Answer to Question #139068 in Inorganic Chemistry for chi

Question #139068
A sample of caustic soda weighing 1.675 g was found to contain 97.25% of total alkali calculated as NaOH and of which is 2.48% was Na2CO3. What volume of 1.15 N acid could have been consumed in the Polypropylene and MO titrations as ordinarily performed?

MW: NaOH = 40.0, Na2CO3 = 106.0
Expert's answer

(Assuming that 1.15 N actually means 1.15 M HCl acid, Polypropylene was meant to be Phenolphthalein indicator and MO is Methyl Orange indicator.)

Mass of NaOH

Mass of Na2CO3

Moles of NaOH

Moles of Na2CO3

Moles of acid required to react with all the NaOH

Mole ratio=1:1

Hence moles of acid

Moles of acid required to react with all Na2CO3

Mole ratio = 2:1

Hence moles of acid

Total moles of acid used = 0.03586 + 1.692x10^-4 = 0.03602915mol.

Volume of acid

= 31.33 cm3 for Phenolphthalein

For MO cm3

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