Question

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

Answer 1

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:

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

Answer 2

There is  5.8 %  of NaOCl in the sample.

Oxidation of hypochlorite

The first reaction is the oxidation of hypochlorite as follows;

OCl^- + 2I^- + 2H^+ -----> I2 + Cl2 + H2O

The second step is the reaction of iodine and sodium thiosulfate as follows;

I2 + 2NaS2O3 -----> Na2S4O6 + 2NaI

Number of moles of thiosulfate = 0.150 M  * 0.890/1000 L = 1.335 * 10^-4 moles

Since 1 mole of I2 reacts with 2 moles of thiosulfate

x moles of I2 reacts with 1.335 * 10^-4 moles of thiosulfate

x =  1 mole * 1.335 * 10^-4 moles/2 moles

= 6.675 * 10^-5 moles of iodine

Since 1 mole of hypochlorite produces 1 mole of iodine, 6.675 * 10^-5 moles of hypochlorite was reacted.

Hence, mass of NaOCl = 6.675 * 10^-5 moles * 74.44 g/mole = 4.97 * 10^-3 g

Percent of NaOCl =  4.97 * 10^-3 g/0.0854 g * 100/1 = 5.8 %

There is  5.8 %  of NaOCl in the sample.

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