A solution is prepared by dissolving 0.300 g of impure phenol and diluting it to 250 mL . A 25.0 mL aliquot of this solution is taken in a flask and treated with 25.0 mL of 0.0200 M KBrO 3 in presen

Question

A solution is prepared by dissolving 0.300 g of impure phenol and diluting it to 250 mL . A 25.0 mL aliquot of this solution is taken in a flask and treated with 25.0 mL of $0.0200 M KBrO_{3}$ in presence of excess KBr and acid, so that bromine is generated in situ. The bromine reacts with the phenol present in the solution. After some time, excess KI is added to react with the unconsumed bromine. The iodine liberated is then titrated with $0.100 M Na_{2} S_{2} O_{3}$ solution, and 18.60 mL of the thiosulfate solution is required for complete titration. Assuming that the only impurity present in the sample is inert and non-reactive, calculate the percentage purity of phenol in the sample.

Reaction involved are :

TestHub · Accepted Answer

Calculation n ( K B r O 3 ​ ) n ( B r 2 ​ ) n ( I 2 ​ ) B r 2 ​ co n s u m e d n ( p h e n o l , 25 m L ) n ( t o t a l ) M a ss % p u r i t y ​ = 0.0200 × 0.0250 = 5.00 × 1 0 − 4 = 3 × 5.00 × 1 0 − 4 = 1.50 × 1 0 − 3 = 2 1.86 × 1 0 − 3 ​ = 9.30 × 1 0 − 4 = 1.50 × 1 0 − 3 − 9.30 × 1 0 − 4 = 5.70 × 1 0 − 4 = 3 5.70 × 1 0 − 4 ​ = 1.90 × 1 0 − 4 = 1.90 × 1 0 − 3 = 1.90 × 1 0 − 3 × 94 = 0.1786 g = 0.300 0.1786 ​ × 100 = 59.53% ​

Chemistry - Redox Reactions Question with Solution | TestHub

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