+2
The chemical symbol for sodium thiosulphate (hypo) is Na2S2O3•5H2O.
Sodium Thiosulphate is Na2S2O3 and water is H2O.
A method of preparing sodium thiosulphate is to react sodium sulphite with sulphur. Method Dissolve 3g sodium sulphite (Na2SO3) in 15 cm3 hot water and add 1g sulphur (S8). boil the suspension until nearly all the sulphur has reacted. Filter hot (using a glass filter funnel plugged with a small piece of cotton. Ensure that the tip of the funnel is heated as well as the container for the filtrate) and evaporate the filtrate until crystallization starts. Cool and filter the crystals by suction. Dry the product in a warm oven (keep below 480C). EquationNa2SO3 (s) + 1/8 S8 (s) -------H2O/1000C--------> Na2S2O3 (s)
The formula for hyposulfite is Na2S2O3.
Mol.Wt =248.17Molarity =( Wt/ Mol.Wt ) /No:of litres of solution0.1 =( Wt / 248.17) / 0.5Wt =0.1 *0.5*248.17Wt = 12.4085g in 500ml of solution.
Oxidation no of sulphur is +2.
In Na2S2O3, the oxidation numbers are +1 for sodium (Na), -2 for sulfur (S), and +2 for oxygen (O). This can be determined by considering the overall charge of the compound and known oxidation number rules.
This compound is sodium thiosulphate. S represents +6 oxidation number.
Na2S2O3(aq) + 2HCL(aq) => 2NaCl(aq) + S(s) + SO2(g) + H2O(l)
The reaction between Na2S2O3 and HCl involves the oxidation of thiosulfate ions by hydrogen ions. The rate of the reaction can be determined by measuring the initial rate of the disappearance of thiosulfate ions. The reaction is typically carried out in acidic conditions to provide protons for the oxidation process.
The equivalent weight of sodium thiosulfate (Na2S2O3) is the molar mass divided by the change in oxidation state per mole of electrons transferred in a reaction. In this case, the equivalents of Na2S2O3 is equal to its molar mass (158.11 g/mol) divided by the change in oxidation state involved in the reaction it participates in.
To prepare a 0.1N Na2S2O3 solution, dissolve 24.98 grams of Na2S2O3·5H2O (sodium thiosulfate pentahydrate) in distilled water and dilute to 1 liter in a volumetric flask. This will give you a 0.1N (normal) solution of Na2S2O3.
first we calculate the molecular weight M.W=158.11 mass of Na2S2O3=M.W*concentration*volume(in litter) for a concentration of 0.001 mass of Na2S2O3=158.11*0.001*1=0.15811 g so we use this mass and continue the volume to be 1 L
The structure of Na2S2O3, also known as sodium thiosulfate, consists of two sodium ions (Na+), two sulfur atoms (S), and three oxygen atoms (O) in its chemical formula. The compound contains a thiosulfate ion (S2O3 2-) where the central sulfur atom is connected to three oxygen atoms and to a second sulfur atom.
The oxidation number of S in S2Cl2 is +1. Each Cl atom has an oxidation number of -1, and since the molecule is neutral, the overall oxidation numbers of S must balance out to zero. Thus, the oxidation number of S in this compound is +1.
S = +4 oxidation state O = -2 oxidation state
Na2S2O3