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A student followed the procedure of this experiment to determine the percent NaOCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-mL aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL

1. Calculate the number of moles of S2O3^-2 on required for the titration

2. Calculate the number of moles of I2 produced in the titration mixture.

3.Calculate the number of OCI- ion present inthe diluted bleaching solution titrated.

4. Calculate the mass of NaOCI present inthe diluted bleaching titrated

5. Calculate the mass of commercial bleaching solution titrated

6. Determine the percent NaOCI in the comercial bleaching solution

7. Calculate the mass of one gallon of the commercial bleaching solution.

8. Calculate the cost of 100g of the commercial bleaching solution.

9. Determine the cost of the amount of bleaching solution required to supply 100g of NaOCI

Explanation / Answer

This reaction must be carried out in a basic solution - It is necessary to add NaOH to the solution before titration.
Equation:
4 NaOCl + Na2S2O3 + 2 NaOH 4 NaCl + 2 Na2SO4 + H2O
4mol NaOCl react with 1 mol Na2S2O3

You ask the question:
*) Calculate the number of mol Na2S2O3:
Mol Na2S2O3 in 35.46mL of 0.152M solution = 35.46/1000*0.152 = 0.00539 mol Na2S2O3 used in the titration.

*) 2 S2O32-+2I- ---> I2 + S4O6

2 mole of S2O32-produce 1 mole I2. So , I2 produced is 0.00539/2=0.002695 mol

*) How many mol of NaOCl is in the 20mL titration sample:
From the balanced equation: 4* 0.00539 = 0.0216 mol NaOCl in 20mL

*) mol NaOCl in the 250mL of solution = 250/20* 0.0216 = 0.269 mol NaOCl

*) Therefore 50mL of the original commercial bleaching solution contains 0.269 mol NaOCl

*) Calculate mol NaOCl in 1 litre of original commercial solution = 1000/50*0.269 = 5.38 mol NaOCl

*) molar mass NaOCl = 23+35.5+16 = 74.5g/mol.
    5.38mol = 5.38*74.5 = 400.81 g NaOCl in 1 litre of solution.

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