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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:
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?