Question

An aqueous solution containing \[{\text{0}}{\text{.10 g KI}}{{\text{O}}_3}\] (formula weight = 214.0) was treated with an excess of \[{\text{KI}}\] solution. The solution was acidified with \[{\text{HCl}}\] . The liberated \[{{\text{I}}_2}\] consumed \[{\text{45}}{\text{.0 mL}}\] of thiosulphate solution to decolourise the blue starch iodine complex. Calculate the molarity of the sodium thiosulphate solution.
A. 0.0623
B. 0.126
C. 1.123
D. None of the above

Answer 1

Hint:Number of moles is the ratio of the mass to the molecular weight.
Molarity is the ratio of the number of moles of solute to the volume (in liters) of the solution.

Complete answer:
Writ down the balanced chemical equation for the given chemical reaction:
\[{\text{KI}}{{\text{O}}_3}{\text{ + 5 KI }} \to {{\text{K}}_2}{\text{O + 3 }}{{\text{I}}_2}\]
Potassium iodate reacts with potassium iodide to form potassium oxide and iodine.
Calculate the number of moles of potassium iodate by dividing its mass with molecular weight
The number of moles of \[{\text{KI}}{{\text{O}}_3} = \dfrac{{0.1}}{{214}}\]
As per the balanced chemical equation, when one mole of potassium iodate reacts, three moles of iodine are formed. Calculate the number of moles of iodine formed.
Number of moles of iodine formed \[ = 3 \times \dfrac{{0.1}}{{214}}\]
Write the balanced chemical equation for the reaction between sodium thiosulphate and liberated iodine.
\[{\text{2 N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{I}}_2}{\text{ }} \to {\text{ 2 NaI + N}}{{\text{a}}_2}{{\text{S}}_4}{{\text{O}}_6}\]
Sodium thiosulphate reacts with iodine to form sodium iodide.
As per the balanced chemical equation, one mole of iodine reacts with two moles of sodium thiosulphate.
Calculate the number of sodium thiosulphate moles that will react with \[ = 3 \times \dfrac{{0.1}}{{214}}\] moles of iodine.
The number of sodium thiosulphate moles \[ = 2 \times 3 \times \dfrac{{0.1}}{{214}}\]
Convert the unit of volume of sodium thiosulphate from mL to L.
\[\Rightarrow \dfrac{{45.0}}{{1000}} = 0.0450 \, L\]
Calculate the molarity of sodium thiosulphate by dividing the number of moles with volume
The molarity of sodium thiosulphate \[ = \dfrac{{2 \times 3}}{{0.0450}} \times \dfrac{{0.1}}{{214}}\]
The molarity of sodium thiosulphate \[ = 0.0623{\text{ M}}\]
Thus, the molarity of sodium thiosulphate solution is \[0.0623{\text{ M}}\]

Hence, the correct option is the option (A). \[0.0623{\text{ M}}\]

Note:
In this titration, first you calculate the number of moles of iodine from the mass and molecular weight of potassium iodate. Then from the number of moles of iodine, you calculate the number of moles and the molarity of sodium thiosulphate solution, with the help of given volume of sodium thiosulphate solution.