The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine, because the reaction goes to completion. The amount of iodine produced is then determined by titration with sodium thiosulphate $N a_{2} S_{2} O_{3},$ which is oxidized to sodium tetrathionte $N a_{2} S_{4} O_{6} .$ Potassium iodide was added in excess to $5.00 m L$ of bleach (density = 1.0 g/cc). This solution containing the iodine released in the reaction, was titrated with $0.1 M N a_{2} S_{2} O_{3} .$ If $40 m L$ of hypo was required to reach the end point, what was the mass percentage of $N a C l O$ in the bleach.
$(A t W t s : N a = 23, S = 32, O = 16, I = 127, C l = 35.5)$

Question

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine, because the reaction goes to completion. The amount of iodine produced is then determined by titration with sodium thiosulphate $N a_{2} S_{2} O_{3},$ which is oxidized to sodium tetrathionte $N a_{2} S_{4} O_{6} .$ Potassium iodide was added in excess to $5.00 m L$ of bleach (density = 1.0 g/cc). This solution containing the iodine released in the reaction, was titrated with $0.1 M N a_{2} S_{2} O_{3} .$ If $40 m L$ of hypo was required to reach the end point, what was the mass percentage of $N a C l O$ in the bleach.
$(A t W t s : N a = 23, S = 32, O = 16, I = 127, C l = 35.5)$

Infinity Learn · Accepted Answer

H2O+NaOCl+2I−→I2+NaCl+2OH− I2+2Na2S2O3→2NaI+Na2S4O6 Mass of NaOCl=40×0.11000×2×74.5=0.149≃0.15∵ The density of bleach is 1g/cc, then 5cc = 5gMass prsenct of NaOCl=0.155×100=3%

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