In this experiment, the amount of sodium hypochlorite in commercial bleach will be determined by reacting it with sodium thiosulfate while in the presence of iodide ions and starch. A solution of sodium thiosulfate of a known concentration will be added to the bleach using a buret in a titration. An average of 17.2 mL of sodium thiosulfate was used in the titration, resulting in a calculation of 4 .
7% of sodium hypochlorite in the commercial bleach. Because the expected value was 6%, the percent error was calculated to be 21 .
67%.
Commercial bleaches are made by bubbling chlorine gas into a sodium hydroxide solution. Some of the chlorine is oxidized to the hypochlorite ion present in a solution, ClO
–
, and some is reduced to the chloride ion, Cl
–
. Therefore, the solution remains strongly basic. The ionic chemical equation for the process is:
Cl
2
(g) + 2 OH
−
(aq) → ClO
−
(aq) + Cl
−
(aq) + H
2
O(l) (1)
The amount of hypochlorite ion present in a solution of bleach is determined by a redox titration. In this experiment, the titration involves iodide and thiosulfate ions. Iodide ion, I
– can be oxidized by almost any oxidizing agent. Three reactions occur that analyze the titration:
2 H
+
(aq) + ClO
−
(aq) + 2 I
−
(aq) → Cl
−
(aq) + I
2
(aq) + H
2
O(l)
I
2
(aq) + I
−
(aq) → I
−
3
(aq)
I
−
3
(aq) + 2 S
2
O
3
2 −
(aq) → 3 I
−
(aq) + S
4
O
6
2 −
(aq)
(2)
(3)
(4)
The last reaction has the triiodide ions (originally red-brown) fade to yellow and eventually clear (iodide ion). Addition of starch turns the solution into a deep blue color which can be reversed. The disappearance of the blue color provides an effective method of determining the endpoint.
Transfer 5.00 mL of commercial bleach solution into a 100 mL volumetric flask. Dilute the solution with distilled water and mix well. Two grams of solid KI is massed and added to the flask. Twenty-five mL of distilled water is also added to the volumetric flask. The flask is then swirled to dissolve the KI, and then 2 mL of 3 M HCl. A buret is set up with 0.100 M sodium thiosulfate solution. The initial buret volume reading is recorded. The solution is then titrated until the iodine color of red-brown fades to light yellow. Then, a dropperful of starch solution is added, which should turn the solution blue. Continue the titration until the blue color disappears. Record the final buret volume reading in the Data Table. Repeat the titration procedure for more results.
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Molarity of Na
2
S
2
O
3
: 0.1 M
Initial buret reading
Final buret reading
Trial 1
0 mL
16.8 mL
Trial 2 Trial 3 Average
0 mL
17.6 mL
Volume of Na
2
S
2
O
3 added 16.8 mL 17.6 mL
-
-
-
-
-
17.2 mL
NOTE: A third trial was not performed due to time restrictions.
2

Post-Lab Calculations and Analysis
1. Determine the number of moles of sodium thiosulfate that are equivalent to one mole of sodium hypochlorite.
2 S
2
O
3
2 −
= I
−
3
= I
2
= ClO
− mole of sodium hypochlorite.
, therefore, for every mole of sodium thiosulfate, we need one half of a
2. Calculate the average volume of Na
2
S
2
O
3
16 .
8 + 17 .
6 to determine the molarity of the diluted bleach.
= 17 .
2
2
3. Use the average volume and molarity of Na
2
S
2
O
3 to determine the molarity of the diluted bleach.
mol Na
2
S
2
O
3
= (0 .
0172)(0 .
1) = 0 .
00172
⇓
(0 .
00172)
= 0 .
000852 mol NaClO
2
⇓
0 .
000852
= 0 .
341 M NaClO
0 .
025
4. Calculate the average molarity of the commercial bleach before it was diluted.
0 .
000852 × 4 = 0 .
0034
0 .
0034
= 0 .
68 M NaClO
0 .
005
5. Assume the density of the commercial bleach is 1.08 g/mL. Calculate the average percent by mass of
NaClO in the commercial bleach.
0 .
0034 × 74 .
4424 = 0 .
2536 g
5mL × 1 .
08 g/mL = 5 .
4g
0 .
2536
= 4 .
70%
5 .
4
6. From the label information of the commercial bleach calculate the average deviation of the three titration values.
| 0 .
047 − 0 .
060 | = 0 .
013
7. Calculate the percent error of your value.
0 .
047 − 0 .
06
0 .
06
≈ 21 .
67%
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In the experiment, a redox titration was performed in order to determine the percent of sodium hypochlorite in commercial bleach. This is done by reacting the bleach with sodium thiosulfate in the presence of iodide ions and starch. The concentration of sodium thiosulfate is already known, and calculations can be done in order to calculate the percent by mass of NaClO in commercial bleach. Approximately 17.2 mL of sodium thiosulfate was used in the titration procedure, which amounts to a percent of 4 .
7%. Because 6% of the commercial bleach is sodium hypochlorite, the percent error was found to be 21 .
67%.
There was the potential of many errors in this experiment. Mainly, these errors are focused on contamination. A total of four different solutions (with varying degrees of molarity) were used, and any mixture between could have resulted in different solutions and concentrations. This could have resulted in a different titration scheme, and a different percent of sodium hypochlorite would have been calculated.
Similarly, the amount of distilled water used in this experiment would have a profound impact on the final percentage calculation. For example, the commercial bleach solution was originally 5 mL, but was diluted in a 100 mL volumetric flask. Any addition of distilled water, whether it was by inaccurate measurement, or because of equipment cleaning, could have skewed the molarity and therefore, the number of moles used in calculations. If there was too much distilled water, then the molarity and moles would seem lower, yielding a lower percentage calculation.
Overall, the results of the experiment are fairly precise and accurate. Multiple trials were performed, so precision was enhanced. In order to improve accuracy and/or precision, better equipment could have been used. The stopcock on the buret was not very accurate in that it did not provide a steady flow of titrant.
Had an improved buret been used, a more accurate reading of the titrant volume could have been used, and a better concentration result could have been calculated.
Pre-Lab Questions
1. What is meant by a “titration?”
Titration is a quantitative chemical analysis that is used to determine the unknown concentration of a solution. A reagent, called the titrant, of known concentration (a standard solution) and volume is used to react with a solution of the titrand, whose concentration is not known.
2. A solution of household vinegar is to be analyzed. A pipet is used to measure out 10.0 mL of the vinegar, which is placed in a 250-mL volumetric flask. Distilled water is added until the total volume of solution is 250 mL. A 25.0 mL portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide.
The neutralization reaction is as follows:
HC
2
H
3
O
2
(aq) + OH
–
(aq) −−→ C
2
H
3
O
2
–
(aq) + H
2
O(l)
It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar.
Calculate the molarity of the diluted vinegar.
V
A
M
A
= V
B
M
B
(0 .
0167)(0 .
05) = (0 .
025) M
B
0 .
0334 = M
B
3. Calculate the molarity of the household vinegar.
V
A
M
A
(0 .
01) M
A
M
A
= V
B
M
B
= (0 .
25)(0 .
0334)
= 0 .
835
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4. The household vinegar has a density of 1.05 g/mL. Calculate the percent by mass of acetic acid in the household vinegar.
g
10 .
0 mL × 1 .
05 mL
⇓
= 10 .
5 g HC
2
H
3
O
2
Molar mass of HC
2
H
3
O
2
: 60 g/mol
⇓
0.835
× 60 = 0 .
501 g HC
2
H
3
O
2
⇓
0 .
501
= 4 .
77%
10 .
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Post-Lab Questions
1. Define oxidation and reduction.
Oxidation is defined as an increase in oxidation number, and reduction as a decrease in oxidation number. This is usually achieved through the transfer of electrons. In oxidation, the reductant yields the product plus an electron. In reduction, the oxidant plus electron yields a product.
2. Write balanced oxidation and reduction half-reactions for the overall redox reactions in equations (1) and (3) in the THEORY section. For each half-reaction, identify which substance is oxidized or reduced.
For equation (1), we have:
2 H
+
+ ClO
−
+ e
−
2 I
−
→ Cl
→ I
2
−
+ H
2
O
+ e
− reduction oxidation where iodide is being oxidized and the hypochlorite ions are being reduced. For equation (3), we have:
I
−
+ I
2
+ 2 e
−
2 S
2
O
3
2 −
→ 3 I
− reduction
→ S
4
O
6
2 −
+ 2 e
− oxidation where the thiosulfate ions are being oxidized while iodine is being reduced.
3. In this analysis, “aliquot,” (diluted fraction) of the initial solution is used for the titration. What advantage is there in diluting the original solution for analysis?
The major advantage of using an aliquot is that it makes the titration more effective by conserving both time and materials. When the substance is diluted, only a small portion is used, so there will be less moles present in the solution, and therefore, it will take less time to neutralize. Also, because of the lower number of moles present, the color change occurs much faster.
4. How many 25.0-mL aliquots can be measured from a 100-mL volumetric flask? Explain.
Four 25.0 mL aliquots can be measured from a 100 mL volumetric flask in theory. In practice, not exactly 25.0 mL of solution can be extracted from the volumetric flask due to inaccuracy errors.
Therefore, it is more likely that one can only obtain 3 full sets of 25 mL.
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5. The reaction of triiodide with thiosulfate produces the dithionate ion, S number of sulfur in this ion. Do you think that the sulfur atoms in S
4
O
4
O
2-
6
6
2. Calculate the oxidation all have the same oxidation number? What might the oxidation numbers be?
No, I do not think that all of the sulfur atoms have the same oxidation numbers. Each oxygen has an oxidation number of -2, with an overall charge of -12. However, the molecule has a charge of -2, meaning that the total charge of the sulfurs must equate to +10. A plus 10 charge does not divide evenly into 4 sulfur atoms (in dithionate) so therefore, the charges are not all equal. Most likely, the charges are +2, +2, +3, +3.
6. How would each of the following laboratory mistakes affect the calculated value of the percent NaClO in the commercial bleach (too high, too low, no change)? Explain.
(a) In step 1, the pipet was rinsed with distilled water immediately before being used to measure the commercial bleach solution.
The calculated percentage will be less than the expected value because the distilled water in the pipet will dilute the commercial bleach solution to an extent. Therefore, the number of moles will also seem to decrease, and when divided by the total mass of the bleach, a smaller percent of
NaClO in the bleach is calculated.
(b) In step 2, three grams of KI were used instead of two grams.
Adding an extra gram of KI will not exhibit any change in the calculated percentage of NaClO.
This is because KI is already added in excess in the procedure, and any more will not effect the calculation of the percentage.
(c) In step 3, some of the iodine that formed sublimed from the solution.
Sublimation of iodine occurs when solid iodine is converted into iodine gas. In this instance, less solid iodine would cause a less (indirect) reaction with sodium thiosulfate. Therefore, a smaller calculation in the number of moles of NaClO will be determined, and a smaller percentage will result.
7. What is the major source of experimental error in this experiment? Explain.
See CONCLUSION .
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