 17.60 ml = 0.01760 L

Molarity = 0.00253 mol/0.01760 L = 0.144 M

Trial 5:-

0.552 g KHP x (1mol KHP/204.22 g KHP) x (1mol NaOH/1mol KHP) = 0.00270 mol NaOH

volume = 10.50 ml = 0.01050 L

molarity = 0.00270 mol / 0.01050 L = 0.257 M

Trial 1 and trial 5 values are closer. So, the average of these two trial values = (0.229 M + 0.257 M)/2 = 0.243 M

Now, the molarity of acetic acid would be calculated using the equation, M1V1 = M2V2

trial 1:-

M1(0.005) = 0.243M(0.02710)

M1 = 1.32 M

Trial 2:-

M1(0.005) = 0.243M(0.0244)

M1 = 1.19 M

Average molarity of acetic acid = (1.32 M + 1.19 M)/2 = 1.26 M

Average moles of acetic acid = 5 ml x (1L/1000ml) x (1.26 mol/L) = 0.0063 mol

molar mass of acetic acid is 60.05 g/mol.

So, mass of acetic acid = 0.0063 mol x (60.05 g/mol) = 0.378 g

Mass percentage of acetic acid(CH3COOH) could be calculated if the mass of Vinegar is given using the formula..

mass percent of acetic acid = (mass of acetic acid/mass of vinegar) x 100

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