Name: Uses of Chlorine
Uploaded: 2019-07-19
Duration: 2 min 45 s
Description: Chlorine is a poisonous pale green gas. This video discusses its usage and applications. Watch this video thoroughly to understand the concept.

Question

A

0.1337 g

sample of an alkali metal iodate,

{XIO}_{3}

, was dissolved in water, acidified, and an excess of potassium iodide,

KI

added. The resulting iodine solution required

36.64 {cm}^{3}

of the sodium thiosulphate pentahydrate solution,

{Na}_{2} S_{2} O_{3} . 5 H_{2} O (25.49 g {dm}^{- 3})

for complete titration using starch solution as an indicator. Calculate the relative atomic mass of

X

and hence identify the metal. Deduce all relevant half-equations involved.

Accepted Answer

Solution:

It is given that in acidified aqueous solutions, iodate's and iodide's react with each other to form iodine. So ,the reaction is, ${IO}_{3}^{-} (aq) + 5 I^{-} + 6 H^{+} \to 3 I_{2} (aq) + 3 H_{2} O (l)$ .................................(i)

From the equation (i), ${IO}_{3}^{-}$ (aq) is the oxidizing agent and $I^{-}$ (aq) is the reducing agent. $2 {IO}_{3}^{-} (aq) + 12 H^{+} (aq) + I_{2} (aq) + 6 H_{2} O (l)$ $2 I^{-} (aq) \to I_{2} (aq) + 2 e^{-}$

The iodine formed in the equation (i) react with thiosulfate ions to give iodate ions and the reaction is, $I_{2} (aq) + 2 S_{2} {O_{3}}^{- 2} (aq) \to 2 I^{-} (aq) + S_{4} {O_{6}}^{- 2} (aq)$ ............................(ii)

In the equation (ii), $36.64 {cm}^{3}$ acts as oxidizing agent and $S_{2} {O_{3}}^{- 2}$ act as reducing agent. $I_{2} (aq) + 2 e^{-} \to 2 I^{-} (aq)$ $2 S_{2} O 3^{- 2} (aq) \to S_{4} {O_{6}}^{- 2} (aq) + 2 e^{-}$ Molecular weight of ${Na}_{2} S_{2} O_{3} . H_{2} O = 248.22$ and the number of moles of thiosulphate ions present in $25.49$ gm of sodium thiosulphate $= \frac{25.49}{248.22} = 0.1027$ mol/L $(Since, Mole' s number = \frac{weight}{molecular weight})$

It is mentioned that, the iodine solution formed, is required $36.64$ cm³of the sodium thiosulphate solution. Then, the amount of thiosulphate ions reacted with iodine $= 0.1027 \times 0.03664 = 0.003763$ mol, from equation (ii), two moles of thiosulphate reacts with one mole of iodine. Then the amount of $I_{2} (aq) = \frac{0.003763}{2} = 0.001882$ mol, and from equation (i) it is cleared that $3$ moles of iodine is generated from $1$ mole of iodate ions. Hence, the amount of ${IO}_{3}^{-} (aq) = \frac{0.001882}{3} = 0.0006273$ mol.

We know that, the amount of ${XIO}_{3} (s)$ is equal to the amount of ${IO}_{3}^{-} (aq)$ . Because, ${XIO}_{3} (s) \to X^{+} (aq) + {IO}_{3}^{-} (aq)$ .

Therefore, moles number of ${XIO}_{3} = 0.0006273$ mol. Molecular weight of ${XIO}_{3} = \frac{Weight}{Mole' s number} = \frac{0.1337}{0.0006273} = 213.14$ .

Finally, relative atomic mass (molecular weight) of Alkali metal, $X = Molecular mass of {XIO}_{3} - Atomic mass of iodine \times (3 \times atomic mass of xygen)$ $= 213.14 - 126.90 - (3 \times 16) = 38.2$ .

Then, the alkali metal with the closest relative atomic mass is $K$ (potassium) with the atomic mass of $39.10$ amu.

Important Questions on Redox Processes

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