Titration of a Sample of Household bleaches:
The oxidizing agent in a household bleach is determined by placing 100 mL of distilled water in a 250 mL Erlenmeyer flask and adding 10 mL of the 10% (w/w) KI solution, swirling the contents of the flask to mix the solutions. Add an accurately measured 0.5 mL of your assigned bleach to the flask, and swirl. Then add 10 mL of 2 M H2S04 and proceed with the titration, using the standardized sodium thiosulfate solution. Add the starch indicator when the solution is pale yellow and continue the titration until the deep blue colour of the starch indicator disappear. Record the volume of thiosulfate solution used in the titration.
From the concentration and volume of the added sodium thiosulfate solution used to titrate the different brands of bleach, calculate the numbers of moles of the oxidizing agent presence, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5 mL of bleach. Express the final result as the mass of sodium hypochlorite per 100 mL of bleach.
The sodium thiosulfate reacts with the sodium hypochlorite according to the following stoichiometric redox equations:
NaClO (aq) + 2H^+ + 2I- ---> I2 + Cl- +H2O + Na^+
Recall:
2S203^ 2- + I2- ---- > 2I- + S4O6 ^ 2-
B. Titration of household bleach: Javex
Calculated Concentration of Na2S2O3- 0.032400M
Trail #
1-Volume of bleach ( 0.5mL)
Volume of Na2S2O3 ( 53 mL)
= mass of NaOCl/100 mL bleach ________________?
2-Volume of bleach ( 0.5mL)
Volume of Na2S2O3 ( 55.4 mL)
= mass of NaOCl/100 mL bleach ________________?
Average mass of NaOCl/100 mL bleach ___________________?
Reaction #1 (properly balanced):
2ClO-(aq) + 4H+(aq) + 6I-(aq) –> 3I2 + 2Cl-(aq) + 2H2O
Reaction #2 (properly balanced):
2S203^2-(aq) + I2 ---- > 2I-(aq) + S4O6^2- (aq)
Moles of S203^2- = (molarity)(volume):
(0.032400 mol/L)(__liters S203^2-) = __moles S203^2-
Moles if I2 based on Equation #2:
Moles I2 = (2)(__moles S203^2-)
Moles of ClO^- based on Equation #1:
Moles ClO-(aq) = (2/3)(__moles I2)
Correction with apologies. The corrected mole ratio is in bold print:
Moles if I2 based on Equation #2:
Moles I2 = (1/2)(__moles S203^2-)
One more comment:
The final answer assumes that:
moles of ClO-(aq) = moles of NaClO
Based on that, convert the number of moles to grams of NaClO by multiplying the number of moles by the formula mass of NaClO. That would give you the grams of NaClO per 0.5 mL of bleach solution.
To get grams NaClO / 100 mLs of bleach, what would you do?
Moles of S203^2- = (molarity)(volume):
=(0.032400 mol/L)(0.053 litres S203^2-)
= 0.03240 moles S203^2-
Moles if I2 based on Equation #2:
Moles I2 = (1/2)( 0.03240 moles S203^2-)
= .00086 moles I2
Moles of ClO^- based on Equation #1:
Moles ClO-(aq) = (2/3)(.00086 moles I2)
= .00057 moles ClO^3-
moles of ClO-(aq) = moles of NaOCl
.00057 moles ClO^3- = .00057 moles NaOCl
NaOCl mass = (molar mass * moles)
= (74) (.00057) = 0.04235 g
Do we assume this gives us grams of NaOCl per 0.5 mL of bleach solution, since we didn’t divide or multiply 0.0005 L with anything? And to make it NaOCl / 100 mLs of bleach do we multiply the by 100?
To calculate the mass of sodium hypochlorite (NaOCl) per 100 mL of bleach, you need to use the stoichiometric relationship between sodium thiosulfate (Na2S2O3) and sodium hypochlorite in the titration equation. Here's how you can calculate the mass:
1. Convert the volume of sodium thiosulfate solution used in each trial to moles. To do this, multiply the volume in mL by the concentration of sodium thiosulfate (0.032400 M).
Trial 1: Moles of Na2S2O3 = 53 mL * 0.032400 M = 1.7172 mmol
Trial 2: Moles of Na2S2O3 = 55.4 mL * 0.032400 M = 1.79576 mmol
2. Use the stoichiometry of the reaction between sodium thiosulfate and sodium hypochlorite to determine the moles of sodium hypochlorite present in each trial. From the balanced equation, we can see that 1 mole of Na2S2O3 reacts with 1 mole of NaOCl.
Trial 1: Moles of NaOCl = Moles of Na2S2O3 = 1.7172 mmol
Trial 2: Moles of NaOCl = Moles of Na2S2O3 = 1.79576 mmol
3. Convert the moles of sodium hypochlorite to grams using the molar mass of NaOCl. The molar mass of NaOCl is approximately 74.44 g/mol.
Trial 1: Mass of NaOCl = Moles of NaOCl * molar mass of NaOCl = 1.7172 mmol * 74.44 g/mol = 127.7852 mg
Trial 2: Mass of NaOCl = Moles of NaOCl * molar mass of NaOCl = 1.79576 mmol * 74.44 g/mol = 133.6987 mg
4. Finally, calculate the mass of NaOCl per 100 mL of bleach by multiplying the mass obtained in each trial by a factor of 20. The factor of 20 is used because the mass was calculated for 0.5 mL of bleach, and we need to express the result per 100 mL.
Trial 1: Mass of NaOCl per 100 mL bleach = Mass of NaOCl * (20/0.5) = 127.7852 mg * 40 = 5111.408 mg/L
Trial 2: Mass of NaOCl per 100 mL bleach = Mass of NaOCl * (20/0.5) = 133.6987 mg * 40 = 5347.948 mg/L
To determine the average mass of NaOCl per 100 mL of bleach, you can take the average of the two trial results:
Average mass of NaOCl per 100 mL bleach = (5111.408 mg/L + 5347.948 mg/L) / 2 = 5229.678 mg/L
Therefore, the average mass of sodium hypochlorite (NaOCl) per 100 mL of bleach (Javex) is approximately 5229.678 mg/L.