Question

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000524 mol K I O 3 were titrated with an unknown solution of N a 2 S 2 O 3 and the endpoint was reached after 17.80 mL . How many moles of N a 2 S 2 O 3 did this require

Answer 1

Answer: The amount of sodium thiosulfate required is $2.62\times 10^{-5}$ moles

Explanation:

Moles of $KIO_3$ solution given = 0.0000524 moles

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.0000524 moles of potassium iodate will react with = $\frac{1}{2}\times 0.0000524=0.0000262mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $2.62\times 10^{-5}$ moles

Answer 2

The reaction required 2.62×10¯⁵ mole of Na₂S₂O₃.

We'll begin by writing the balanced equation for the reaction. This is given below:

2KIO₃ + Na₂S₂O₃ —> K₂S₂O₃ + 2NaIO₃

From the balanced equation above,

2 moles of KIO₃ required 1 mole of Na₂S₂O₃.

Finally, we shall determine the number of mole of Na₂S₂O₃ required to react with 0.0000524 mole of KIO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KIO₃ required 1 mole of Na₂S₂O₃.

Therefore,

0.0000524 mole of KIO₃ will require = $\frac{0.0000524 * 1}{2}$ = 2.62×10¯⁵ mole of Na₂S₂O₃.

Thus,  2.62×10¯⁵ mole of Na₂S₂O₃ is required for the reaction.

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