A 10 gm sample of NaCN is dissolved in 50 ml of 0.3 M mild alkaline solution of KMnO 4 and heated strongly to convert all CN − to OCN − . The left solution is then acidified with H 2 SO 4 which

Question

A 10 gm sample of NaCN is dissolved in 50 ml of 0.3 M mild alkaline solution of KMnO 4 ​ and heated strongly to convert all CN − to OCN − . The left solution is then acidified with H 2 ​ SO 4 ​ which require 500 ml of 0.05 M FeSO 4 ​ . If percentage purity of NaCN in sample is ' Y '.

TestHub · Accepted Answer

For II titration Milligram equivalents of $FeSO_{4} = KMnO_{4}$ (in acidic medium)

$0.05 \times 500 \times 1 = n_{KMnO_{4}} \times 5$

$n_{KMnO_{4}} = 5$

Used

Total millimoles of $KMnO_{4}$ initially taken $= 15$ ,

so millimoles of $KMnO_{4}$ used in I titration $= 15 - 5 = 10$

For I titration

Milligram equivalents of $NaCN = KMnO_{4}$ (in basic medium)

$n_{NaCN} \times 2 = 10 \times 3$

$n_{NaCN} = 15$

$% NaCN = \frac{15 \times 1 0 ^{- 13} \times 49}{10} \times 100 = 7.35$

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