CHEM 110L - Experiment 10 It's Back... Titration
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Experiment 10 It’s Back… Titration Total Points: 40 Student’s Name:
Lab Section: NOTE: You must show your work for all calculations; no work, no credit. Mass of Na
2
S
2
O
3
______2.503g_______ (2 pts) Molarity of Na
2
S
2
O
3
_____0.010073M_________ (Show your calculation) 2.503g x 1mol Na2S2O3/248.18g/1L = 0.010073 (1 pt) Density of Bleach _____0.10262_________ g/mL (Show your calculation) 31.625g-21.363g/100mL = 0.10262 Initial Samples 20 mL Diluted Bleach Treated with KI Volume of Na
2
S
2
O
3 Used During Sample Titration (3 pts) Titrant Added (mL) to: After Starch Added Titrant Added (mL) to: Total Volume Color: Brown Initial Buret Volume, mL Light Orange/ Dark Yellow Dark Blue End Point: Colorless (mL) Sample 1 0 10.15 18.86 29.01 Sample 2 0 8.820 8.080 16.90 1.
Calculate the following for each trial. Make sure to show all your work to receive credit. a.
(2 pts) Moles of Na
2
S
2
O
3 used to titrate the bleach solution. S1: 0.010073M x 0.2901L = 0.00292 moles S2: 0.010073M x 0.1690L = 0.001702 moles b.
(2 pts) Moles of I
2
formed by reaction with bleach. S1: 0.00292mol/2 = 0.001461 moles S2: 0.001702mol/2 = 8.512E-4 moles c.
(4 pts) Molarity of OCl
-
present in the bleach solution. (Remember to account for the dilution of the original sample.) S1: (0.001461mol)(1mol OCl-/1mol I2)/(0.02L)(10)= 0.73054M S2: (8.512E-4)(1mol OCl-/1mol I2)/(0.02L)(10)= 0.42558M d.
(2 pts) Grams of NaOCl present in the bleach (in g/mL): S1: (0.73054M)(74.44g/mol) = 54.382g/L -> 0.054382g/mL S2: (0.42558M)(74.44g/mol)= 31.680g/L -> 0.031680g/mL e.
(2 pts) Percent NaOCl present in the bleach: S1: (0.054382g/mL)/(0.10262g/mL) x 100 = 52.99% S2: (0.031680g/mL)/(0.10262g/mL) x 100 = 30.87%
Experiment 10 It’s Back… Titration Total Points: 40 2.
(6 pts) Use the following equation to answer the following questions: I
2
-starch complex
(aq)
+ 2 S
2
O
2-
3(aq)
→
starch + 2 I
-
(aq)
+ S
4
O
6
2-
(aq) What is being reduced? ____Iodine I2_____________________________ What is the oxidizing agent? ______Iodine I2_______________________ How many electrons are being gamed by each iodine atom? _1 e-__ Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit. 1.
(6 pts) Balance the redox reaction first (it is in basic medium). OCl
-
(aq)
+ CrO
2
-
(aq)
→
Cl
-
(aq)
+ CrO
4
2-
(aq)
2OH- + 3ClO- + 2CrO
2
2- →
3Cl- + 2CrO
4
2- + H2O 2.
(4 pts) A titration experiment is set up to use 0.777M bleach (NaOCl) to analyze CrO
2
-
. If 50.0 mL of the CrO
2
-
(aq)
solution required 12.27 mL of bleach to react to completion, what would you calculation as the molarity of the CrO
2
-
solution? 0.777M x 0.1227L = 0.93379 mol NaCl 2mol/3mol x 0.93379 mol NaCl = 0.635586 mol/0.50L = 0.127M 3.
(6 pts) Excess I
2
can be used to determine vitamin C levels in solution. Vitamin C (ascorbic acid) reacts in the following manner: C
6
H
8
O
6(aq)
+ I
2
→
C
6
H
6
O
6(aq)
+ 2 H
+
(aq)
+ 2 I
-
(aq)
The I
2
that did not
react with ascorbic acid is then determined in a thiosulfate reaction as you did in this experiment. If 22.07 mL of 0.8310 M I
2
was added to a sample containing vitamin C, the unused I
2
required 20.00 mL of 0.143 M S
2
O
3
2-
to react. How many moles of vitamin C were originally present? I2 + 2 S2O3 -> 2I + S4O6 I2= 0.8310M x 0.2207L = 0.1834017 mol S203= 0.143M x 0.2000L = 0.0286mol /2mol = 0.0143mol 0.1834017mol –
0.0143mol = 0.1691017 moles of Vitamin C
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