to the flask. Titrated the vinegar sample until the end point was reached. Recorded the final
reading and prepared a new sample for the second trial.
Results
Standardizing a Sodium Hydroxide Solution
Data:
I.
Delivering Water from a Buret
II. Preparing the KHP Sample
Determination 1
Determination 2
III. Analyzing an Unknown Permanganate Ion Solution
IV. Titrating the KHP
Initial Buret Reading, mL
Final Buret Reading, mL
Volume of H2o delivered, mL
0.40mL
11.50mL
11.10mL
mass of weighing paper,
g
mass of weighing paper
and KHP, g
mass of weighing paper,
g
mass of weighing paper
and KHP, g
2.85g
4.83g
2.86g
4.58g
code number of
unknown
%T
absorbance, A
molarity, M
#1
80.6
0.094
4.6x10^-4
#2
58.8
0.231
1.16x10^-4

Determination 1
Determination 2
Calculations:
1. mass of KHP titrated, determination 1: 1.98g
mass of KHP titrated, determination 2: 1.72g
2. Calculate number of moles of KHP titrated
1.98g(1mol/204.4g)= .0096mol KHP
moles of KHP titrated, determination1: 0.009696mol KHP
moles of KHP titrated, determination2: 0.008423mol KHP
3. Calculate the number of moles of NaOH required to neutralize the KHP
0.009696(1molNaOH/1molKHP)= 0.009696mol NaOH
moles of NaOH required, determination1: 0.009696mol NaOH
moles of NaOH required, determination2: 0.008423mol NaOH
4. Calculate the volume (L) of NaOH solution required to neutralize the KHP
19.80mL-0.02mL= 19.78mL
19.78ml(10^3/ml)= 0.01978L
volume of NaOH solution, determination1: 0.01978L
volume of NaOH solution, determination2: 0.0171L
5. Calculate the molarity of the NaOH solution
(0.009696molNaOH)/((19.78ml)(10^3L/mL))= 0.490M
initial buret reading, mL
0.02mL
0.00mL
finial buret reading, mL
19.80mL
17.10mL

6. Find the average molarity of the NaOH solution (0.490M + 0.493M)/2=0.4915M average molarity of NaOH solution: 0.4915M
Analyzing the Acetic Acid Content of Vinegar
Data:
Trial 1
Trial 2
Average Calculated Molarity of NaOH from Standardization Lab, mol/L: 0.4915mol/L
Calculations:
1.
Moles of NaOH delivered, mol
00.4915mL(0.0098mL)= 0.0048167mol
Trial1: 0.0048167mol
Trial 2: 0.0041286mol
2. Moles of acetic acid
(1mol NaOH=1mol CH3COOH)
Trial1: 0.0048167mol
Initial buret reading, mL
0.00mL
9.8mL
Final buret reading, mL
9.8mL
18.2mL
Volume of NaOH delivered, mL
9.8mL
8.4mL
Volume of acid in titrated sample,
mL
10.00mL
10.00mL

Trial 2: 0.0041286mol
3. Molarity of acetic acid. mol/L
(0.0048167molCH3COOH)/((10ml)((1L/1,000mL))= 0.48167M
Trial 1: 0.48167M
Trial 2: 0.41286M
4. Average molarity of acetic acid, mol/L
(0.48167M+0.41286M)/2=0.4473M
Average: 0.4473M
5. Mass of acetic acid in sample, g
0.004867molCH3COOH(60.05g/mol)=0.2922g
Trial1: 0.2922g
Trial2: 0.2479g
6. Mass percent of acetic acid
((0.2922g)/(10.00g))x100= 2.922%
Trial1: 2.922%
Trial2: 2.479%
7. Average mass percent
(2.922%+2.479%)/2=2.7005%
Average mass percent: 2.7005%
Percent error for the standardized NaOH
((0.5M - 0.4473M)/(0.5))x1005= 10.54%

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