Solveeit Logo
Login

Question

Question: What is \( N{a_2}{S_2}{O_3} \) and find its n-factor in a basic medium?...

What is Na2S2O3N{a_2}{S_2}{O_3} and find its n-factor in a basic medium?

Explanation

Solution

Hint : Gold mining, water treatment, analytical chemistry, the development of silver-based photographic film and prints, and medicine all employ sodium thiosulfate. Treatment of cyanide poisoning and pityriasis are two medicinal applications for sodium thiosulfate. It is on the WHO's List of Essential Medications, which includes the safest and most effective medicines required in a health system.

Complete Step By Step Answer:
There are numerous polymorphs of the anhydrous salt. The thiosulfate anion is tetrahedral in shape in the solid state and is thought to be formed by substituting one of the oxygen atoms in a sulphate anion with a sulphur atom. The S-S distance implies a single bond, suggesting that the terminal sulphur is negatively charged while the S-O interactions are more double-bond in nature. In the Winkler test for dissolved oxygen, this specific application may be set up to assess the oxygen concentration in water through a long sequence of reactions. It's also used to estimate the chlorine content in commercial bleaching powder and water, as well as the volumetric concentrations of specific chemicals in solution (hydrogen peroxide, for example).
Because the thiosulfate anion interacts stoichiometrically with iodine in aqueous solution, reducing it to iodide when the thiosulfate is oxidised to tetrathionate, the most important usage is in analytical chemistry:
2 S2O32−  + I2  → S4O62−  + 2 I−2{\text{ }}{S_2}{O_3}^{2 - }\; + {\text{ }}{I_{2\;}} \to {\text{ }}{S_4}{O_6}^{2 - }\; + {\text{ }}2{\text{ }}{I^ - }
The oxidation state of S changes from +2 to +2.5 in the described process. Because each formula unit contains two S atoms, the change for each formula unit is 1.
ENa2  S2O3= Molecular weight n - factor =M1=M{{\text{E}}_{{\text{Na}}2\;{{\text{S}}_2}{{\text{O}}_3}}} = \frac{{{\text{ Molecular weight }}}}{{{\text{n - factor }}}} = \frac{{\text{M}}}{{\mathbf{1}}} = {\text{M}}
Hence the n – factor is 1.

Note :
The mass of one equivalent, that is, the mass of a given material that will mix with or displace a specified quantity of another substance, is known as equivalent weight. The mass of an element that combines with or displaces 1.008 grams of hydrogen, 8.0 grams of oxygen, or 35.5 grams of chlorine is its equivalent weight.