What volume of 1M NaOH(in ml) will be required to react completely with 100 g of oleum which is 109% labelled

Asked by Anil | 16th May, 2017, 10:18: PM

Expert Answer:

109% oleum means that 9g of H2O is required to convert all SO3 present in 100g of it to H2SO4.

SO3 + H2O → H2SO4

1 mole (80g) of SO3 reacts with 1 mole (18g) of H2O

So 40g of SO3 will react with 9g of H2O.

 

According to the following equation, 80g of NaOH will react with 80g and 98g of SO3 and H2SO4 respectively.

SO3 + 2NaOH →  Na2SO4 + H2O

H2SO4 + 2NaOH →  Na2SO4 + 2H2O

Wt. of SO3 present in 100g of the oleum sample = 40g

Wt. of H2SO4 present in 100g of the oleum sample = (100 – 40)g = 60g

Wt. of NaOH required to completely neutralize 40g of SO3 = (80x40)/80 = 40g

Wt.  of NaOH required to completely neutralize 60g of H2SO4 = (80x60)/98 = 48.98g

Total Wt.  of NaOH required for complete neutralization = 40 + 48.98 = 88.98g

Volume of 1M NaOH solution containing 88.98g of NaOH = (1000x88.98)/40 = 2224.5mL

 

Answered by Vaibhav Chavan | 17th May, 2017, 12:46: PM

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