A student followed the procedure of this experiment to determine the percent NaO

Question

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 mL of the 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 0.1502M Na2S2O3 solution. A faded price lable on the gallon battle read $0.79. The density of the bleaching solution was 1.10 gmL-1. Calculate the mass of commercial bleaching solution titrated

1. calculate the number of moles of s2032- on required for the titration

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

3. calculate the number of moles of OCL- ion present in the diluted bleaching solution titrated

4. calculate the mass of naocl present in the diluted bleaching solution titrated

5. determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated

6. calculae the mass of commercial bleaching solution titrated

7. determine the percent naocl in the commercial bleaching solution

8. calculate the mass of one gallon of the commercial bleaching solution

9.. calculate the cost of 100 g of the commercial bleaching solution

Explanation / Answer

Solution.

1) 4NaOCl + Na2S2O3 + 2 NaOH 4 NaCl + 2 Na2SO4 + H2O

4 OCl- + S2O32- + 2 OH- = 2 SO42- + 4Cl- + H2O;

n(S2O32-) = cV = 0.1502*35.46=5.326 mmoles;

2) There were no I2 produced in the titration mixture.

3) The number of moles of OCl- ion present in the diluted bleaching solution titrated aliquote is

n(OCl- ) = 4n(S2O32-) = 4(5.326) = 21.304 mmoles;

4) The mass of NaOCl in the aliquote is m = nM = 21.304*74.44 =1586 mg, or 1.586 g.

5)  The volume of commercial bleaching solution present in the diluted bleaching solution titrated is

(50/250)*20 = 4 mL.

6) The mass of commercial bleaching solution titrated is 4*1.10 =4.4 g.

7) The percent NaOCl in the commercial bleaching solution is 1.586/4.4 = 0.36, or 36%;

8) One gallon is 3.785 liters, it welghts 3.785*1.10 = 4.164 kg;

9) The cost of 100 g of the commercial bleaching solution is (0.79/4164)*100=0.019 $.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.