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Question: 12 g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of sodium bicarb...

12 g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of sodium bicarbonate and the resulting solution was diluted by 250 ml. 25 ml of this solution was completely oxidized by 22.4 ml of a solution of iodine. 25 ml of this iodine solution reacted with same volume of a solution containing 24.8 g of sodium thiosulphate solution (Na2S2O3.5H2ON{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O) in one litre. Calculate the percentage of arsenious oxide in the sample. (Atomic mass of As = 75)

Explanation

Solution

Normality is generally used to measure the concentration of a solution. It is represented by the letter N and sometimes referred to as the equivalent concentration of a solution. Units of normality expressed as eqL1eq{{L}^{-1}} or meqL1meq{{L}^{-1}}.

Complete Solution :
Normality can also be defined as the number of gram or mole equivalents of solute present in one litre of a solution. Where equivalent represents the number of moles of reactive units in a compound.
Normality is calculated by the formula: N=Weight of soluteequivalent weight×volumeN=\dfrac{Weight~of~solute}{equivalent~weight\times volume}
Normality of Na2S2O3.5H2ON{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O solution = 24.8248×1=0.1N\dfrac{24.8}{248\times 1}=0.1N

Applying the normality equation:
N1V1=N2V2{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}
Volume of AS2O3solution in NaHCO3×NormalityVolume~of~A{{S}_{2}}{{O}_{3}}solution~in~NaHC{{O}_{3}}\times Normality = Volume of iodine solution×NormalityVolume~of~iodine~solution\times Normality

- We have to find out normality of AS2O3solutioninNaHCO3A{{S}_{2}}{{O}_{3}} solution in NaHC{{O}_{3}}, let us suppose it as N
25×N=22.4×0.125\times N = 22.4\times 0.1
N=22.4×0.125=0.0896N=\dfrac{22.4\times 0.1}{25} = 0.0896
Amount of As2O3A{{s}_{2}}{{O}_{3}} present in 250 ml of solution is given by:
N×Equivalent mass of As2O31000×250N\times \dfrac{Equivalent~mass~of~A{{s}_{2}}{{O}_{3}}}{1000}\times 250
0.0896×1984×1000×250=1.1088g0.0896\times \dfrac{198}{4\times 1000}\times 250 = 1.1088g
Percentage of As2O3A{{s}_{2}}{{O}_{3}} = 1.108812×100=9.24\dfrac{1.1088}{12}\times 100=9.24
Percentage of arsenious oxide in the sample is 9.24%.

Note: Arsenious oxide is an inorganic compound. It is generally used as a medication to treat a type of cancer called acute promyelocytic leukemia in which it is by injection directly to the vein. It is also used in the manufacture of wood preservatives, pesticides etc.