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Rashid [163]
3 years ago
7

In part 2 of the experiment, you will be analyzing a sample of household bleach. A 0.0854 g sample of household bleach is comple

tely reacted with KI(s). The resulting solution is then titrated with 0.150 M NaS2O3 solution. 0.890 mL of the solution is required to reach the colorless endpoint. What is the mass percent of NaOCl (MM = 74.44 g/mole) in the bleach? a. 30.1% b. 96.5% c. 2.23% d. 5.82%
Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

The correct answer is option d.

Explanation:

NaClO(aq)+2KI(s)+H_2O(l)\rightarrow NaCl(aq)+I_2(aq)+2KOH(aq)..[1]

I_2(aq)+2Na_2S_2O_3(aq)\rightarrow 2NaI(aq)+Na_2S_4O_6(aq)..[2]

moles of sodium thiosulfate solution = n

Molarity of the sodium thiosulfate solution = 0.150 M

Volume of sodium thiosulfate solution used = 0.890 mL = 0.00089 L

n=Molarity\times Volume (L)

n=0.150 M\times 0.00089 L=0.0001335 mol

According to reaction [2], 2 moles of sodium thiosulfate reacts with 1 mol I_2.

Then , 0.0001335 mol of sodium thiosulfate will reacts with:

\frac{1}{2}\times 0.0001335 mol=0.00006675 mol of I_2

Moles of I_2 = 0.00006675 mol

According to reaction [1], 1 mol of I_2 is obatnied from 1 mol of NaClO.

Then 0.00006675 mol of I_2 will be obtained from:

\frac{1}{1}\times 0.00006675 mol=0.00006675 mol

Mass of 0.00006675 mol of NaClO=

0.00006675 mol × 74.44 g/mol = 0.00496887 g

Mass of sample = 0.0854 g

Percentage of NaClO in sample:

\frac{0.00496887 g}{0.0854 g}\times 100=5.82\%

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