1 TITLE:
Analysis of Bleach DATE(S):
March 29, 2017 PURPOSE:
The purpose of this lab is to determine the percent composition of sodium hypoc hlorite in a commercial bleach solution by titrating it with sodium thiosulfate. PRE-LAB:
2
PROCEDURE:
• First, you would dilute a 5 mL bleach solution • Next, you would take 25 mL and add 2 grams of potassium iodide and 25 mL of distilled water • Then, you would add 2 mL of 3 M hydrochloric acid to that same solution • Next, you would titrate the solution with sodium thiosulfate until it turns light yellow. • After that, you would add a drop of starch and wait for a blue color to appear • Then, you would keep titrating the solution until the blue color disappears • Next, you would repeat the second step through the sixth step • Finally, you would trade data with another group MEASUREMENTS TO BE TAKEN:
• 5 mL of commercial bleach • 95 mL of distilled water • Approximately 2 grams of potassium iodide • 25 mL of dilute bleach (twice) • 25 mL of distilled water (twice) • 2 mL 3 M HCl (twice) • Volume of Na2S2O3 titrated (twice) MATERIALS:
• 5 mL 5% commercial bleach • 6 mL 3 M HCl • 6 g KI
3 • 70 mL 0.100 M Na2S2O3 • 3 mL 2% starch solution • 0.1 g-precision balance • Buret • Buret clamp • Distilled Water • 250 mL Erlenmeyer flask • Pipet Bulb • Ring Stand • 5 mL Transfer Pipet • 25 mL Transfer Pipet • 100 mL Volumetric Flask • Volumetric flask stopper • Wash Battle • Weighing Disk DATA: Molarity of Na2S2O3
Trial 1-Us
Initial Buret Reading
Trial 2-Us
Trial 3-Group 2
Trial 4- Group 2
0.00
3.00
50.00
50.00
Final Buret Reading
32.00*
29.80
26.00
28.00
Volume of Na2S2O3 added
32.00*
26.80
24.00
22.00
4 * = we over-titrated the first round, and so we couldn’t get the second indicator to work and thus we had to stop titrating CALCULATIONS:
We had to calculate the difference between the final and initial buret reading: • Example: Final-Initial = Volume 29.8-3=26.8 mL We had to convert moles of ClO- to moles of S2O32-: • Example: 1 mol ClO-
1 mol I2
I mol I3-
2 mol S2O32-
1 mol ClO-
I mol I2
I mol I3-
= 2 mol S2O32We had to average out our values for the volume of Na2S2O3 added to the titration: • Example: (32 + 26.8 + 24 + 22)/4 = 104.8/4 = 26.2 We had to convert liters of Na 2S2O3 to the molarity of NaClO: • Example: 0.032 L Na2S2O3
0.100 mol Na2S2O3
1 mol NaClO
1 M NaClO
1 L Na2S2O3
2 mol Na2S2O3
0.025 L NaClO
= 0.064 M NaClO We had to calculate the molarity of the concentrated bleach based on the molarity of the dilute bleach: • Example:
5 M1V1 = M2V2 (0.064 M)(0.1 L) = M2 x (0.005 L) 0.0064 = M2 x 0.005 M2 = 1.28 M We had to calculate the percent mass of bleach: • Example: 1.28 mol NaClO
1 L NaClO
1 mL NaClO
74.4 g NaClO
1 L NaClO
1000 mL NaClO
1.08 g NaClO
1 mol NaClO
x 100% = 8.82 g/mol We had to calculate the average deviation: • Example:
"((8.25-8.82)2 + (8.25-7.38)2 + (8.25-6.61)2 + (8.25-6.06)2) "(0.3249 + 0.7469 + 2.6896 + 4.7961)/4 "(8.5675)/4 2.927/4 = 0.73% average deviation We had to calculate the percent error: • Example: (theoretical value-experimental value)/theoretical value x 100% (8.25-8.82)/8.25 x 100% 0.0691 x 100% = 6.91% Na2S2O3 RESULTS:
6 The concentration of sodium hypochlorite in the bleach sample is 8.184% based on the average of our set of data combined with the data of Group 2, which was slightly off from the actual value of 8.25%. ANALYSIS/CONCLUSIONS:
The point of this lab was to get practice with titrations and to determine the c oncentration of sodium hypochlorite in a commercial solution of bleach. This lab taught us how to analyze the percent composition of a compound in a solution and how to find it based on titrations and calculations. The experiment worked fairly well, since it called for us to collaborate with other groups, thus giving us someone else’s work for comparison. It also called for data from multiple titrations, which canceled out many major human errors such as over-titrating. Our overall percent error was 1.53% when accounting for the mean of all four titrations of our group and Group 2, which is fairly small. However, each individual titration was blatantly incorrect. The individual titrations had a percent error of 6.91% and 10.54% for our two and 19.88% and 26.55% for the two of Group 2. Plus, the average deviation was 0.73%, which is fairly large. The most likely source of this error is that we titrated it very quickly a t some points. This caused us to accidentally over-titrate at one point, which would skew the results lower due to how we couldn’t add enough moles to complete the titration since the indicator wouldn’t work past the equivalence point. Moreover, rinsing out the transfer pipets with distilled water before filling them with bleach likely lowered our calculated percent composition. This is because the extra water would have diluted the bleach and decreased the amount that would fit up until the marker for the given amount of milliliters. Thus, the amount of moles of sodium hypochlorite would have decreased
7 and consequently decreased the percent composition. To get better results, we would avoid overtitrating the first titration and not rinse out the pipet before titrating. The lab is worth repeating, since our individual titration results were so far off. POST-LAB QUESTIONS:
1. Oxidation is a state in which an atom loses electrons, whereas reduction is a state in which an atom gains electrons 2.
3. Diluting the original sample makes it less sensitive to the titrant and thus easier. Since the molarity is smaller, the titration can be done in a way wh ich will get more precise results and not go over the equivalence point as quickly. 4. 3 25.0 mL aliquots can be taken from a 100 mL volumetric flask. This is because it can be ensured that at least three can be taken out, but you can’t ensure that there will be enough in the volumetric flask for a fourth aliquot, since the amount has to be exact and it is always safe to have some left over. 5. O = -2, -2 x 6 = -12, -12 + 2 = -10, -10/4 = -2.5, -2.5 x -1 = 2.5
8 The oxidation number of all of the combined sulfurs is +10, which translates to about +2.5 each. Because it appears in fraction form, and oxidation numbers can’t be fractions, the oxidation numbers are different for different sulfurs. Because it is in Group 16, and the oxidation number of sulfur is normally -2, it is likely that two sulfurs are +2 and the remaining two are +3, since those are closest in absolute value to the regular oxidation number. 6. a) The leftover water could get inside the pipet, thus messing up the measurement of the bleach by taking up space and resulting in less than the desired amount of bleach in moles, which would cause an overall lower percent composition b) An extra gram of KI would have increased the amount of iodine in the solution. This would cause the solution to change before the equivalence point was reached, due to how the iodine helps the solution change color. c) The iodine helps the solution change blue whenever the end point is reached. Thus, if some of sublimed and not enough was available to change the color, then the risk of overtitrating would be higher, since the iodine wouldn’t change color as strongly. 7.
The major source of experimental error is rising out the bulbs before measuring the solution.
Not only does this dilute the solution further, but also it decreases the amount of bleach put in the solution. Thus, it made the amount of number of moles appear lower, and so the percent composition appeared lower than the actual value of 8.25%.
